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## How do you solve for #x# and #y# in the simultaneous equations #x + 2y = 3# and #x^2 - y^2 = 24#?

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## How to Solve Systems of Algebraic Equations Containing Two Variables

Last Updated: February 10, 2023 References

## Using the Substitution Method

- You know that x = 2 - ½y .
- Your second equation, that you haven't yet altered, is 5x + 3y = 9 .
- In the second equation, replace x with "2 - ½y": 5(2 - ½y) + 3y = 9 .

- 5(2 - ½y) + 3y = 9
- 10 – (5/2)y + 3y = 9
- 10 – (5/2)y + (6/2)y = 9 (If you don't understand this step, learn how to add fractions . This is often, but not always, necessary for this method.)
- 10 + ½y = 9

- You know that y = -2
- One of the original equations is 4x + 2y = 8 . (You can use either equation for this step.)
- Plug in -2 instead of y: 4x + 2(-2) = 8 .

- If you end up with an equation that has no variables and isn't true (for instance, 3 = 5), the problem has no solution . (If you graphed both of the equations, you'd see they were parallel and never intersect.)
- If you end up with an equation without variables that is true (such as 3 = 3), the problem has infinite solutions . The two equations are exactly equal to each other. (If you graphed the two equations, you'd see they were the same line.)

## Using the Elimination Method

- You have the system of equations 3x - y = 3 and -x + 2y = 4 .
- Let's change the first equation so that the y variable will cancel out. (You can choose x instead, and you'll get the same answer in the end.)
- The - y on the first equation needs to cancel with the + 2y in the second equation. We can make this happen by multiplying - y by 2.
- Multiply both sides of the first equation by 2, like this: 2(3x - y)=2(3) , so 6x - 2y = 6 . Now the - 2y will cancel out with the +2y in the second equation.

- Your equations are 6x - 2y = 6 and -x + 2y = 4 .
- Combine the left sides: 6x - 2y - x + 2y = ?
- Combine the right sides: 6x - 2y - x + 2y = 6 + 4 .

- You have 6x - 2y - x + 2y = 6 + 4 .
- Group the x and y variables together: 6x - x - 2y + 2y = 6 + 4 .
- Simplify: 5x = 10
- Solve for x: (5x)/5 = 10/5 , so x = 2 .

- You know that x = 2 , and one of your original equations is 3x - y = 3 .
- Plug in 2 instead of x: 3(2) - y = 3 .
- Solve for y in the equation: 6 - y = 3
- 6 - y + y = 3 + y , so 6 = 3 + y

- If your combined equation has no variables and is not true (like 2 = 7), there is no solution that will work on both equations. (If you graph both equations, you'll see they're parallel and never cross.)
- If your combined equation has no variables and is true (like 0 = 0), there are infinite solutions . The two equations are actually identical. (If you graph them, you'll see that they're the same line.)

## Graphing the Equations

- Your first equation is 2x + y = 5 . Change this to y = -2x + 5 .
- Your second equation is -3x + 6y = 0 . Change this to 6y = 3x + 0 , then simplify to y = ½x + 0 .
- If both equations are identical , the entire line will be an "intersection". Write infinite solutions .

- If you don't have graph paper, use a ruler to make sure the numbers are spaced precisely apart.
- If you are using large numbers or decimals, you may need to scale your graph differently. (For example, 10, 20, 30 or 0.1, 0.2, 0.3 instead of 1, 2, 3).

- In our examples from earlier, one line ( y = -2x + 5 ) intercepts the y-axis at 5 . The other ( y = ½x + 0 ) intercepts at 0 . (These are points (0,5) and (0,0) on the graph.)
- Use different colored pens or pencils if possible for the two lines.

- In our example, the line y = -2x + 5 has a slope of -2 . At x = 1, the line moves down 2 from the point at x = 0. Draw the line segment between (0,5) and (1,3).
- The line y = ½x + 0 has a slope of ½ . At x = 1, the line moves up ½ from the point at x=0. Draw the line segment between (0,0) and (1,½).
- If the lines have the same slope , the lines will never intersect, so there is no answer to the system of equations. Write no solution .

- If the lines are moving toward each other, keep plotting points in that direction.
- If the lines are moving away from each other, move back and plot points in the other direction, starting at x = -1.
- If the lines are nowhere near each other, try jumping ahead and plotting more distant points, such as at x = 10.

## Practice Problems and Answers

## Community Q&A

## Video . By using this service, some information may be shared with YouTube.

- You can check your work by plugging the answers back into the original equations. If the equations end up true (for instance, 3 = 3), your answer is correct. ⧼thumbs_response⧽ Helpful 2 Not Helpful 1
- In the elimination method, you will sometimes have to multiply one equation by a negative number in order to get a variable to cancel out. ⧼thumbs_response⧽ Helpful 1 Not Helpful 1

## You Might Also Like

- ↑ https://www.mathsisfun.com/definitions/system-of-equations.html
- ↑ https://calcworkshop.com/systems-equations/substitution-method/
- ↑ https://www.cuemath.com/algebra/substitution-method/
- ↑ http://tutorial.math.lamar.edu/Classes/Alg/SystemsTwoVrble.aspx
- ↑ http://www.purplemath.com/modules/systlin2.htm
- ↑ http://www.virtualnerd.com/algebra-2/linear-systems/graphing/solve-by-graphing/equations-solution-by-graphing
- ↑ https://www.khanacademy.org/math/algebra/multiplying-factoring-expression/factoring-quadratics-in-two-vari/v/factoring-quadratics-with-two-variables

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## Introduction to Equations

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## Algebra Solutions

We start here with a very simple technique.

## Let us say that there is an equation x + 3 = 9 and we need to solve it. Solving an equation means that we need to know the possible values of the variables that when put into the equation will satisfy it. We present the solution next.

So x = 6 is the required solution to the equation.

## You have an equation 3x = 9. Find x.

So x = 3 is the required solution to the equation.

## We have two simultaneous equations in two variables x and y. Find x and y. x – y = 10 --------(1) x + y = 15 --------(2)

The solution below will make the idea of Substitution clear.

x – y = 10 -----(1) x = 10 + y

Now we put this value of x into the 2 nd equation.

x + y = 15 -----(2) (10 + y) + y = 15 10 + 2y = 15 2y = 15 – 10 = 5 y = 5/2

Putting this value of y into any of the two equations will give us the value of x.

x + y = 15 x + 5/2 = 15 x = 15 – 5/2 x = 25/2

Hence (x , y) = (25/2, 5/2) is the solution to the given system of equations.

## We have two simultaneous equations in two variables x and y and we need to find x and y. x – y = 10 -----(1) x + y = 15 -----(2)

x – y = 10 x + y = 15 2x = 25 x = 25/2

Putting the value of x into any of the two equations will give y = 5/2

Elimination Method - By Equating Coefficients:

## Given two equations in two variables x and y. Find the values of x and y that satisfy these equations simultaneously. 2x – y = 10 ------(1) x + 2y = 15 ------(2)

Next we subtract this equation (2)’ from equation (1)

2x – y = 10 2x + 4y = 30 –5y = –20 y = 4

Putting this value of y into equation (1) will give us the correct value of x.

2x – y = 10 ------(1) 2x – 4 = 10 2x = 10 + 4 = 14 x = 14/2 = 7

Hence (x , y) =( 7, 4) gives the complete solution to these two equations.

## Different Equation Types

## How to find system of linear equations using Simple Algebric techniques

Find a value of x that satisfies this equation: 4x – 3 = 3x + 8.

Separating the variables and the coefficients gives:

Simplifying the above equation on the L.H.S (Left Hand Side) and the R.H.S (Right Hand Side) gives

Hence x = 11 is the required solution to the above equation.

## Solve x + 4 = 11 to find the value of x.

Hence x = 7 is the solution to the given equation.

## How to solve system of Linear Equations by Elimination

Solve the following system using elimination. 2x + y = 15 3x – y = 10, (5, 5).

2x + y = 15 ------(1) 3x – y = 10 ------(2) ______________ 5x = 25

What we are left with is a simplified equation in x alone. i.e., 5x = 25

Which is another equation in a single variable y.

Hence the solution to the system of equations is (x , y) = (5, 5)

## Solve the following system using Elimination x + 2y = 15 x – y = 10

x + 2y = 15 ------(1) x – y = 10 ------(2) ______________ 2x + y = 25

– ( x + 2y ) = – 15 – x – 2y = – 15 ------(1’)

– x – 2y = – 15 ------(1’) x – y = + 10 ------(2) ______________ – 3y = – 5

x – 5/3 + 5/3 = 10 + 5/3 x = (30+ 5)/3 = 35/3

Hence the solution to the given system of equations is (x , y) = ( 35/3 , 5/( 3 ))

## Solve the following system of linear equations using Elimination. 8x – 13y = 2 –4x + 6.5y = –2

8x – 13y = 2 ------(1) –4x + 6.5y = –2 ------(2)

2(–4x + 6.5y ) = 2(–2) –8x + 13y = –4 ------(2’)

8x – 13y = 2 ------(1) –8x + 13y = –4 ------(2’) ______________ 0 = –2

## How to solve system of Linear Equations by Substitution

2x – 2y = –2 ------(1) x + y = 24 ------(2)

2(24 – y) – 2y = –2 48 – 2y – 2y = –2 48 – 4y = – 2

48 – 4y – 48 = –2 –48 –4y = –50

-4y/-4 = -50/-4 y = 50/4 = 25/2

x + 25/2 - 25/2 = 24 - 25/2 x = (48 - 25)/2 = 23/2

Hence the solution to the given system of equations is (x , y) = ( 23/2 , 25/( 2 ))

x + y = 24 24 – y + y = 24 ∵ (x = 24 – y) 24 = 24

## Solve the following system of linear equations by method of substitution. y = 24 – 4x 2x + y/2 = 12

y = 24 – 4x ------(1) 2x + y/2 = 12 ------(2)

(Putting this value of y into equation (2) and then solving for x gives)

2x + (24-4x)/2 = 12 ------(2) (∵ y = 24 – 4x) 2x + 24/2- 4x/2 = 12 2x + 12 – 2x = 12 12 = 12

Hence the solution to the given system of equations is the entire line: y = 24 – 4x

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Addition and subtraction equations

Some equations involve only addition and/or subtraction.

To check your answer, simply plug your answer into the equation:

To solve this equation, you must get y by itself on one side. Therefore, add 9 to both sides.

To check, simply replace y with 34:

To solve, subtract 15 from both sides.

To check, simply replace x with –9 :

Multiplication and division equations

Divide each side of the equation by 3.

To solve, multiply each side by 5.

Subtract 4 from both sides to get 2 x by itself on one side.

Then divide both sides by 2 to get x .

To check, substitute your answer into the original equation:

Subtract 2 from both sides (which is the same as adding –2).

Note that 3 x – x is the same as 3 x – 1 x .

Previous Quiz: Variables and Algebraic Expressions

Next Quiz: Solving Simple Equations

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## IMAGES

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## COMMENTS

Learn how to solve literal equations. A literal equation is an equation where the unknown values are represented by variables.

Algebra 1 lesson about Linear Equations (Equations with x and y). This lesson shows how to transform and make a table of values for a linear

Solving for two variables (normally denoted as "x" and "y") requires two sets of equations. Assuming you have two equations

Solution x + 2y = 4 2x + y = 5. Multiply the second equation by 2 x + 2y = 4 4x + 2y = 10. Subtract the equations to eliminate y. · x + 2y = 4 - [ 4x + 2y = 10 ]

Solves algebra problems and walks you through them. ... y=x^2+1. Disclaimer: This calculator is not perfect. ... problems? Try MathPapa Math Practice.

The two solutions are (−7,5) and (5,−1) . Explanation: To solve the system of equations: {x+2y=3 (1)x2−y2=24 (2).

Using the Elimination Method · You have the system of equations 3x - y = 3 and -x + 2y = 4. · Let's change the first equation so that the y variable will cancel

Solve an equation, inequality or a system. Example: 2x-1=y,2y+3=x. 1

Hence (x , y) = (25/2, 5/2) is the solution to the given system of equations. Elimination Method - By Equating Coefficients: In Elimination Method, our aim is

To solve this equation, you must get y by itself on one side. Therefore, add 9 to both sides. To check, simply replace y with 34: Example 3.