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!
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