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To answer this, I'll use an example:

Say you're equations are both in slope-intercept form:

y = 3x + 5

y = x + 7

You substitute one equation into another (since y = y, you can set it up in the same equation).

x + 7 = 3x + 5

Subtract variables.

7 = 2x +5

Subtract numbers.

2 = 2x

Divide.

1 = x

You could also have a problem with one or two standard form equations. First, make sure you have at least one slope-intercept equation. Here's an example:

3x - 2y = 20

y = x + 15

Substitute the slope-intercept equation into y (because y = y).

3x - 2(x + 15) = 20

Distribute.

3x- 2x - 30 = 20

Like terms!

x - 30 = 20

Add 30 to both sides.

x = 50

For both equations, you can substitute your solution into one of the equations to find the other variable.

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Q: How do you solve linear systems with substitutions?
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