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Valentino Patton
Mathematics
1 January, 20:14
Solving systems by substitutions
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Muhammad Donaldson
1 January, 20:39
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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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