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maks197457 [2]
3 years ago
12

Solving systems by substitutions

Mathematics
1 answer:
svp [43]3 years ago
8 0
Let's take two different equations.

4x+2y=8
7x+3y=-10
(I just came up with these on the spot, so the answers might be strange but i should be able to explain nevertheless)
First, let's solve the first equation for "y".
4x+2y=8
-4x      -4x
2y=8-4x
/2
y=4-2x
Now that we have "y", let's plug it into the second equation.
7x+3(4-2x)=-10
7x+12-6x=-10
x+12=-10
-12    -12
x=-22
Alright, now we have "x" definitively, let's plug it in to find "y".
4(-22)+2y=8
-88+2y=8
+88       +88
2y=96
/2
y=48
So from this example you can see. You solve one variable in terms of another (solve y as being 4-2x), plug it into the second equation to find one variable's value, then plug that variable's value back in again later to get the first variable's value.
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