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Online Systems of Equations Solver

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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.

• solve y = 2x, y = x + 10
• solve system of equations {y = 2x, y = x + 10, 2x = 5y}
• y = x^2 - 2, y = 2 - x^2
• solve 4x - 3y + z = -10, 2x + y + 3z = 0, -x + 2y - 5z = 17
• solve system {x + 2y - z = 4, 2x + y + z = -2, z + 2y + z = 2}
• solve 4 = x^2 + y^2, 4 = (x - 2)^2 + (y - 2)^2
• x^2 + y^2 = 4, y = x
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• x+y+z=25,\:5x+3y+2z=0,\:y-z=6
• x+2y=2x-5,\:x-y=3
• 5x+3y=7,\:3x-5y=-23
• x^2+y=5,\:x^2+y^2=7
• xy+x-4y=11,\:xy-x-4y=4
• 3-x^2=y,\:x+1=y
• xy=10,\:2x+y=1
• How do you solve a system of equations by substitution?
• To solve a system of equations by substitution, solve one of the equations for one of the variables, and substitute this expression into the other equation. Then, solve the resulting equation for the remaining variable and substitute this value back into the original equation to find the value of the other variable.
• How do you solve a system of equations by graphing?
• To solve a system of equations by graphing, graph both equations on the same set of axes and find the points at which the graphs intersect. Those points are the solutions.
• How do you solve a system of equations by elimination?
• To solve a system of equations by elimination, write the system of equations in standard form: ax + by = c, and multiply one or both of the equations by a constant so that the coefficients of one of the variables are opposite. Then, add or subtract the two equations to eliminate one of the variables. Solve the resulting equation for the remaining variable.
• What are the solving methods of a system of equations?
• There are several methods for solving a system of equations, including substitution, elimination, and graphing.
• What is a system of linear equations?
• A system of linear equations is a system of equations in which all the equations are linear and in the form ax + by = c, where a, b, and c are constants and x and y are variables.

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• High School Math Solutions – Systems of Equations Calculator, Elimination A system of equations is a collection of two or more equations with the same set of variables. In this blog post,...

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System of equations

A system of equations is a set of equations which share the same variables . Below is an example of a system of equations.

• 1 Solve 2 variable equations in less than 5 seconds!!!
• 2.1.1 Problem
• 2.1.2 Solution
• 2.2.1 Problem
• 2.2.2 Solution
• 2.3.1 Problem
• 2.3.2 Solution
• 2.4 Advanced Methods
• 3.1 Symmetry
• 3.2 Clever Substitution
• 4.1 Introductory
• 4.2 Intermediate

Solve 2 variable equations in less than 5 seconds!!!

Video Link: https://youtu.be/pSYT95hSH6M

Solving Linear Systems

A system of linear equations is where all of the variables are to the power 1. There are three elementary ways to solve a system of linear equations.

Gaussian Elimination

Gaussian elimination involves eliminating variables from the system by adding constant multiples of two or more of the equations together. Let's look at an example:

• Substitution

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We'll show how to solve the same problem from the elimination section using substitution.

Plugging this into the second equation yields

The third method for solving a system of linear equations is to graph them in the plane and observe where they intersect. We'll go back to our same example to illustrate this.

We graph the two lines as follows:

Advanced Methods

Matrices can also be used to solve systems of linear equations. In fact, they provide a way to make much broader statements about systems of linear equations.

There is a whole field of mathematics devoted to the study of linear equations called linear algebra .

Convenient Systems

Some systems can be solved by taking advantage of specific forms. Such systems can often seem tough to solve at first, however.

Consider the below system.

Clever Substitution

Introductory

• 2002 AMC 8 Problems/Problem 17
• 2007 iTest Problems/Problem 2

Intermediate

• 1989 AIME Problems/Problem 8
• 1993 AIME Problems/Problem 3

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Systems of Equations - Problems & Answers

Systems of 2 linear equations - problems with solutions Test

Problem 2 Are the systems equivalent(check if solutions of the both systems are the same): $\begin{array}{|l} 4x + 5y = 11 \\ x - y = 5 \end{array}$ and $\begin{array}{|l} 4x - 5y = 11 \\ 2x + y = 9? \end{array}$ Answer: No.

(3-32) Solve the system of equations:

Problem 3 $\begin{array}{|l} 2x - y = -5 \\ y = 1-3x \end{array}$ Answer: x = 1, y = -2.

Problem 4 $\begin{array}{|l} 3x - y = 13 \\ 3y - 2x = -4 \end{array}$ Answer: x = 5, y = 2.

Problem 5 $\begin{array}{|l} 6x - y = 11 \\ 12x - 2y - 22 = 0 \end{array}$ Answer: The solution is every couple of numbers, which is solution of the equation $6x - y = 11$.

Problem 6 $\begin{array}{|l} 5u - 6v = -2 \\ 7u + 18v = 2 \end{array}$ Answer: $x = -1, y = \frac{1}{2}$.

Problem 7 $\begin{array}{|l} 8x - 5y + 16 = 0 \\ 1x + 3y - 17 = 0 \end{array}$ Answer: $x = \frac{1}{2}, y = 4$.

Problem 8 $\begin{array}{|l} 4(x + 2) - 7(x - y) = 7 \\ 7(x + y) + 10(x - 2) = 79 \end{array}$ Answer: $x = 5, y = 2.$

Problem 9 $\begin{array}{|l} 3x + 4(x - 3) = 3(2y - 3) - 3y \\ 3y + 2(x - 4) = 5(y + 2) - 28\end{array}$ Answer: (-4; 1).

Problem 10 $\begin{array}{|l} (x + 3)(x-1) = 4y + x^2 + 5 \\ (x - 3)(3x + 2) = 3x^2 - 14y + 15\end{array}$ Answer: There's no solution.

Problem 11 $\begin{array}{|l} (x - 1)(y + 2) - (x - 2)(y + 5) = 0 \\ (x + 4)(y - 3) - (x + 7)(y - 4) = 0\end{array}$ Answer: x = 5, y = 7.

Problem 12 $\begin{array}{|l} (x + 2)^2 - (x + 3)(x - 3) - 3(y + 5) = 0 \\ (2y - 3)^2 - y(4y - 3) + 12x - 15 = 0\end{array}$ Answer: The solution is every couple, which is solution to the equation 4x - 3y - 2 = 0.

Problem 13 $\begin{array}{|l} \frac{y + 2}{6} - \frac{y-4}{2} = \frac{x}{3} \\ \frac{4}{3}(y - 1) - 2x = -2\end{array}$ Answer: x = 3, y = 4.

Problem 14 $\begin{array}{|l} 0.25x - 0.04y = 1 \\ 0.4x + 1.5y = 40.7\end{array}$ Answer: x = 8, y = 25.

Problem 15 $\begin{array}{|l} \frac{5x-3y}{4} = \frac{x-5y}{3} \\ 7x + y = 12\end{array}$ Answer: x = 2, y = -2

Problem 16 $\begin{array}{|l} \frac{3x+1}{5}+2y-3 = 0 \\ \frac{4y-5}{6}+3y-9 = -\frac{1}{2}\end{array}$ Answer: $x = -\frac{42}{11}, y = \frac{28}{11}$

Problem 17 $\begin{array}{|l} \frac{3x-1}{5}+3y-4 = 15 \\ \frac{3y-5}{6}+2x-8 = \frac{23}{3}\end{array}$ Answer: $x = 7, y = 5$

Problem 18 $\begin{array}{|l} \frac{2x-z}{6}+\frac{2x-z}{9} = 3 \\ \frac{x+z}{3}-\frac{x-z}{4} = 4\end{array}$ Answer: x = y = 6

Problem 19 $\begin{array}{|l} \frac{x-1}{3} + \frac{5y+1}{2} = \frac{x+10y-8}{6} \\ \frac{(x+2)(5y-2)}{2} = 5+\frac{5xy}{2}-2(x+1)\end{array}$ Answer: There's no solution.

Problem 20 $\begin{array}{|l} \frac{5x-1}{6} + \frac{3y-1}{10} = 3 \\ \frac{11-x}{6} + \frac{11+y}{4} = 3\end{array}$ Answer: x = 5, y = -3.

Problem 21 $\begin{array}{|l}y-0.2(x - 2) = 1.4\\ \frac{5}{2} - \frac{2y - 3}{4} = \frac{4x - y}{8}\end{array}$ Answer: $x = 5, y = 2.$

Problem 22 $\begin{array}{|l}\frac{x}{5} + 0.03(10y - 20) = 0.8\\ \frac{2x + 4.5}{20} - 0.75 = \frac{y - 3}{8}\end{array}$ Answer: $x = 4, y = 2.$

Problem 23 $\begin{array}{|l}y-x-\frac{5x-4}{2}=3-\frac{11y+17}{4}\\ x+\frac{9y+11}{4}-\frac{3y+4}{7}=6\end{array}$ Answer: $x = 2, y = 1.$

Problem 24 $\begin{array}{|l}\frac{5x-3y}{3}-\frac{2y-3x}{5}=x+1\\ \frac{2x-3y}{3}-\frac{3y-4x}{2}=y+1\end{array}$ Answer: $x = 3, y = 2.$

Problem 25 $\begin{array}{|l}\frac{x-1}{4}\frac{1+y}{2}=\frac{1}{6}-\frac{x+2y}{6}\\ \frac{x-2}{3}+\frac{x}{15}=\frac{y+4}{5}-\frac{4x-y}{15}\end{array}$ Answer: The solution is every couple, which is solution of the equation $5x - 2y = 11.$

Problem 26 $\begin{array}{|l}\frac{x+2y}{4}-\frac{x-2y}{2}=1-\left[x-\frac{7-2y}{3}\right]\\ 3x-2y=8\end{array}$ Answer: $x = 3, y = \frac{1}{2}.$

Problem 27 $\begin{array}{|l}\frac{7+x}{5}-\frac{2x-y}{4}-3y=-5\end{array}$ $\begin{array}{|l}\frac{5y-7}{2}+\frac{4x-3}{6}-18=-5x\end{array}$ Answer: x = 3, y = 2.

Problem 28 $\begin{array}{|l}\frac{11y}{20}-0.8\left(\frac{x}{4}+2.5\right)=\frac{5}{2}\end{array}$ $\begin{array}{|l}\frac{6x-0.3y}{2}-\frac{3}{2}=2(1+x)\end{array}$ Answer: x = 5, y = 10.

Problem 29 $\begin{array}{|l}0.5x-\frac{y-4}{5}=0.3x-\frac{y-4}{2}\end{array}$ $\begin{array}{|l}0.5y-\frac{x-4}{6}=\frac{7y}{12}-\frac{x-3}{3}\end{array}$ Answer: x = 3, y = 2.

Problem 30 $\begin{array}{|l}\frac{2(x-y)}{3}+1.6=\frac{8x}{15}-\frac{3y-10}{5}\end{array}$ $\begin{array}{|l}\frac{3x+4}{4}+\frac{y}{8}=\frac{5x}{6}-\frac{y-17}{12}\end{array}$ Answer: x = 5, y = 4.

Problem 31 $\begin{array}{|l}\frac{(2+x)(5y-2)}{2}=5+\frac{5xy}{2}-2(1+x)\end{array}$ $\begin{array}{|l}(x-1)^2+(2y+1)^2=2(1+2y)(x-1)\end{array}$ Answer: The solution is every couple, which is solution of the equation x + 5y = 5

System of Equations

In mathematics, a system of equations, also known as a set of simultaneous equations or an equation system, is a finite set of equations for which we sought common solutions. In systems of equations, variables are related in a specific way in each equation. i.e., the equations can be solved simultaneously to find a set of values of variables that satisfies each equation.

System of linear equations finds applications in our day-to-day lives in modelling problems where the unknown values can be represented in form of variables. Solving system of equations involves different methods such as substitution, elimination, graphing, etc. Let us look into each method in detail.

What is a System of Equations?

In algebra, a system of equations comprises two or more equations and seeks common solutions to the equations. "A system of linear equations is a set of equations which are satisfied by the same set of values of variables."

System of Equations Example

A system of equations as discussed above is a set of equations that seek a common solution for the variables included. The following set of linear equations is an example of the system of equations:

• 2x - y = 12
• x - 2y = 48

Note that the values x = -8 and y = -28 satisfy each of these equations and hence the pair (x, y) = (-8, -28) is the solution of the above system of equations. But how to solve system of equations? Let us see.

Solutions of System of Equations

Solution of system of equations is the set of values of variables that satisfies each linear equation in the system. The main reason behind solving an equation system is to find the value of the variable that satisfies the condition of all the given equations true. There systems of equations are classified into 3 types depending on their number of solutions:

• Linear System with "Unique solution"
• Linear System with "No solution"
• Linear System with "Infinitely many solutions"

We know that every linear equation represents a line on the coordinate plane. In this perception, the above figure should give more sense to understanding the different types of solutions of system of equations.

Unique Solution of a System of Equations

A system of equations has unique solution when there is only set of variables exist that satisfy each equation in the system. In terms of graphs, a system with a unique solution has lines (representing the equations) that intersect (at one point).

No Solution

A system of equations has no solution when there exists no set of variables that satisfy each linear equation in the system. If we graph that kind of system, the resulting lines will be parallel to each other.

Infinite Many Solutions

A system of equations can have infinitely many solutions when there exist infinite sets of variables that satisfy each equation. In such cases, the lines corresponding to the linear equations would overlap each other on the graph. i.e., both equations represent the same line. Since a line has infinite points on it, each point on it becomes a solution of the system.

Solving System of Equations

Solving a system of equations means finding the values of the variables used in the set of equations. Any system of equations can be solved in different methods.

• Substitution Method
• Elimination Method
• Graphical Method
• Cross-multiplication method

To solve a system of equations in 2 variables, we need at least 2 equations. Similarly, for solving a system of equations in 3 variables, we will require at least 3 equations. Let us understand 3 ways to solve a system of equations given the equations are linear equations in two variables.

☛ Also Check: Solving System of Linear Equations

Solving System of Linear Equations By Substitution Method

For solving the system of equations using the substitution method given two linear equations in x and y, in one of the equations, express y in terms x in one of the equations and then substitute it in the other equation.

Example: Solve the system of equations using the substitution method.

3x − y = 23 → (1)

4x + 3y = 48 → (2)

From (1), we get:

y = 3x − 23 → 3

Plug in y in (2),

4x + 3 (3x − 23) = 48

13x − 69 = 48

Now, plug in x = 9 in (1)

y = 3 × 9 − 23 = 4

Hence, x = 9 and y = 4 is the solution of the given system of equations.

Solving System of Equations By Elimination Method

Using the elimination method to solve the system of equations, we eliminate one of the unknowns, by multiplying equations by suitable numbers, so that the coefficients of one of the variables become the same.

Example: Solve the following system of linear equations by elimination method.

2x + 3y = 4 → (1) and 3x + 2y = 11 → (2)

The coefficients of y are 3 and 2; LCM (3, 2) = 6

Multiplying Equation (1) by 2 and Equation (2) by 3, we get

4x + 6y = 8 → (3)

9x + 6y = 33 → (4)

On subtracting (3) from (4), we get

Plugging in x = 5 in (2) we get

15 + 2y = 11

Hence, x = 5, y = −2 is the solution.

Solving System of Equations by Graphing

In this method, solving system of linear equations is done by plotting their graphs. "The point of intersection of the two lines is the solution of the system of equations using graphical method ."

Example: 3x + 4y = 11 and -x + 2y = 3

Find at least two values of x and y satisfying equation 3x + 4y = 11

So we have 2 points A (1, 2) and B (3, 0.5).

Similarly, find at least two values of x and y satisfying the equation -x + 2y = 3

We have two points C(-3, 0) and D( 3, 3). Plotting these points on the graph we can get the lines in a coordinate plane as shown below.

We observe that the two lines intersect at (1, 2). So, x = 1, y = 2 is the solution of given system of equations. Methods I and II are the algebraic way of solving simultaneous equations and III is the graphical method .

Solving System of Equations Using Cross-multiplication method

In this method, we solve the systems of equations a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 using the cross-multiplication formula:

x / (b 1 c 2 - b 2 c 1 ) = -y / (a 1 c 2 - a 2 c 1 ) = 1 / (a 1 b 2 - a 2 b 1 )

For more detailed information of this method, click here .

Solving System of Equations Using Matrices

The solution of a system of equations can be solved using matrices . In order to solve a system of equations using matrices, express the given equations in standard form , with the variables and constants on respective sides. for the given equations,

a\(_1\)x + \(b_1\)y + \(c_1\)z = \(d_1\)

a\(_2\)x + \(b_2\)y + \(c_2\)z = \(d_2\)

a\(_3\)x + \(b_3\)y + \(c_3\)z = \(d_3\)

we can express them in the form of matrices as,

\(\left[\begin{array}{ccc} a_1x + b_1y + c_1 z \\ a_2x + b_2y + c_2 z \\ a_3x + b_3y + c_3 z \end{array}\right] = \left[\begin{array}{ccc} d_1 \\ d_2 \\ d_3 \end{array}\right]\)

⇒\(\left[\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right] + \left[\begin{array}{ccc} x \\ y \\ z \end{array}\right] = \left[\begin{array}{ccc} d_1 \\ d_2 \\ d_3 \end{array}\right]\)

A = \(\left[\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right]\), X = \(\left[\begin{array}{ccc} x \\ y \\ z \end{array}\right]\), B = \(\left[\begin{array}{ccc} d_1 \\ d_2 \\ d_3 \end{array}\right]\)

⇒ The solution of the system is given by the formula X = A -1 B, where A -1 = inverse of the matrix A. To understand this method more in detail, click here .

Alternatively, Cramer's rule can also be used to solve the system of equations using determinants .

Applications of System of Equations

Systems of equations are a very useful tool and find application in our day-to-day lives for modeling real-life situations and analyzing questions about them.

• System of Equations Word Problems
• Systems of Equations Worksheet

For applying the concept of the system of equations, we need to translate the given situation into two linear equations in two variables , then further solve to find the solution of linear programming problems. Any method to solve the system of equations, substitution, elimination, graphical, etc methods. Follow the below-given steps to apply the system of equations to solve problems in our daily lives,

• To translate and represent the given situation in form of a system of equations, identify unknown quantities in a problem and represent them with variables.
• Write a system of equations modelling the conditions of the problem.
• Solve the system of equations.
• Check and express the obtained solution in terms of the given context.

☛Related Articles:

Check these articles related to the concept of the system of equations.

• Solutions of a Linear Equation
• Simultaneous Linear Equations
• Solving Linear Equations Calculator
• Equation Calculator
• System of Equations Calculator

System of Equations Examples

Example 1: In a ΔLMN, ∠N = 3∠M = 2(∠L+∠M). Calculate the angles of ΔLMN using this information.

Let ∠L = x° and ∠M = y°.

∠N = 3∠M = (3y)°.

∠L + ∠M + ∠N = 180° ∴ x + y + 3y = 180 x + 4y = 180 ... (1) From the given equation, 3∠M = 2(∠L+∠M) ∴ 3y = 2(x + y) ⇒ 2x - y = 0 ... (2) On multiplying (2) by 4 and adding the result to (1), we get 9x = 180 x = 20 Substitute this in (2): 40 - y = 0 ⇒ y = 40 Therefore, ∠L = 20° and ∠M = 40°. Therefore, the third angle is, ∠N = 180 - (20 + 40) = 120°. Answer: ∠L = 20°, ∠M = 40°, and ∠N = 120°.

Example 2: Peter is three times as old as his son. 5 years later, he shall be two and half times as old as his son. What's Peter's present age?

Let Peter's age be x years and his son's age be y years. Then by given info,

x = 3y ... (1)

5 years later,

Peter's age = x + 5 years and his son's age = y + 5 years.

By the given condition,

x + 5 = 5/2 (y + 5)

2x - 5y - 15 = 0 ... (2)

From (1), we have x = 3y. Substitute this in (2):

2(3y) - 5y - 15 = 0

Substitute this in (1): x = 3(15) = 45

Thus, the solution of the given system is (x, y) = (45, 15).

Answer: Present age of Peter = 45 years; Present age of son = 15 years.

Example 3: Tressa starts her job with a certain monthly salary and earns a fixed increment every year. If her salary was $1500 after 4 years and $1800 after 10 years of service, find her starting salary and annual increment.

Let her starting salary be x and annual increment be y.

After 4 years, her salary was $1500

x + 4y = 1500 ... (1)

After 10 years, her salary was $1800

x + 10y = 1800 ... (2)

Subtracting Eqn (1) from (2), we get

On putting y = 50 in (1), we get x = 1300.

Answer: Starting salary was $1300 and annual increment is $50.

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Practice Questions on System of Equations

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FAQs on System of Equations

What is a system of equations in mathematics.

A system of equations in mathematics is a set of linear equations that need to be solved to find a common solution. A real-life problem with two or more unknowns can be converted into a system of equations and can be solved to find a set of values of variables that satisfy all equations.

How Do You Solve a System of Equations?

Solving a system of equations is computing the unknown variables still balancing the equations on both sides. We solve an equation system to find the values of the variables that satisfy the condition of all the given equations. There are different methods to solve a system of equations,

• Graphical method
• Substitution method
• Elimination method

How Do You Create a System of Equations with Two Variables?

To create a system of equations with two variables:

• First, identify the two unknown quantities in the given problem.
• Next, find out the two conditions given and frame equations for each of them.

How to Solve a System of Equations by Substitution Method?

The substitution method is one of the ways to solve a system of equations in two variables, given the set of linear equations. In this method, we substitute the value of a variable found by one equation in the second equation.

How to Solve a System of Equations Using the Elimination Method?

The elimination method is used to solve a system of linear equations. In the elimination method, we eliminate one of the two variables by multiplying each equation with the required numbers and try to solve equations with another variable. In this process, finding LCM of coefficients would be helpful.

What is the Purpose of Graphing Systems of Equations?

To solve the system of equations, given a set of linear equations graphically, we need to find at least two solutions for the respective equations. We observe the pattern of lines after plotting the points to infer it is consistent, dependent, or inconsistent.

• If the two lines are intersecting at the same point, then the intersection point gives a unique solution for the system of equations.
• If the two lines coincide, then in this case there are infinitely many solutions.
• If the two lines are parallel, then in this case there is no solution.

What are Homogeneous System of Linear Equations?

The homogeneous system of linear equations is a set of linear equation each one of which has its constant term to be 0. The process of solving these kind of systems can be learnt in detailed by clicking here .

How to Solve a System of Equations With Cross Multiplication Method?

While solving a system of equations using the cross-multiplication method , we use the formula x / (b 1 c 2 - b 2 c 1 ) = -y / (a 1 c 2 - a 2 c 1 ) = 1 / (a 1 b 2 - a 2 b 1 ) to solve the system a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0.

How do you Solve a System of Equations Using 2 Equations with 3 Variables?

An equation with 3 variables represents a plane .

• Step 1) To solve a system of 2 equations with 3 variables say x, y, and z, we will consider the 1st two equations and eliminate one of the variables, say x, to obtain a new equation.
• Step 2) Next, we write the 2nd variable, y in terms of z from the new equation and substitute it in the third equation.
• Step 3) Assuming z = a, we will obtain values of x and y also in terms of 'a'.
• Step 4) Once, we know the value of a, we can find the values of x and y in terms of 'a'.

What is the Solution to the System of Equations and How to Find It?

Solving system of equations is finding the values of variables that satisfy each equation in that system. There are three main methods to solving system of equations, they are:

Where Can I Find System of Linear Equations Calculator?

The system of linear equations calculator is available here . This allows us to enter the linear equations. Then it will show the solution along with step-by-step solution.

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How to Solve Systems of Equations

Last Updated: March 26, 2024 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 8 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 243,390 times.

Solving a system of equations requires you to find the value of more than one variable in more than one equation. You can solve a system of equations through addition, subtraction, multiplication, or substitution. If you want to know how to solve a system of equations, just follow these steps.

Solve by Subtraction

• For example, if both equations have the variable positive 2x, you should use the subtraction method to find the value of both variables.
• Write one equation above the other by matching up the x and y variables and the whole numbers. Write the subtraction sign outside the quantity of the second system of equations.
• 2x + 4y = 8
• -(2x + 2y = 2)

• 2x - 2x = 0
• 4y - 2y = 2y
• 2x + 4y = 8 -(2x + 2y = 2) = 0 + 2y = 6

• Divide 2y and 6 by 2 to get y = 3

• Plug y = 3 into the equation 2x + 2y = 2 and solve for x.
• 2x + 2(3) = 2
• You have solved the system of equations by subtraction. (x, y) = (-2, 3)

• 2(-2) + 4(3) = 8
• -4 + 12 = 8
• 2(-2) + 2(3) = 2

Solve by Addition

• Write one equation above the other by matching up the x and y variables and the whole numbers. Write the addition sign outside the quantity of the second system of equations.
• 3x + 6y = 8
• +(x - 6y = 4)

• 3x + x = 4x
• 6y + -6y = 0
• = 4x + 0 = 12

• 4x + 0 = 12
• Divide 4x and 12 by 3 to get x = 3

• Plug x = 3 into the equation x - 6y = 4 to solve for y.
• You have solved the system of equations by addition. (x, y) = (3, -1/6)

• 3(3) + 6(-1/6) = 8
• 3 - (6 * -1/6) =4
• 3 - - 1 = 4

Solve by Multiplication

• 3x + 2y = 10

• 2 (2x - y = 2)
• 4x - 2y = 4

• + 4x - 2y = 4
• 7x + 0 = 14

• x = 2 ---> 2x - y = 2
• You have solved the system of equations by multiplication. (x, y) = (2, 2)

• Plug (2, 2) in for (x, y) in the equation 3x + 2y = 10.
• 3(2) + 2(2) = 10
• Plug (2, 2) in for (x, y) in the equation 2x - y = 2.
• 2(2) - 2 = 2

Solve by Substitution

• If you're working with the equations 2x + 3y = 9 and x + 4y = 2, you should isolate x in the second equation.

• x = 2 - 4y --> 2x + 3y = 9
• 2(2 - 4y) + 3y = 9
• 4 - 8y + 3y = 9
• -5y = 9 - 4

• y = -1 --> x = 2 - 4y
• x = 2 - 4(-1)
• You have solved the system of equations by substitution. (x, y) = (6, -1)

• 2(6) + 3(-1) = 9
• Plug (6, -1) in for (x, y) in the equation x + 4y = 2.
• 6 + 4(-1) = 2

Community Q&A

• You should be able to solve any linear system of equations using the addition, subtraction, multiplication, or substitution method, but one method is usually the easiest depending on the equations. Thanks Helpful 0 Not Helpful 0

You Might Also Like

• ↑ https://content.nroc.org/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U14_L2_T2_text_final.html
• ↑ https://www.mathsisfun.com/definitions/system-of-equations.html
• ↑ https://www.cliffsnotes.com/study-guides/algebra/algebra-i/equations-with-two-variables/solving-systems-of-equations-simultaneous-equations
• ↑ https://www.purplemath.com/modules/systlin5.htm
• ↑ https://flexbooks.ck12.org/cbook/ck-12-interactive-middle-school-math-8-for-ccss/section/5.5/related/lesson/solving-systems-of-equations-by-multiplication-and-addition-c-alg/
• ↑ https://virtualnerd.com/sat-math/algebra/systems-equations/equations-solution-by-elimination-multiplication
• ↑ https://www.khanacademy.org/math/algebra/systems-of-linear-equations/solving-systems-of-equations-with-substitution/v/solving-systems-with-substitution
• ↑ https://opentextbc.ca/businesstechnicalmath/chapter/solve-systems-of-equations-by-substitution/

About This Article

To solve a system of equations by elimination, make sure both equations have one variable with the same coefficient. Subtract the like terms of the equations so that you’re eliminating that variable, then solve for the remaining one. Plug the solution back into one of the original equations to solve for the other variable. To solve by substitution, solve for 1 variable in the first equation, then plug the value into the second equation and solve for the second variable. Finally, solve for the first variable in either of the first equations. Write your answer by placing both terms in parentheses with a comma between. If you want to learn how to check your answers, keep reading the article! Did this summary help you? Yes No

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4.4 Solve Systems of Equations with Three Variables

Learning objectives.

By the end of this section, you will be able to:

• Determine whether an ordered triple is a solution of a system of three linear equations with three variables
• Solve a system of linear equations with three variables
• Solve applications using systems of linear equations with three variables

Be Prepared 4.10

Before you get started, take this readiness quiz.

Evaluate 5 x − 2 y + 3 z 5 x − 2 y + 3 z when x = −2 , x = −2 , y = −4 , y = −4 , and z = 3 . z = 3 . If you missed this problem, review Example 1.21 .

Be Prepared 4.11

Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. { − 2 x + y = −11 x + 3 y = 9 . { − 2 x + y = −11 x + 3 y = 9 . If you missed this problem, review Example 2.6 .

Be Prepared 4.12

Classify the equations as a conditional equation, an identity, or a contradiction and then state the solution. { 7 x + 8 y = 4 3 x − 5 y = 27 . { 7 x + 8 y = 4 3 x − 5 y = 27 . If you missed this problem, review Example 2.8 .

Determine Whether an Ordered Triple is a Solution of a System of Three Linear Equations with Three Variables

In this section, we will extend our work of solving a system of linear equations. So far we have worked with systems of equations with two equations and two variables. Now we will work with systems of three equations with three variables. But first let's review what we already know about solving equations and systems involving up to two variables.

We learned earlier that the graph of a linear equation , a x + b y = c , a x + b y = c , is a line. Each point on the line, an ordered pair ( x , y ) , ( x , y ) , is a solution to the equation. For a system of two equations with two variables, we graph two lines. Then we can see that all the points that are solutions to each equation form a line. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions

We know when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Similarly, for a linear equation with three variables a x + b y + c z = d , a x + b y + c z = d , every solution to the equation is an ordered triple, ( x , y , z ) ( x , y , z ) , that makes the equation true.

Linear Equation in Three Variables

A linear equation with three variables, where a, b, c, and d are real numbers and a, b , and c are not all 0, is of the form

Every solution to the equation is an ordered triple, ( x , y , z ) ( x , y , z ) that makes the equation true.

All the points that are solutions to one equation form a plane in three-dimensional space. And, by finding what the planes have in common, we’ll find the solution to the system.

When we solve a system of three linear equations represented by a graph of three planes in space, there are three possible cases.

To solve a system of three linear equations, we want to find the values of the variables that are solutions to all three equations. In other words, we are looking for the ordered triple ( x , y , z ) ( x , y , z ) that makes all three equations true. These are called the solutions of the system of three linear equations with three variables .

Solutions of a System of Linear Equations with Three Variables

Solutions of a system of equations are the values of the variables that make all the equations true. A solution is represented by an ordered triple ( x , y , z ) . ( x , y , z ) .

To determine if an ordered triple is a solution to a system of three equations, we substitute the values of the variables into each equation. If the ordered triple makes all three equations true, it is a solution to the system.

Example 4.31

Determine whether the ordered triple is a solution to the system: { x − y + z = 2 2 x − y − z = −6 2 x + 2 y + z = −3 . { x − y + z = 2 2 x − y − z = −6 2 x + 2 y + z = −3 .

ⓐ ( −2 , −1 , 3 ) ( −2 , −1 , 3 ) ⓑ ( −4 , −3 , 4 ) ( −4 , −3 , 4 )

Try It 4.61

Determine whether the ordered triple is a solution to the system: { 3 x + y + z = 2 x + 2 y + z = −3 3 x + y + 2 z = 4 . { 3 x + y + z = 2 x + 2 y + z = −3 3 x + y + 2 z = 4 .

ⓐ ( 1 , −3 , 2 ) ( 1 , −3 , 2 ) ⓑ ( 4 , −1 , −5 ) ( 4 , −1 , −5 )

Try It 4.62

Determine whether the ordered triple is a solution to the system: { x − 3 y + z = −5 − 3 x − y − z = 1 2 x − 2 y + 3 z = 1 . { x − 3 y + z = −5 − 3 x − y − z = 1 2 x − 2 y + 3 z = 1 .

ⓐ ( 2 , −2 , 3 ) ( 2 , −2 , 3 ) ⓑ ( −2 , 2 , 3 ) ( −2 , 2 , 3 )

Solve a System of Linear Equations with Three Variables

To solve a system of linear equations with three variables, we basically use the same techniques we used with systems that had two variables. We start with two pairs of equations and in each pair we eliminate the same variable. This will then give us a system of equations with only two variables and then we know how to solve that system!

Next, we use the values of the two variables we just found to go back to the original equation and find the third variable. We write our answer as an ordered triple and then check our results.

Example 4.32

How to solve a system of equations with three variables by elimination.

Solve the system by elimination: { x − 2 y + z = 3 2 x + y + z = 4 3 x + 4 y + 3 z = −1 . { x − 2 y + z = 3 2 x + y + z = 4 3 x + 4 y + 3 z = −1 .

Try It 4.63

Solve the system by elimination: { 3 x + y − z = 2 2 x − 3 y − 2 z = 1 4 x − y − 3 z = 0 . { 3 x + y − z = 2 2 x − 3 y − 2 z = 1 4 x − y − 3 z = 0 .

Try It 4.64

Solve the system by elimination: { 4 x + y + z = −1 − 2 x − 2 y + z = 2 2 x + 3 y − z = 1 . { 4 x + y + z = −1 − 2 x − 2 y + z = 2 2 x + 3 y − z = 1 .

The steps are summarized here.

Solve a system of linear equations with three variables.

• If any coefficients are fractions, clear them.
• Decide which variable you will eliminate.
• Work with a pair of equations to eliminate the chosen variable.
• Multiply one or both equations so that the coefficients of that variable are opposites.
• Add the equations resulting from Step 2 to eliminate one variable
• Step 3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
• Step 4. The two new equations form a system of two equations with two variables. Solve this system.
• Step 5. Use the values of the two variables found in Step 4 to find the third variable.
• Step 6. Write the solution as an ordered triple.
• Step 7. Check that the ordered triple is a solution to all three original equations.

Example 4.33

Solve: { 3 x − 4 z = 0 3 y + 2 z = −3 2 x + 3 y = −5 . { 3 x − 4 z = 0 3 y + 2 z = −3 2 x + 3 y = −5 .

We can eliminate z z from equations (1) and (2) by multiplying equation (2) by 2 and then adding the resulting equations.

Notice that equations (3) and (4) both have the variables x x and y y . We will solve this new system for x x and y y .

To solve for y , we substitute x = −4 x = −4 into equation (3).

We now have x = −4 x = −4 and y = 1 . y = 1 . We need to solve for z . We can substitute x = −4 x = −4 into equation (1) to find z .

We write the solution as an ordered triple. ( −4 , 1 , −3 ) ( −4 , 1 , −3 )

We check that the solution makes all three equations true.

3 x − 4 z = 0 ( 1 ) 3 ( −4 ) − 4 ( −3 ) = ? 0 0 = 0 ✓ 3 y + 2 z = −3 ( 2 ) 3 ( 1 ) + 2 ( −3 ) = ? − 3 −3 = −3 ✓ 2 x + 3 y = −5 ( 3 ) 2 ( −4 ) + 3 ( 1 ) = ? − 5 −5 = −5 ✓ The solution is ( −4 , 1 , −3 ) . 3 x − 4 z = 0 ( 1 ) 3 ( −4 ) − 4 ( −3 ) = ? 0 0 = 0 ✓ 3 y + 2 z = −3 ( 2 ) 3 ( 1 ) + 2 ( −3 ) = ? − 3 −3 = −3 ✓ 2 x + 3 y = −5 ( 3 ) 2 ( −4 ) + 3 ( 1 ) = ? − 5 −5 = −5 ✓ The solution is ( −4 , 1 , −3 ) .

Try It 4.65

Solve: { 3 x − 4 z = −1 2 y + 3 z = 2 2 x + 3 y = 6 . { 3 x − 4 z = −1 2 y + 3 z = 2 2 x + 3 y = 6 .

Try It 4.66

Solve: { 4 x − 3 z = −5 3 y + 2 z = 7 3 x + 4 y = 6 . { 4 x − 3 z = −5 3 y + 2 z = 7 3 x + 4 y = 6 .

When we solve a system and end up with no variables and a false statement, we know there are no solutions and that the system is inconsistent. The next example shows a system of equations that is inconsistent.

Example 4.34

Solve the system of equations: { x + 2 y − 3 z = −1 x − 3 y + z = 1 2 x − y − 2 z = 2 . { x + 2 y − 3 z = −1 x − 3 y + z = 1 2 x − y − 2 z = 2 .

Use equation (1) and (2) to eliminate z .

Use (2) and (3) to eliminate z z again.

Use (4) and (5) to eliminate a variable.

There is no solution.

We are left with a false statement and this tells us the system is inconsistent and has no solution.

Try It 4.67

Solve the system of equations: { x + 2 y + 6 z = 5 − x + y − 2 z = 3 x − 4 y − 2 z = 1 . { x + 2 y + 6 z = 5 − x + y − 2 z = 3 x − 4 y − 2 z = 1 .

Try It 4.68

Solve the system of equations: { 2 x − 2 y + 3 z = 6 4 x − 3 y + 2 z = 0 − 2 x + 3 y − 7 z = 1 . { 2 x − 2 y + 3 z = 6 4 x − 3 y + 2 z = 0 − 2 x + 3 y − 7 z = 1 .

When we solve a system and end up with no variables but a true statement, we know there are infinitely many solutions. The system is consistent with dependent equations. Our solution will show how two of the variables depend on the third.

Example 4.35

Solve the system of equations: { x + 2 y − z = 1 2 x + 7 y + 4 z = 11 x + 3 y + z = 4 . { x + 2 y − z = 1 2 x + 7 y + 4 z = 11 x + 3 y + z = 4 .

Use equation (1) and (3) to eliminate x .

Use equation (1) and (2) to eliminate x again.

Use equation (4) and (5) to eliminate y y .

The true statement 0 = 0 0 = 0 tells us that this is a dependent system that has infinitely many solutions. The solutions are of the form ( x , y , z ) ( x , y , z ) where x = 5 z − 5 ; y = −2 z + 3 x = 5 z − 5 ; y = −2 z + 3 and z is any real number.

Try It 4.69

Solve the system by equations: { x + y − z = 0 2 x + 4 y − 2 z = 6 3 x + 6 y − 3 z = 9 . { x + y − z = 0 2 x + 4 y − 2 z = 6 3 x + 6 y − 3 z = 9 .

Try It 4.70

Solve the system by equations: { x − y − z = 1 − x + 2 y − 3 z = −4 3 x − 2 y − 7 z = 0 . { x − y − z = 1 − x + 2 y − 3 z = −4 3 x − 2 y − 7 z = 0 .

Solve Applications using Systems of Linear Equations with Three Variables

Applications that are modeled by a systems of equations can be solved using the same techniques we used to solve the systems. Many of the application are just extensions to three variables of the types we have solved earlier.

Example 4.36

The community college theater department sold three kinds of tickets to its latest play production. The adult tickets sold for $15, the student tickets for $10 and the child tickets for $8. The theater department was thrilled to have sold 250 tickets and brought in $2,825 in one night. The number of student tickets sold is twice the number of adult tickets sold. How many of each type did the department sell?

Try It 4.71

The community college fine arts department sold three kinds of tickets to its latest dance presentation. The adult tickets sold for $20, the student tickets for $12 and the child tickets for $10.The fine arts department was thrilled to have sold 350 tickets and brought in $4,650 in one night. The number of child tickets sold is the same as the number of adult tickets sold. How many of each type did the department sell?

Try It 4.72

The community college soccer team sold three kinds of tickets to its latest game. The adult tickets sold for $10, the student tickets for $8 and the child tickets for $5. The soccer team was thrilled to have sold 600 tickets and brought in $4,900 for one game. The number of adult tickets is twice the number of child tickets. How many of each type did the soccer team sell?

Access this online resource for additional instruction and practice with solving a linear system in three variables with no or infinite solutions.

• Solving a Linear System in Three Variables with No or Infinite Solutions
• 3 Variable Application

Section 4.4 Exercises

Practice makes perfect.

In the following exercises, determine whether the ordered triple is a solution to the system.

{ 2 x − 6 y + z = 3 3 x − 4 y − 3 z = 2 2 x + 3 y − 2 z = 3 { 2 x − 6 y + z = 3 3 x − 4 y − 3 z = 2 2 x + 3 y − 2 z = 3

ⓐ ( 3 , 1 , 3 ) ( 3 , 1 , 3 ) ⓑ ( 4 , 3 , 7 ) ( 4 , 3 , 7 )

{ − 3 x + y + z = −4 − x + 2 y − 2 z = 1 2 x − y − z = −1 { − 3 x + y + z = −4 − x + 2 y − 2 z = 1 2 x − y − z = −1

ⓐ ( −5 , −7 , 4 ) ( −5 , −7 , 4 ) ⓑ ( 5 , 7 , 4 ) ( 5 , 7 , 4 )

{ y − 10 z = −8 2 x − y = 2 x − 5 z = 3 { y − 10 z = −8 2 x − y = 2 x − 5 z = 3

ⓐ ( 7 , 12 , 2 ) ( 7 , 12 , 2 ) ⓑ ( 2 , 2 , 1 ) ( 2 , 2 , 1 )

{ x + 3 y − z = 15 y = 2 3 x − 2 x − 3 y + z = −2 { x + 3 y − z = 15 y = 2 3 x − 2 x − 3 y + z = −2

ⓐ ( −6 , 5 , 1 2 ) ( −6 , 5 , 1 2 ) ⓑ ( 5 , 4 3 , −3 ) ( 5 , 4 3 , −3 )

In the following exercises, solve the system of equations.

{ 5 x + 2 y + z = 5 − 3 x − y + 2 z = 6 2 x + 3 y − 3 z = 5 { 5 x + 2 y + z = 5 − 3 x − y + 2 z = 6 2 x + 3 y − 3 z = 5

{ 6 x − 5 y + 2 z = 3 2 x + y − 4 z = 5 3 x − 3 y + z = −1 { 6 x − 5 y + 2 z = 3 2 x + y − 4 z = 5 3 x − 3 y + z = −1

{ 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3 { 2 x − 5 y + 3 z = 8 3 x − y + 4 z = 7 x + 3 y + 2 z = −3

{ 5 x − 3 y + 2 z = −5 2 x − y − z = 4 3 x − 2 y + 2 z = −7 { 5 x − 3 y + 2 z = −5 2 x − y − z = 4 3 x − 2 y + 2 z = −7

{ 3 x − 5 y + 4 z = 5 5 x + 2 y + z = 0 2 x + 3 y − 2 z = 3 { 3 x − 5 y + 4 z = 5 5 x + 2 y + z = 0 2 x + 3 y − 2 z = 3

{ 4 x − 3 y + z = 7 2 x − 5 y − 4 z = 3 3 x − 2 y − 2 z = −7 { 4 x − 3 y + z = 7 2 x − 5 y − 4 z = 3 3 x − 2 y − 2 z = −7

{ 3 x + 8 y + 2 z = −5 2 x + 5 y − 3 z = 0 x + 2 y − 2 z = −1 { 3 x + 8 y + 2 z = −5 2 x + 5 y − 3 z = 0 x + 2 y − 2 z = −1

{ 11 x + 9 y + 2 z = −9 7 x + 5 y + 3 z = −7 4 x + 3 y + z = −3 { 11 x + 9 y + 2 z = −9 7 x + 5 y + 3 z = −7 4 x + 3 y + z = −3

{ 1 3 x − y − z = 1 x + 5 2 y + z = −2 2 x + 2 y + 1 2 z = −4 { 1 3 x − y − z = 1 x + 5 2 y + z = −2 2 x + 2 y + 1 2 z = −4

{ x + 1 2 y + 1 2 z = 0 1 5 x − 1 5 y + z = 0 1 3 x − 1 3 y + 2 z = −1 { x + 1 2 y + 1 2 z = 0 1 5 x − 1 5 y + z = 0 1 3 x − 1 3 y + 2 z = −1

{ x + 1 3 y − 2 z = −1 1 3 x + y + 1 2 z = 0 1 2 x + 1 3 y − 1 2 z = −1 { x + 1 3 y − 2 z = −1 1 3 x + y + 1 2 z = 0 1 2 x + 1 3 y − 1 2 z = −1

{ 1 3 x − y + 1 2 z = 4 2 3 x + 5 2 y − 4 z = 0 x − 1 2 y + 3 2 z = 2 { 1 3 x − y + 1 2 z = 4 2 3 x + 5 2 y − 4 z = 0 x − 1 2 y + 3 2 z = 2

{ x + 2 z = 0 4 y + 3 z = −2 2 x − 5 y = 3 { x + 2 z = 0 4 y + 3 z = −2 2 x − 5 y = 3

{ 2 x + 5 y = 4 3 y − z = 3 4 x + 3 z = −3 { 2 x + 5 y = 4 3 y − z = 3 4 x + 3 z = −3

{ 2 y + 3 z = −1 5 x + 3 y = −6 7 x + z = 1 { 2 y + 3 z = −1 5 x + 3 y = −6 7 x + z = 1

{ 3 x − z = −3 5 y + 2 z = −6 4 x + 3 y = −8 { 3 x − z = −3 5 y + 2 z = −6 4 x + 3 y = −8

{ 4 x − 3 y + 2 z = 0 − 2 x + 3 y − 7 z = 1 2 x − 2 y + 3 z = 6 { 4 x − 3 y + 2 z = 0 − 2 x + 3 y − 7 z = 1 2 x − 2 y + 3 z = 6

{ x − 2 y + 2 z = 1 − 2 x + y − z = 2 x − y + z = 5 { x − 2 y + 2 z = 1 − 2 x + y − z = 2 x − y + z = 5

{ 2 x + 3 y + z = 12 x + y + z = 9 3 x + 4 y + 2 z = 20 { 2 x + 3 y + z = 12 x + y + z = 9 3 x + 4 y + 2 z = 20

{ x + 4 y + z = −8 4 x − y + 3 z = 9 2 x + 7 y + z = 0 { x + 4 y + z = −8 4 x − y + 3 z = 9 2 x + 7 y + z = 0

{ x + 2 y + z = 4 x + y − 2 z = 3 − 2 x − 3 y + z = −7 { x + 2 y + z = 4 x + y − 2 z = 3 − 2 x − 3 y + z = −7

{ x + y − 2 z = 3 − 2 x − 3 y + z = −7 x + 2 y + z = 4 { x + y − 2 z = 3 − 2 x − 3 y + z = −7 x + 2 y + z = 4

{ x + y − 3 z = −1 y − z = 0 − x + 2 y = 1 { x + y − 3 z = −1 y − z = 0 − x + 2 y = 1

{ x − 2 y + 3 z = 1 x + y − 3 z = 7 3 x − 4 y + 5 z = 7 { x − 2 y + 3 z = 1 x + y − 3 z = 7 3 x − 4 y + 5 z = 7

In the following exercises, solve the given problem.

The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is twice the measure of the first angle. The third angle is twelve more than the second. Find the measures of the three angles.

The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is three times the measure of the first angle. The third angle is fifteen more than the second. Find the measures of the three angles.

After watching a major musical production at the theater, the patrons can purchase souvenirs. If a family purchases 4 t-shirts, the video, and 1 stuffed animal, their total is $135.

A couple buys 2 t-shirts, the video, and 3 stuffed animals for their nieces and spends $115. Another couple buys 2 t-shirts, the video, and 1 stuffed animal and their total is $85. What is the cost of each item?

The church youth group is selling snacks to raise money to attend their convention. Amy sold 2 pounds of candy, 3 boxes of cookies and 1 can of popcorn for a total sales of $65. Brian sold 4 pounds of candy, 6 boxes of cookies and 3 cans of popcorn for a total sales of $140. Paulina sold 8 pounds of candy, 8 boxes of cookies and 5 cans of popcorn for a total sales of $250. What is the cost of each item?

Writing Exercises

In your own words explain the steps to solve a system of linear equations with three variables by elimination.

How can you tell when a system of three linear equations with three variables has no solution? Infinitely many solutions?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

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Bibliometrics & citations, view options, recommendations, regularization methods for solving differential-algebraic equations.

In [M. Knorrenschild, SIAM J. Numer. Anal. 29 (1992) 1694-1715], a class of regularization methods for solving autonomous index-1 DAEs (differential-algebraic equations) has been proposed. In this paper, we generalize it to non-autonomous index-1 and ...

The Averaged Kaczmarz Iteration for Solving Inverse Problems

We introduce a new iterative regularization method for solving inverse problems that can be written as systems of linear or nonlinear equations in Hilbert spaces. The proposed averaged Kaczmarz (AVEK) method can be seen as a hybrid method between the ...

The prox-Tikhonov regularization method for the proximal point algorithm in Banach spaces

It is known, by Rockafellar (SIAM J Control Optim 14:877---898, 1976), that the proximal point algorithm (PPA) converges weakly to a zero of a maximal monotone operator in a Hilbert space, but it fails to converge strongly. Lehdili and Moudafi (...

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Preconditioned flow as a solution to the hierarchical growth problem in the generalized Lefschetz thimble method

• Regular Article - Theoretical Physics
• Open access
• Published: 19 July 2024
• Volume 2024 , article number  174 , ( 2024 )

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• Jun Nishimura 1 , 2 ,
• Katsuta Sakai 1 , 3 &
• Atis Yosprakob 1 , 2 , 4  

A preprint version of the article is available at arXiv.

The generalized Lefschetz thimble method is a promising approach that attempts to solve the sign problem in Monte Carlo methods by deforming the integration contour using the flow equation. Here we point out a general problem that occurs due to the property of the flow equation, which extends a region on the original contour exponentially to a region on the deformed contour. Since the growth rate for each eigenmode is governed by the singular values of the Hessian of the action, a huge hierarchy in the singular value spectrum, which typically appears for large systems, leads to various technical problems in numerical simulations. We solve this hierarchical growth problem by preconditioning the flow so that the growth rate becomes identical for every eigenmode. As an example, we show that the preconditioned flow enables us to investigate the real-time quantum evolution of an anharmonic oscillator with the system size that can hardly be achieved by using the original flow.

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Acknowledgments

We would like to thank Yuhma Asano, Genki Fujisawa, Masafumi Fukuma and Nobuyuki Matsumoto for valuable discussions. The computations were carried out on the PC clusters in KEK Computing Research Center and KEK Theory Center. K.S. was supported by the Grant-in-Aid for JSPS Research Fellow, No. 20J00079. A.Y. is supported by a Grant-in-Aid for Transformative Research Areas “The Natural Laws of Extreme Universe — A New Paradigm for Spacetime and Matter from Quantum Information” (KAKENHI Grant No. JP21H05191) from JSPS of Japan.

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KEK Theory Center, Institute of Particle and Nuclear Studies, High Energy Accelerator Research Organization, 1-1 Oho, Tsukuba, Ibaraki, 305-0801, Japan

Jun Nishimura, Katsuta Sakai & Atis Yosprakob

Graduate Institute for Advanced Studies, SOKENDAI, 1-1 Oho, Tsukuba, Ibaraki, 305-0801, Japan

Jun Nishimura & Atis Yosprakob

College of Liberal Arts and Sciences, Tokyo Medical and Dental University, 2-8-30 Kounodai, Ichikawa, Chiba, 272-0827, Japan

Katsuta Sakai

Department of Physics, Niigata University, 8050 Igarashi 2-no-cho, Nishi-ku, Niigata-shi, Niigata, 950-2181, Japan

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Nishimura, J., Sakai, K. & Yosprakob, A. Preconditioned flow as a solution to the hierarchical growth problem in the generalized Lefschetz thimble method. J. High Energ. Phys. 2024 , 174 (2024). https://doi.org/10.1007/JHEP07(2024)174

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Received : 02 May 2024

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DOI : https://doi.org/10.1007/JHEP07(2024)174

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