Here is the algorithm for substitution.
- Solve one of the equations for one variable.
- Plug this variable into the other equation.
- Solve the second equation, which now has only one variable.
- Finally, use the equation you found in step (1) to find the other variable.
3x+4y=13x+4y=1 size 12{3x+4y=1} {}
2x−y=82x−y=8 size 12{2x - y=8} {}
- The easiest variable to solve for here is the yy size 12{y} {} in the second equation.
- −y=−2x+8−y=−2x+8 size 12{ - y= - 2x+8} {}
- y=2x−8y=2x−8 size 12{y=2x - 8} {}
- Now, we plug that into the other equation:
- 3x+4(2x−8)=13x+4(2x−8)=1 size 12{3x+4 \( 2x - 8 \) =1} {}
- We now have an equation with only xx size 12{y} {} in it, so we can solve for xx size 12{y} {}.
- 3x+8x−32=13x+8x−32=1 size 12{3x+8x - "32"=1} {}
- 11x=3311x=33 size 12{"11"x="33"} {}
- x=3x=3 size 12{x=3} {}
- Finally, we take the equation from step (1),
y=2x−8y=2x−8 size 12{y=2x - 8} {}, and use it to find
yy size 12{y} {}.
- y=2(3)−8=−2y=2(3)−8=−2 size 12{y=2 \( 3 \) - 8= - 2} {}
So
(3,−2)(3,−2) size 12{ \( 3, - 2 \) } {} is the solution. You can confirm this by plugging this pair into both of the original equations.
Why does substitution work?
We found in the first step that
y=2x−8y=2x−8 size 12{y=2x - 8} {}. This means that
yy size 12{y} {} and
2x−82x−8 size 12{2x - 8} {} are equal in the sense that we discussed in the first chapter on functions—they will always be the same number, in these equations—they are the same. This gives us permission to simply replace one with the other, which is what we do in the second (“substitution”) step.
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