Question

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Answer 1

@bibby can you help?

Answer 2

never mind, I did some research
\(\huge \frac{1}{j}+\frac{1}{j-8}=\frac{1}{3}\)

Answer 3

"Harry can rake the leaves in the yard 8 hours faster than his little brother Jimmy can."
"jimmy+harry=3"
\(\large \frac{1}{x}+\frac{1}{x+8}=\frac{1}{3}\)
\(\large(3)(x+8)\cancel{(x)}\frac{1}{\cancel{(x)}}+(3)(x)\cancel{(x+8)}\frac{1}{\cancel{(x+8)}}=\cancel{(3)}(x)(x+8)\frac{1}{\cancel{(3)}}\)
\(\large 3x+24+3x=x^2+8x\)
\(\large 6x+24=x^2+8x\)
\(\large 0=x^2+8x-6x-24\)
\(\large 0=x^2+2x-24\)
\(\large 0=(x+6)(x-4)\)
x=4, we reject x=-6 because you can't have negative time

since harry can do it in 4 hours, and we know that h=j-8
4=j-8
12=jimmy