Intermediate Algebra for College Students (7th Edition)

$20$ hours.
Say the experienced bricklayer needs x hours to complete the job, working alone. Then, the apprentice needs $x+10$ hours.\\\\ In 1 hour, the bricklayer does $\displaystyle \frac{1}{x}$ of the full job, the apprentice does $\displaystyle \frac{1}{x+10}$ of the full job, In 12 hours, together they complete $( \displaystyle \frac{12}{x}+\frac{12}{x+10} )$ of the job, which, by the text, is $\displaystyle \frac{1}{1},$ the whole job. $\displaystyle \frac{12}{x}+\frac{12}{x+10}=1\qquad$ ... LCD=$x(x+10)$ ... multiply with $x(x+10)$ $12(x+10)+12x=x(x+10)$ $12x+120+12x=x^{2}+10x\qquad$... add $-24x-120$ $0=x^{2}-14x-120$ ... two factors of $-120$ whose sum is $-14$ ... are $-20$ and $+6$ $0=(x-20)(x+6)$ ... $x=20$ or $x=-3$ ... negative time makes no sense, so we discard that solution. $x$ = $20$ hours.