How do you solve $y = \frac{5 (10 - y)}{3 (y + 1)}$ $y=(5(10-y))/(3(y+1))$ using the quadratic formula?

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Answer 1 · 2016-07-03T01:34:23

$y = \frac{50 - 5 y}{3 y + 3}$ $y = (50 - 5y)/(3y + 3)$

$y (3 y + 3) = 50 - 5 y$ $y(3y + 3) = 50 - 5y$

$3 y^{2} + 3 y = 50 - 5 y$ $3y^2 + 3y = 50 - 5y$

$3 y^{2} + 8 y - 50 = 0$ $3y^2 + 8y - 50 = 0$

$y = \frac{- 8 \pm \sqrt{8^{2} - 4 \times 3 \times - 50}}{2 \times 3}$ $y = (-8 +- sqrt(8^2 - 4 xx 3 xx -50))/(2 xx 3)$

$y = \frac{- 8 \pm \sqrt{664}}{6}$ $y = (-8 +- sqrt664)/6$

$y = \frac{- 8 \pm 4 \sqrt{166}}{6}$ $y= (-8 +- 4sqrt166)/6$

$y = \frac{- 4 \pm 2 \sqrt{166}}{3}$ $y = (-4 +- 2sqrt166)/3$

Hopefully this helps!

How do you solve y=(5(10-y))/(3(y+1))y=5(10−y)3(y+1)y=(5(10-y))/(3(y+1)) using the quadratic formula?