Solving systems using substitution and elimination?    3x-2y=0 x+y=-5 4x+2y=8 y=-2x+4

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nathanshields | High School Teacher | (Level 1) Associate Educator

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3x-2y=0
x+y=-5

We are almost ready for substitution; we just have to get one variable alone.  Since x + y = -5, then x = -5 - y.

Substitute this into the first equation and you have
3(-5 - y) - 2y = 0

Distribute
-15 - 3y - 2y = 0

Combine like terms
-15 - 5y = 0

-5y = 15

y = -3

Now plug back into the expression you substituted to find x:

x = -5 - y ...so... x = -5 - -3 ... x = -5+3 ... x = -2

Now double-check the first equation (we know the second equation works) with the values we found:

3(-2) - 2(-3) = 0

-6 - -6 = 0

-6 + 6 = 0

CHA-CHING!

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3x-2y=0
x+y=-5

To solve with elimination, double the second equation (the -2y and the positive y are just aching to be eliminated):

3x - 2y = 0
2x + 2y = -10

Add 'em up:

5x     = -10

So x = -2, you plug it into any equation to find y, check your answer, and

CHA-CHING!

 

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4x+2y=8
y=-2x+4

This one's all set for substitution.  Take the second equation (already solved for y) and plug it into the first equation:

4x + 2(-2x + 4) = 8

4x + -4x + 8 = 8

4x - 4x = 0

4x = 4x

x = x

OH SNAP!  Any x value will work, so there are INFINITELY MANY SOLUTIONS.

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If we try elimination (but who would? substitution was all set up for us), we need the equation to line up

4x+2y=8
y=-2x+4

Add 2x to both sides of the second equation so that all the variables are on the left side:

4x + 2y = 8
2x + y = 4

Now let's double the second equation

4x + 2y = 8
4x + 2y = 8

OH SNAP! Same equation, so really we just have one equation, and there are infinitely many solutions.

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