Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
3. Solving Linear Systems by Adding or Subtracting
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Exercise 10 Page 409

When adding or subtracting the equations, the aim is to eliminate one of the variables.

See solution.

Practice makes perfect

To decide whether to add or subtract the equations to eliminate a variable, we have to consider the coefficients. If a variable has the same coefficient in both equations, we should subtract. Conversely, if a variable has opposite coefficients, we should add the equations. Let's see two examples to fully understand.

Example 1

Suppose we are given a system of linear equations, and want to find the solution. 2x+3y=2 & (I) 2x-y=2 & (II) Note that the x-variable has the same coefficient in both equations. Thus, we will subtract Equation (II) from Equation (I).

2x+3y=2 2x-y=2
2x+3y-( 2x-y)=2- 2 2x-y=2
2x+3y-2x+y=2-2 2x-y=2
4y=0 2x-y=2

We have eliminated the x-variable in Equation (I). Now, we can find y by using the Division Property of Equality.

4y=0 2x-y=2 (I) ÷ 4 y=0 2x-y=2 Finally, to find the value of x we will substitute 0 for y in Equation (II).

y=0 2x-y=2
y=0 2x- 0=2
y=0 2x=2
y=0 x=1

Example 2

Let's now suppose we are given another system, and want to find its solution. 3x+2y=40 & (I) 5x-2y=40 & (II) This time, the y-variable has opposite coefficients. Thus, we will add Equation (I) to Equation (II).

3x+2y=40 5x-2y=40
3x+2y=40 5x-2y+( 3x+2y)=40+ 40
3x+2y=40 5x-2y+3x+2y=40+40
3x+2y=40 8x=80

We have eliminated the y-variable in Equation (II). Now, we can find x by using the Division Property of Equality. 3x+2y=40 8x=80 (II) ÷ 8 3x+2y=40 x=10 Finally, to find the value of y, we will substitute 10 for x in Equation (I).

3x+2y=40 x=10
3( 10)+2y=40 x=10
30+2y=40 x=10
2y=10 x=10
y=5 x=10