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Coolmath Algebra
Systems of Equations 2x2's Lesson 2 - Solving by Substitution  
(page 2 of 4)

---- This algebra lesson explains how to solve a 2x2 system of equations by substitution.

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Let's do another one where we get a bit of a head start:

3x - 7y = -14

x = 2y - 3
Notice that one equation is solved for
x...

Let's stick that x blob into the other equation in place of x:

3x - 7y = -14 ... x = 2y - 3 ... plug the 2y - 3 into the first equation for x

This gives us

3 ( 2y - 3 ) - 7y = -14 ... solve for y ... 6y - 9 - 7y = -14 ... -y - 9 = -14 ... add 9 to both sides, which gives -y = -5 ... y = 5

OK, we've got y...  Now, we need x...  See the circled blob above?

Stick it in there!  ( That's why I circled it! )

x = 2y - 3 ... y = 5 ... stick the 5 into the first equation for y

x = 2 ( 5 ) - 3 = 7 ... x = 7

Is our answer (7, 5) ?  Let's check... 
and you have to check it in BOTH equations!

3x - 7y = -14 ... 3 ( 7 ) - 7 ( 5 ) = -14 ... yep

x = 2y - 3 ... 7 = 2 ( 5 ) - 3 ... yep

So, the answer is (7, 5).


TRY IT:

Solve by substitution:

4x - 3y = 25 and y = 6x - 13


Continued on the next page

 The printing and distribution and/or downloading of these lessons is strictly prohibited.

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