Same concept as the last question. To solve this by substitution, you can choose either of the equations and substitute in for y. I'll substitute y = 2x + 5 into the 2nd equation:

y = 3x - 1

2x + 5 = 3x - 1 (Substitute 2x + 5 in for y)

-x + 5 = -1 (Subtract 3x from both sides)

-x = -6 (Subtract 5 from both sides)

x = 6 (Divide each side by -1 to get rid of the negative).

Hope this helps!

Nov 13th, 2014

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