If m∠2 = 5x-8, m∠3 = 3y, and m∠4 = 132, find the values of x and y.

x = 132, y = 48
x = 28, y = 16
x = 16, y = 28
x = 48, y = 132

Answer is x = 28, y = 16

m<3 + m<4 = 180 (plug in m∠3 = 3y, and m∠4 = 132)

3y + 132 = 180 ...solve for y
3y = 180 - 132
3y = 48
y = 48/3
y = 16

m∠2 + m∠3 = 180 (plug in m∠2 = 5x-8, m∠3 = 3y)
5x - 8 + 3y = 180
5x -8 + 3(16) = 180
5x - 8 + 48 = 180
5x + 40 = 180
5x = 180 -40
5x =140
x = 140/5
x =28

so x = 28 and y = 16

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