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Choose the most convenient method to solve a system of linear equations

When you will have to solve a system of linear equations in a later math class, you will usually not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

This table has two rows and three columns. The first row labels the columns as “Graphing,” “Substitution,” and “Elimination.” Under “Graphing” it says, “Use when you need a picture of the situation.” Under “Substitution” it says, “Use when one equation is already solved for one variable.” Under “Elimination” it says, “Use when the equations are in standard form.”

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

{ 3 x + 8 y = 40 7 x 4 y = −32 { 5 x + 6 y = 12 y = 2 3 x 1

Solution

{ 3 x + 8 y = 40 7 x 4 y = −32
Since both equations are in standard form, using elimination will be most convenient.
{ 5 x + 6 y = 12 y = 2 3 x 1
Since one equation is already solved for y , using substitution will be most convenient.

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For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

{ 4 x 5 y = −32 3 x + 2 y = −1 { x = 2 y 1 3 x 5 y = 7

Since both equations are in standard form, using elimination will be most convenient. Since one equation is already solved for x , using substitution will be most convenient.

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For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

{ y = 2 x 1 3 x 4 y = 6 { 6 x 2 y = 12 3 x + 7 y = −13

Since one equation is already solved for y , using substitution will be most convenient; Since both equations are in standard form, using elimination will be most convenient.

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Access these online resources for additional instruction and practice with solving systems of linear equations by elimination.

Key concepts

  • To Solve a System of Equations by Elimination
    1. Write both equations in standard form. If any coefficients are fractions, clear them.
    2. Make the coefficients of one variable opposites.
      • Decide which variable you will eliminate.
      • Multiply one or both equations so that the coefficients of that variable are opposites.
    3. Add the equations resulting from Step 2 to eliminate one variable.
    4. Solve for the remaining variable.
    5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
    6. Write the solution as an ordered pair.
    7. Check that the ordered pair is a solution to both original equations.

Practice makes perfect

Solve a System of Equations by Elimination

In the following exercises, solve the systems of equations by elimination.

{ 5 x + 2 y = 2 −3 x y = 0

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{ −3 x + y = −9 x 2 y = −12

(6, 9)

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{ 6 x 5 y = −1 2 x + y = 13

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{ 3 x y = −7 4 x + 2 y = −6

( −2 , 1 )

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{ x + y = −1 x y = −5

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{ x + y = −8 x y = −6

( −7 , −1 )

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{ 3 x 2 y = 1 x + 2 y = 9

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{ −7 x + 6 y = −10 x 6 y = 22

( −2 , −4 )

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{ 3 x + 2 y = −3 x 2 y = −19

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{ 5 x + 2 y = 1 −5 x 4 y = −7

( −1 , 3 )

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{ 6 x + 4 y = −4 −6 x 5 y = 8

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{ 3 x 4 y = −11 x 2 y = −5

( −1 , 2 )

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{ 5 x 7 y = 29 x + 3 y = −3

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{ 6 x 5 y = −75 x 2 y = −13

( −5 , 9 )

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{ x + 4 y = 8 3 x + 5 y = 10

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{ 2 x 5 y = 7 3 x y = 17

(6, 1)

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{ 5 x 3 y = −1 2 x y = 2

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{ 7 x + y = −4 13 x + 3 y = 4

( −2 , 10 )

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{ −3 x + 5 y = −13 2 x + y = −26

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{ 3 x 5 y = −9 5 x + 2 y = 16

(2, 3)

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{ 4 x 3 y = 3 2 x + 5 y = −31

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{ 4 x + 7 y = 14 −2 x + 3 y = 32

( −7 , 6 )

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{ 5 x + 2 y = 21 7 x 4 y = 9

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{ 3 x + 8 y = −3 2 x + 5 y = −3

( −9 , 3 )

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{ 11 x + 9 y = −5 7 x + 5 y = −1

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{ 3 x + 8 y = 67 5 x + 3 y = 60

(9, 5)

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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