Solve x^2+2x+3x+4=x+5for x.

jmc

algebra intermediate

Problem

Solve

\frac{x ^{2} + 2 x + 3}{x + 4} = x + 5

for

x

Solution

Cross-multiplication gives

x^{2} + 2 x + 3 = (x + 4) (x + 5) = x^{2} + 9 x + 20.

Therefore

0 = 7 x + 17

and

x = - \frac{17}{7}

Final answer

-\frac{17}7