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## Solving Linear Equations Calculator

Instructions: Use this linear equation calculator to solve any linear equation you want, of one or more variables . Please type in the linear equation you want to solve.

## This Linear Equation Calculator

This linear equation calculator will allow you solve linear equations you provide, showing all the steps. For example, you may be interested in solving something like '1/3 x +1/4 y = 1/6', which is a linear equation with two variables, x and y.

Once you specified a valid linear equation you want to solve, then you can click on "Calculate" and you will be provided with corresponding steps needed to arrive to the solution.

Solving linear equation is the easiest among the broader task of solving polynomial equations , which can be a lot harder, especially for polynomial with a higher degree.

## What is a Linear Equation

A linear equation is a math equation in which you have that both sides of the equation are linear expressions. A linear expression is the sum or subtraction of constants or constant multiplied by a variable.

For example, '2x + 3y = 1' is a linear equation , but '2x = cos(x)' is not. It is important to distinguish between a linear expression and a linear equation.

Following the same example, '2x + 3y' is a linear expression, but it is not a linear equation, because there is no equality involved. In order to have a linear equation, YOU NEED to have an equality sign in it.

## Linear Equations Formula

A linear equation formula will depend on the number of variables we use. For example, the general linear equation formula for one variable x is:

Some will argue that there is no need to have a constant on the left side, and they would write:

Now, the general linear equation formula for two variables x and y is:

In general, the general linear equation formula for \(n\) variables is:

Notice that we put a "+" in general, but the constants \(a_1\), ..., \(a_n\) can also be negative.

## How to Solve Linear Equations

- Step 1: Make sure you are dealing with an actual linear equation. Then, identify how many variables are involved in the equation
- Step 2: If you have only one variable, say x, you can solve for x, manipulating the terms of the equation, putting x on one side and then solving for x. Solving for x in this case is expected to lead to a numeric solution
- Step 3: If you have more than one variable, then you choose one variable, say x, and then solve for x , in terms of the other variables. Here you don't get a numeric solution, but instead, you get x (or whatever variable you chose) in terms of other variables

Observe that we are dealing here with one linear equation. You can use this system of equations calculator if you are dealing with multiple linear equations.

Having an equation calculator with steps can prove to be extremely useful, as it is sometimes hard to find the correct strategy to use for certain equations. Of course linear equations are simple, but we can find that solving polynomial equations , or solving trigonometric equations , for example, can be tremendously laborious and challenging.

## How do you find the linear equation?

Linear equations appear naturally in algebra problems and all sorts of algebra equations. Linear functions are extremely common both in Algebra and Calculus and will appear literally EVERYWHERE.

You can, for example, use the slope intercept form or the point slope form to calculate a linear function. Usually, you will work the linear equations in standard form , which the way we presented before:

Normally, we don't work with n generic variables, we work with two or three variables, which would look like:

respectively.

## Advantages of working with Linear Equations

- Step 1: Linear equations are simple! They are easy to calculate and easy to interpret
- Step 2: There are no tricks needed to solve a linear equation: pass the terms with to one side, group them and simplify
- Step 3: Linear equations are very common, and have a clear graphical interpretation

Naturally, if we could choose, we would always work with linear equation, but unfortunately reality is not that generous, as it the case that very frequently we will need to deal with equations more difficult than linear equations.

## How do you know if a function is linear?

Fractions are one of the corner stones of algebra and of any general algebraic expression to calculate . Fractions are simple operands, but which can be compounded into more complicated terms using operations like sum, multiplication, etc., and then using functions we can construct even more advanced expressions.

The center of all algebraic calculator starts with the power of basic numbers of fractions.

## Example: Solving linear equations one variable

Solve the following: \(\frac{1}{3} x + \frac{5}{4} = \frac{5}{6}\)

We need to solve the following given linear equation:

The linear equation has only one variable, which is \(x\), so the objective is to solve for it.

Putting \(x\) on the left hand side and the constant on the right hand side we get

Now, solving for \(x\), by dividing both sides of the equation by \(\frac{1}{3}\), the following is obtained

and simplifying we finally get the following

Therefore, the solving for \(x\) for given linear equation leads to \(x=-\frac{5}{4}\). This concludes the solution calculation.

## Other useful equation calculators

Using an equation solver can completely come in handy, especially when dealing with difficult equations. The case of linear equations is truly reduced to a class of simple equations to solve, and you will find equations that will be a lot more challenging.

Coming next in terms of difficulty you will find the polynomial equations , for which you can use a methodology which ensures you have the best chance of finding as many solutions as possible, but you are not guaranteed to find all of them at times. This polynomial calculator will guarantee you get as many solutions as possible.

Then you have the even more complicated of non-polynomial non-linear equations, for which you need to come up with an astute approach usually, if you want to come closer to the solution. Trigonometric equations are notorious for being difficult and reliant on a precise substitution.

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## Linear Equations

6.2 Solving Linear Equations

Equations of the form ax+b=0 are called linear equations in the variable x . In this section we will be concerned with the problem of solving linear equations, and equations that reduce to linear equations.

We deﬁne two equations as equivalent if they have the same solution set. The following two operations on an equation always result in a new equation which is equivalent to the original one. These operations, sometimes called elementary transformations, are:

T.1 The same expression representing a real number may be added to both sides of an equation.

T.2 The same expression representing a nonzero real number may be multiplied into both sides of an equation.

Using these operations we may transform an equation whose solution set is not obvious through a series of equivalent equations to an equation that has an obvious solution set.

Example 1. Solve the equation

(a) 2x-3=4+x

Add -x to both sides to obtain

-x+2x-3=-x+4+x (T.1)

or x-3=4

Add 3 to both sides to obtain

x-3+3=4+3 (T.1)

or x=7

Since 2x-3=4+x is equivalent to x-3=4 , which, in turn, is equivalent to x=7 , whose solution set is obviously {7} , we know that the solution set of (a) is {7} .

Let’s see how our Linear equation solver solves this and similar problems. Click on "Solve Similar" button to see more examples.

Example 2. Solve the equation

(b) 1/2x+2/3=5/2x-1

Add -(1/2)x to both sides to obtain

2/3=5/2x-1/2x-1 (T.1)

or 2/3=2x-1

Add 1 to both sides to obtain

1+2/3=2x (T.1)

or 5/3=2x

Multiply both sides by 1/2 to obtain

5/6=x (T.2)

Thus, the solution set of (b) is {5/6} .

Every linear equation can be solved in the same way as in the above examples. In fact, let us consider the general linear equation

ax+b=0

Add -b to both sides to obtain

ax=-b

Multiply both sides by 1/a to obtain

x=-(b/a)

if a a!=0 . The general linear equation, therefore, has as its solution set {b/a} , if a!=0 . Thus each linear equation has at most one solution.

The next two examples are of equations that reduce to linear equations.

Example 3. Solve the equation

23+4y(5y+4)=9+10y(2y+3)

We expand both sides to obtain

23+20y^2+16y=9+20y^2+30y

Add -20y^2 to both sides to obtain

23+16y=9+30y

We now solve as in the previous examples.

23-9=30y-16y

14=14y

y=1

Thus the solution set is {1} .

Let’s see how our step by step math solver solves this and similar problems. Click on "Solve Similar" button to see more examples.

Example 4. Solve the equation

(c) (2x)/(x-1)=2/(x-1)+1

The replacement set of (c) is all real numbers except 1. Assuming that x!=1 , we multiply both sides of (c) by x-1 to obtain

(d) 2x=2+x-1,x!=1

Solving the equation 2x =2+x- 1 , we obtain 1 as the only solution Since 1 is not in the replacement of (d), (d) has no solution. Furthermore, (c) is equivalent to (d), therefore (c) has no solution.

6.3 Solving Literal Equations

An equation containing more than one variable, or containing symbols representing constants such as a,b , and c , can be solved for one of the symbols in terms of the remaining symbols by applications of the operations T.1 and T.2 in the preceding section. The student will encounter such problems in other courses.

Example 1. Solve cx-3a=b for x .

Add 3a to both sides.

cx=b+3a

Multiply both sides by 1/c .

x=(b+3a)/c

This last equation expresses x in terms of the other symbols.

Example 2. Solve 3ay-2b=2cy for y .

Add 2b to both sides.

3ay=2cy+2b

Add -2cy to both sides.

3ay-2cy=2b

Factor out y .

(3a-2c)y=2b

Multiply both sides by 1/((3a-2c))

y=(2b)/(3a-2c)

Example 3. Solve a/x+b/(2x)=c for x .

Multiply both sides by 2x .

2a+b=2cx

2cx=2a+b

Multiply by 1/(2c) .

x=(2a+b)/(2c)

We conclude this section by including two more examples similar to those that the student may encounter in other areas.

Let’s see how our math solver solves this and similar problems. Click on "Solve Similar" button to see more examples.

Example 4. Solve A=P(1+rt) for r .

Apply the distributive law.

A=P+Prt

Add -P to both sides.

A-P=Prt

Multiply both sides by 1/(Pt) .

(A-P)/(Pt)=r

Example 5. Solve 1/R=1/r_1+1/r_2 for r_1 .

Add the two terms on the right—hand side.

1/(R)=(r_2+r_1)/(r_1r_2)

Multiply by Rr_1r_2 .

r_1r_2=R(r_2+r_1)

r_1r_2=Rr_2+Rr_1

Add -Rr_1 to both sides.

r_1r_2-Rr_1=Rr_2

Factor out r_1 .

r_1(r_2-R)=Rr_2

Multiply by 1/(r_2-R) .

r_1=(Rr_2)/(r_2-R)

6.4 Solving Statement Problems

One of the fundamental applications of algebra is solving problems that have been stated in words. A statement problem is a word description of a situation that involves both known and unknown quantities. In this section each problem will be solvable by means of one equation involving one unknown.

Our problem is to choose the unknown and to determine the equation that it must satisfy. Although there is no single approach to all of the problems, the following suggestions are sometimes helpful:

1. Read the problem carefully until the situation is thoroughly understood.

2. Determine what quantities are asked for, then choose the one that seems to be the best to use as the unknown.

3. Establish the relationship between the unknown and the other quantities in the problem.

4. Find the information that tells which two quantities are equal.

5. Use the information in (4) to write the equation.

6. Solve the equation and check the solution to see that it satisﬁes the original problem.

At this stage, the emphasis will be on translating statement problems into equations. Although some of the problems can be solved almost by inspection, the practice that we obtain in setting up equations will prove helpful in working more difficult problems.

Example 1. If 2 times a certain integer is added to the next consecutive integer the result is 34 . Find the integers. Step 1. Reread! Step 2. Let x be the ﬁrst integer. Step 3. Then x+ 1 is the next consecutive integer. Step 4. 2 times a certain integer plus the next consecutive integer is 34 . Step 5. 2x+(x+1)=34

Step 6. Solve.

2x+(x+1)=34

3x+1=34

3x=33

x=11

Check. 2*11+(11+1)=34

Example 2. Bob and Joe together earned $ 60 . Both were paid at the same rate, but Bob worked three times as long as Joe. How much did each receive?

Step 1. Reread!

Step 2. Let x be the number of dollars that Joe received.

Step 3. Then 3x is the number of dollars that Bob received

Step 4. Bob and Joe together earned $ 60 . Step 5. 3x+x=60

3x+x=60

4x=60

x=15

3x=45

Check 3*15+15=60

Example 3. The sum of the digits of a two digit number is 12 . If the digits are reversed the number is decreased by 36 . What is the number?

Step 2. Let x be the tens digit.

Step 3. Then 12 - x is the units digit.

Step 4. If the digits are reversed then the number is decreased by 36

Step 5. 10(12-x)+x = 10x+ (12-x) -36

Step 6. Solve.

10(12-x)+x = 10x+ (12-x) -36

= 120-10x+x=10x+12-x-36

= 120-9x=9x-24

= 144=18x

= x=8

= 12-x=4

Therefore the number is 84 .

Check. 84-36=48

Example 4. How many pounds of candy valued at 48 ¢ per pound should be added to 50 pounds of candy valued at 80 ¢ per pound in order for the store owner to be able to sell the candy at 60 ¢ per pound? Step 1. Reread! Step 2. Let x be the number of pounds of 48 ¢ per pound candy. Step 3. Then 50+x would be the pounds of candy he would have at 60 ¢ per pound. Step 4. The amount of candy at 48 ¢ per pound times 48 ¢, plus the amount of candy at 80 ¢ per pound times 80 ¢, must be equal to the amount of candy at 60 ¢ per pound times 60 ¢. Step 5. ( 48 ¢/lb)( x lbs) + ( 80 ¢/lb) ( 50 lbs) = ( 60 ¢/lb) [( 50+x )lbs]

48x+80*50=60(50+x)

48x+4000=3000+60x

1000=12x

x=(83(1)/3) lbs

Check. (83+1/3)48+80*50=60(50+83+1/3)

Problems involving velocities (or speeds) will use the formula

d=rt

where d is the distance traveled, r is the rate, and t is the time. When the formula is used, d and r must be expressed in the same unit of distance, while r and t must be expressed in the same unit of time.

Example 5. A group of students drove to a lake in the north woods to ﬁsh. They traveled 380 miles in 7 hours, of which 4 hours were on a paved highway and the remaining time was on a dirt road. If the average speed on the dirt road was 25 miles per hour less than the average speed on the highway, then ﬁnd for each part of the trip the average speed and the distance traveled.

Step 1 . Reread!

Step 2. Let x be the speed on the dirt road.

Step 3. Then x+25 is the speed on the highway.

Step 4. The distance traveled on the highway plus the distance traveled on the dirt road is equal to 380 miles.

Step 5. Since d=rt , we have

[(x+25)(mi)/(hr)](4hrs)+[x(mi)/(hr)](3hrs)=380mi

(x+25)4+3x=380

4x+100+3x=380

7x=280

x=40 miles per hour

x+25=65 miles per hour

Check. (40+25)4+40*3=380

Work problems which involve the rate of performance can often be solved by ﬁrst ﬁnding the fractional part of the task done by each person or machine in one unit of time, and then ﬁnding an equation that relates these various fractional parts.

Example 6. A boy can cut a lawn in 4 hours while the father can cut it in 3 hours. How long would it take them to cut the same lawn working together?

Step 2. Let x be the number of hours that it would take them to cut the lawn Working together.

Step 3 . Choose one hour as our unit of time. Now the boy can cut 1/4 of the lawn in one hour, the father can cut 1/3 of the lawn in one hour, and togeﬂner they can cut 1/x of the lawn in one hour.

Step 4. The amount cut by the boy in one hour plus the amount cut by the father in one hour is equal to the amount they can cut together in one hour.

Step 5. 1/3+1/4=1/x

1/3+1/4=1/x

7/12=1/x

x=12/7 hours

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Solve equations and systems of equations with wolfram|alpha, a powerful tool for finding solutions to systems of equations and constraints.

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Enter your queries using plain English. To avoid ambiguous queries, make sure to use parentheses where necessary. Here are some examples illustrating how to ask about solving systems of equations.

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- solve 4x - 3y + z = -10, 2x + y + 3z = 0, -x + 2y - 5z = 17
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## What are systems of equations?

A system of equations is a set of one or more equations involving a number of variables..

The solutions to systems of equations are the variable mappings such that all component equations are satisfied—in other words, the locations at which all of these equations intersect. To solve a system is to find all such common solutions or points of intersection.

Systems of linear equations are a common and applicable subset of systems of equations. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. The system is said to be inconsistent otherwise, having no solutions. Systems of linear equations involving more than two variables work similarly, having either one solution, no solutions or infinite solutions (the latter in the case that all component equations are equivalent).

More general systems involving nonlinear functions are possible as well. These possess more complicated solution sets involving one, zero, infinite or any number of solutions, but work similarly to linear systems in that their solutions are the points satisfying all equations involved. Going further, more general systems of constraints are possible, such as ones that involve inequalities or have requirements that certain variables be integers.

Solving systems of equations is a very general and important idea, and one that is fundamental in many areas of mathematics, engineering and science.

## System of equations calculator

Enter coefficients of your system into the input fields. Leave cells empty for variables, which do not participate in your equations. To input fractions use / : 1/3 .

- Show how to input the following system: 2x-2y+z=-3 x+3y-2z=1 3x-y-z=2

This calculator solves Systems of Linear Equations with steps shown, using Gaussian Elimination Method , Inverse Matrix Method , or Cramer's rule . Also you can compute a number of solutions in a system (analyse the compatibility) using Rouché–Capelli theorem .

- Leave extra cells empty to enter non-square matrices.
- decimal (finite and periodic) fractions: 1/3 , 3.14 , -1.3(56) , or 1.2e-4
- mathematical expressions : 2/3+3*(10-4) , (1+x)/y^2 , 2^0.5 (= 2 ) , 2^(1/3) , 2^n , sin(phi) , cos(3.142rad) , a_1 , or (root of x^5-x-1 near 1.2)
- matrix literals: {{1,3},{4,5}}
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- units: rad , deg
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- To learn more about matrices use Wikipedia .

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Get detailed solutions to your math problems with our equations step-by-step calculator . practice your math skills and learn step by step with our math solver. check out all of our online calculators here ., example, solved problems, difficult problems.

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We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $1$ from both sides of the equation

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$x+0=x$, where $x$ is any expression

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## Step by Step Solution

Reformatting the input :.

Changes made to your input should not affect the solution: (1): "a8" was replaced by "a^8".

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 2/(a^8)-(9/(a-2))=0

## Step 1 :

Equation at the end of step 1 :, step 2 :, equation at the end of step 2 :, step 3 :, calculating the least common multiple :.

3.1 Find the Least Common Multiple The left denominator is : a 8 The right denominator is : a-2

Least Common Multiple: a 8 • (a-2)

## Calculating Multipliers :

3.2 Calculate multipliers for the two fractions Denote the Least Common Multiple by L.C.M Denote the Left Multiplier by Left_M Denote the Right Multiplier by Right_M Denote the Left Deniminator by L_Deno Denote the Right Multiplier by R_Deno Left_M = L.C.M / L_Deno = a-2 Right_M = L.C.M / R_Deno = a 8

## Making Equivalent Fractions :

3.3 Rewrite the two fractions into equivalent fractions Two fractions are called equivalent if they have the same numeric value. For example : 1/2 and 2/4 are equivalent, y/(y+1) 2 and (y 2 +y)/(y+1) 3 are equivalent as well. To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

## Adding fractions that have a common denominator :

3.4 Adding up the two equivalent fractions Add the two equivalent fractions which now have a common denominator Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

## Step 4 :

Pulling out like terms :.

4.1 Pull out like factors : -9a 8 + 2a - 4 = -1 • (9a 8 - 2a + 4)

## Polynomial Roots Calculator :

4.2 Find roots (zeroes) of : F(a) = 9a 8 - 2a + 4 Polynomial Roots Calculator is a set of methods aimed at finding values of a for which F(a)=0 Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers a which can be expressed as the quotient of two integers The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient In this case, the Leading Coefficient is 9 and the Trailing Constant is 4. The factor(s) are: of the Leading Coefficient : 1,3 ,9 of the Trailing Constant : 1 ,2 ,4 Let us test ....

Note - For tidiness, printing of 13 checks which found no root was suppressed Polynomial Roots Calculator found no rational roots

## Equation at the end of step 4 :

Step 5 :, when a fraction equals zero :.

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero. Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator. Here's how:

Now, on the left hand side, the a 8 • a-2 cancels out the denominator, while, on the right hand side, zero times anything is still zero. The equation now takes the shape : -9a 8 +2a-4 = 0

## Equations of order 5 or higher :

5.2 Solve -9a 8 +2a-4 = 0 Handling of functions of an even degree greater than 6 is not implemented yet

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## Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

## Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

## Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

## Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

## Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

## Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

## Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

- Linear Equations
- Application of Linear Equations
- Two-Variable Linear Equations
- Linear Equations and Half Planes
- One Variable Linear Equations and Inequations

## Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

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## Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

## How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

## How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

## What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

## What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

## What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

## What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

- Substitution method
- Elimination method
- Cross multiplication method
- Graphical method

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## 7.6: Solving Trigonometric Equations

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## Learning Objectives

In this section, you will:

- Solve linear trigonometric equations in sine and cosine.
- Solve equations involving a single trigonometric function.
- Solve trigonometric equations using a calculator.
- Solve trigonometric equations that are quadratic in form.
- Solve trigonometric equations using fundamental identities.
- Solve trigonometric equations with multiple angles.
- Solve right triangle problems.

Figure 1 Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill)

Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles , which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.

In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.

## Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is 2π.2π. In other words, every 2π2π units, the y- values repeat. If we need to find all possible solutions, then we must add 2πk,2πk, where kk is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is 2π:2π:

sinθ=sin(θ±2kπ)sinθ=sin(θ±2kπ)

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

## Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation cosθ=12.cosθ=12.

## Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation sint=12.sint=12.

Given a trigonometric equation, solve using algebra .

- Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
- Substitute the trigonometric expression with a single variable, such as xx or u.u.
- Solve the equation the same way an algebraic equation would be solved.
- Substitute the trigonometric expression back in for the variable in the resulting expressions.
- Solve for the angle.

## Solve the Trigonometric Equation in Linear Form

Solve the equation exactly: 2cosθ−3=−5,0≤θ<2π.2cosθ−3=−5,0≤θ<2π.

Solve exactly the following linear equation on the interval [0,2π):2sinx+1=0.[0,2π):2sinx+1=0.

## Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see Figure 2 ). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is π,π, not 2π.2π. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of π2,π2, unless, of course, a problem places its own restrictions on the domain.

## Solving a Problem Involving a Single Trigonometric Function

Solve the problem exactly: 2sin2θ−1=0,0≤θ<2π.2sin2θ−1=0,0≤θ<2π.

## Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: cscθ=−2,0≤θ<4π.cscθ=−2,0≤θ<4π.

As sinθ=−12,sinθ=−12, notice that all four solutions are in the third and fourth quadrants.

## Solving an Equation Involving Tangent

Solve the equation exactly: tan(θ−π2)=1,0≤θ<2π.tan(θ−π2)=1,0≤θ<2π.

Find all solutions for tanx=3–√.tanx=3.

## Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation 2(tanx+3)=5+tanx,0≤x<2π.2(tanx+3)=5+tanx,0≤x<2π.

## Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

## Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation sinθ=0.8,sinθ=0.8, where θθ is in radians.

Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using π−θ.π−θ.

## Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation secθ=−4,secθ=−4, giving your answer in radians.

Solve cosθ=−0.2.cosθ=−0.2.

## Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as xx or u.u. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

## Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: cos2θ+3cosθ−1=0,0≤θ<2π.cos2θ+3cosθ−1=0,0≤θ<2π.

## Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: 2sin2θ−5sinθ+3=0,0≤θ≤2π.2sin2θ−5sinθ+3=0,0≤θ≤2π.

Make sure to check all solutions on the given domain as some factors have no solution.

Solve sin2θ=2cosθ+2,0≤θ≤2π.sin2θ=2cosθ+2,0≤θ≤2π. [Hint: Make a substitution to express the equation only in terms of cosine.]

## Solving a Trigonometric Equation Using Algebra

Solve exactly:

2sin2θ+sinθ=0;0≤θ<2π2sin2θ+sinθ=0;0≤θ<2π

We can see the solutions on the graph in Figure 3 . On the interval 0≤θ<2π,0≤θ<2π, the graph crosses the x- axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

We can verify the solutions on the unit circle in Figure 2 as well.

## Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: 2sin2θ−3sinθ+1=0,0≤θ<2π.2sin2θ−3sinθ+1=0,0≤θ<2π.

Solve the quadratic equation 2cos2θ+cosθ=0.2cos2θ+cosθ=0.

## Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

## Use Identities to Solve an Equation

Use identities to solve exactly the trigonometric equation over the interval 0≤x<2π.0≤x<2π.

cosxcos(2x)+sinxsin(2x)=3–√2cosxcos(2x)+sinxsin(2x)=32

## Solving the Equation Using a Double-Angle Formula

Solve the equation exactly using a double-angle formula: cos(2θ)=cosθ.cos(2θ)=cosθ.

## Solving an Equation Using an Identity

Solve the equation exactly using an identity: 3cosθ+3=2sin2θ,0≤θ<2π.3cosθ+3=2sin2θ,0≤θ<2π.

## Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as sin(2x)sin(2x) or cos(3x).cos(3x). When confronted with these equations, recall that y=sin(2x)y=sin(2x) is a horizontal compression by a factor of 2 of the function y=sinx.y=sinx. On an interval of 2π,2π, we can graph two periods of y=sin(2x),y=sin(2x), as opposed to one cycle of y=sinx.y=sinx. This compression of the graph leads us to believe there may be twice as many x -intercepts or solutions to sin(2x)=0sin(2x)=0 compared to sinx=0.sinx=0. This information will help us solve the equation.

## Solving a Multiple Angle Trigonometric Equation

Solve exactly: cos(2x)=12cos(2x)=12 on [0,2π).[ 0,2π ).

## Solving Right Triangle Problems

We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem. We begin with the familiar Pythagorean Theorem, a2+b2=c2,a2+b2=c2, and model an equation to fit a situation.

## Using the Pythagorean Theorem to Model an Equation

Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem.

One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure 4 .

## Using the Pythagorean Theorem to Model an Abstract Problem

OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.

Access these online resources for additional instruction and practice with solving trigonometric equations.

- Solving Trigonometric Equations I
- Solving Trigonometric Equations II
- Solving Trigonometric Equations III
- Solving Trigonometric Equations IV
- Solving Trigonometric Equations V
- Solving Trigonometric Equations VI

## 7.5 Section Exercises

Will there always be solutions to trigonometric function equations? If not, describe an equation that would not have a solution. Explain why or why not.

When solving a trigonometric equation involving more than one trig function, do we always want to try to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not?

When solving linear trig equations in terms of only sine or cosine, how do we know whether there will be solutions?

For the following exercises, find all solutions exactly on the interval 0≤θ<2π.0≤θ<2π.

2sinθ=−2–√2sinθ=−2

2sinθ=3–√2sinθ=3

2cosθ=12cosθ=1

2cosθ=−2–√2cosθ=−2

tanθ=−1tanθ=−1

tanx=1tanx=1

cotx+1=0cotx+1=0

4sin2x−2=04sin2x−2=0

csc2x−4=0csc2x−4=0

For the following exercises, solve exactly on [0,2π).[0,2π).

2cosθ=2–√2cosθ=2

2cosθ=−12cosθ=−1

2sinθ=−12sinθ=−1

2sinθ=−3–√2sinθ=−3

2sin(3θ)=12sin(3θ)=1

2sin(2θ)=3–√2sin(2θ)=3

2cos(3θ)=−2–√2cos(3θ)=−2

cos(2θ)=−3√2cos(2θ)=−32

2sin(πθ)=12sin(πθ)=1

2cos(π5θ)=3–√2cos(π5θ)=3

For the following exercises, find all exact solutions on [0,2π).[ 0,2π ).

sec(x)sin(x)−2sin(x)=0sec(x)sin(x)−2sin(x)=0

tan(x)−2sin(x)tan(x)=0tan(x)−2sin(x)tan(x)=0

2cos2t+cos(t)=12cos2t+cos(t)=1

2tan2(t)=3sec(t)2tan2(t)=3sec(t)

2sin(x)cos(x)−sin(x)+2cos(x)−1=02sin(x)cos(x)−sin(x)+2cos(x)−1=0

cos2θ=12cos2θ=12

sec2x=1sec2x=1

tan2(x)=−1+2tan(−x)tan2(x)=−1+2tan(−x)

8sin2(x)+6sin(x)+1=08sin2(x)+6sin(x)+1=0

tan5(x)=tan(x)tan5(x)=tan(x)

For the following exercises, solve with the methods shown in this section exactly on the interval [0,2π).[0,2π).

sin(3x)cos(6x)−cos(3x)sin(6x)=−0.9sin(3x)cos(6x)−cos(3x)sin(6x)=−0.9

sin(6x)cos(11x)−cos(6x)sin(11x)=−0.1sin(6x)cos(11x)−cos(6x)sin(11x)=−0.1

cos(2x)cosx+sin(2x)sinx=1cos(2x)cosx+sin(2x)sinx=1

6sin(2t)+9sint=06sin(2t)+9sint=0

9cos(2θ)=9cos2θ−49cos(2θ)=9cos2θ−4

sin(2t)=costsin(2t)=cost

cos(2t)=sintcos(2t)=sint

cos(6x)−cos(3x)=0cos(6x)−cos(3x)=0

For the following exercises, solve exactly on the interval [0,2π).[ 0,2π ). Use the quadratic formula if the equations do not factor.

tan2x−3–√tanx=0tan2x−3tanx=0

sin2x+sinx−2=0sin2x+sinx−2=0

sin2x−2sinx−4=0sin2x−2sinx−4=0

5cos2x+3cosx−1=05cos2x+3cosx−1=0

3cos2x−2cosx−2=03cos2x−2cosx−2=0

5sin2x+2sinx−1=05sin2x+2sinx−1=0

tan2x+5tanx−1=0tan2x+5tanx−1=0

cot2x=−cotxcot2x=−cotx

−tan2x−tanx−2=0−tan2x−tanx−2=0

For the following exercises, find exact solutions on the interval [0,2π).[0,2π). Look for opportunities to use trigonometric identities.

sin2x−cos2x−sinx=0sin2x−cos2x−sinx=0

sin2x+cos2x=0sin2x+cos2x=0

sin(2x)−sinx=0sin(2x)−sinx=0

cos(2x)−cosx=0cos(2x)−cosx=0

2tanx2−sec2x−sin2x=cos2x2tanx2−sec2x−sin2x=cos2x

1−cos(2x)=1+cos(2x)1−cos(2x)=1+cos(2x)

sec2x=7sec2x=7

10sinxcosx=6cosx10sinxcosx=6cosx

−3sint=15costsint−3sint=15costsint

4cos2x−4=15cosx4cos2x−4=15cosx

8sin2x+6sinx+1=08sin2x+6sinx+1=0

8cos2θ=3−2cosθ8cos2θ=3−2cosθ

6cos2x+7sinx−8=06cos2x+7sinx−8=0

12sin2t+cost−6=012sin2t+cost−6=0

tanx=3sinxtanx=3sinx

cos3t=costcos3t=cost

For the following exercises, algebraically determine all solutions of the trigonometric equation exactly, then verify the results by graphing the equation and finding the zeros.

6sin2x−5sinx+1=06sin2x−5sinx+1=0

8cos2x−2cosx−1=08cos2x−2cosx−1=0

100tan2x+20tanx−3=0100tan2x+20tanx−3=0

2cos2x−cosx+15=02cos2x−cosx+15=0

20sin2x−27sinx+7=020sin2x−27sinx+7=0

2tan2x+7tanx+6=02tan2x+7tanx+6=0

130tan2x+69tanx−130=0130tan2x+69tanx−130=0

For the following exercises, use a calculator to find all solutions to four decimal places.

sinx=0.27sinx=0.27

sinx=−0.55sinx=−0.55

tanx=−0.34tanx=−0.34

cosx=0.71cosx=0.71

For the following exercises, solve the equations algebraically, and then use a calculator to find the values on the interval [0,2π).[0,2π). Round to four decimal places.

tan2x+3tanx−3=0tan2x+3tanx−3=0

6tan2x+13tanx=−66tan2x+13tanx=−6

tan2x−secx=1tan2x−secx=1

sin2x−2cos2x=0sin2x−2cos2x=0

2tan2x+9tanx−6=02tan2x+9tanx−6=0

4sin2x+sin(2x)secx−3=04sin2x+sin(2x)secx−3=0

For the following exercises, find all solutions exactly to the equations on the interval [0,2π).[0,2π).

csc2x−3cscx−4=0csc2x−3cscx−4=0

sin2x−cos2x−1=0sin2x−cos2x−1=0

sin2x(1−sin2x)+cos2x(1−sin2x)=0sin2x(1−sin2x)+cos2x(1−sin2x)=0

3sec2x+2+sin2x−tan2x+cos2x=03sec2x+2+sin2x−tan2x+cos2x=0

sin2x−1+2cos(2x)−cos2x=1sin2x−1+2cos(2x)−cos2x=1

tan2x−1−sec3xcosx=0tan2x−1−sec3xcosx=0

sin(2x)sec2x=0sin(2x)sec2x=0

sin(2x)2csc2x=0sin(2x)2csc2x=0

2cos2x−sin2x−cosx−5=02cos2x−sin2x−cosx−5=0

1sec2x+2+sin2x+4cos2x=41sec2x+2+sin2x+4cos2x=4

## Real-World Applications

An airplane has only enough gas to fly to a city 200 miles northeast of its current location. If the pilot knows that the city is 25 miles north, how many degrees north of east should the airplane fly?

If a loading ramp is placed next to a truck, at a height of 4 feet, and the ramp is 15 feet long, what angle does the ramp make with the ground?

If a loading ramp is placed next to a truck, at a height of 2 feet, and the ramp is 20 feet long, what angle does the ramp make with the ground?

A woman is watching a launched rocket currently 11 miles in altitude. If she is standing 4 miles from the launch pad, at what angle is she looking up from horizontal?

An astronaut is in a launched rocket currently 15 miles in altitude. If a man is standing 2 miles from the launch pad, at what angle is she looking down at him from horizontal? (Hint: this is called the angle of depression.)

A woman is standing 8 meters away from a 10-meter tall building. At what angle is she looking to the top of the building?

A man is standing 10 meters away from a 6-meter tall building. Someone at the top of the building is looking down at him. At what angle is the person looking at him?

A 20-foot tall building has a shadow that is 55 feet long. What is the angle of elevation of the sun?

A 90-foot tall building has a shadow that is 2 feet long. What is the angle of elevation of the sun?

A spotlight on the ground 3 meters from a 2-meter tall man casts a 6 meter shadow on a wall 6 meters from the man. At what angle is the light?

A spotlight on the ground 3 feet from a 5-foot tall woman casts a 15-foot tall shadow on a wall 6 feet from the woman. At what angle is the light?

For the following exercises, find a solution to the word problem algebraically. Then use a calculator to verify the result. Round the answer to the nearest tenth of a degree.

A person does a handstand with his feet touching a wall and his hands 1.5 feet away from the wall. If the person is 6 feet tall, what angle do his feet make with the wall?

A person does a handstand with her feet touching a wall and her hands 3 feet away from the wall. If the person is 5 feet tall, what angle do her feet make with the wall?

A 23-foot ladder is positioned next to a house. If the ladder slips at 7 feet from the house when there is not enough traction, what angle should the ladder make with the ground to avoid slipping?

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## Mathematics > Numerical Analysis

Title: new preconditioner strategy for solving block four-by-four linear systems: an application to the saddle-point problem from 3d stokes equation.

Abstract: We have presented a fast method for solving a specific type of block four-by-four saddlepoint problem arising from the finite element discretization of the generalized 3D Stokes problem. We analyze the eigenvalue distribution and the eigenvectors of the preconditioned matrix. Furthermore, we suggested utilizing the preconditioned global conjugate gradient method (PGCG) as a block iterative solver for handling multiple right-hand sides within the sub-system and give some new convergence results. Numerical experiments have shown that our preconditioned iterative approach is very efficient for solving the 3D Stokes problem

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- \mathrm{Lauren's\:age\:is\:half\:of\:Joe's\:age.\:Emma\:is\:four\:years\:older\:than\:Joe.\:The\:sum\:of\:Lauren,\:Emma,\:and\:Joe's\:age\:is\:54.\:How\:old\:is\:Joe?}
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- Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
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We have presented a fast method for solving a specific type of block four-by-four saddlepoint problem arising from the finite element discretization of the generalized 3D Stokes problem. We analyze the eigenvalue distribution and the eigenvectors of the preconditioned matrix. Furthermore, we suggested utilizing the preconditioned global conjugate gradient method (PGCG) as a block iterative ...

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