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Posted Thu Nov 7 04:06:48 PST 2002 by Randall Hudson (Rlostsoul53@aol.com).

Rogers, Rogers, USA

Activity Time: 15 mins

Concepts Taught: Solving for a variable to substitute its value into a second equation.

Algebra I

Rearranging Before SubstitutionRemember:

In substitution problems we have two equations that each have the same two variables.Ex. 2x+3y=4

x=2y-7We then substitute what x is equal to in the bottom equation in place of the x in the top equation.

Ex. 2(2y-7)+3y=4

Then we solve:

4y-14+3y=4

7y-14=4

7y=18

y=18/7=2.571Then we plug our value of y into either of the two original equations and solve for x.

Ex. 2x+3y=4

2x+3(18/7)=4

2x+7.714=4

2x=-3.714

x=-1.857

So our ordered pair (x,y) is: (-1.857,2.571)

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But what if neither of our two original equations is solved for one of our variables? Then we have to solve one of the equations for one of the variables to use substitution.Ex.: 3x-2y=5

2x-4y=8We need to solve one of the equations for one of our variables. It doesn't matter which one we choose.

The bottom one looks easier so let's pick it.2x-4y=8 Then solve it for x.

2x=4y+8

x=2y+4Now our system of equations is:

3x-2y=5

x=2y+4We plug in 2y+4 for the x in the top equation.

3(2y+4)-2y=5 and solve for y.

6y+12-2y=5

4y+12=5

4y=-7

y=-7/4=-1.75Then we plug our value for y into either of our two original equations and solve for x.

x=2y+4

x=2(-1.75)+4

x=-3.5+4

x=.5

So our ordered pair (x,y) is (.5,-1.75)