$\displaystyle d+8/d-2=9/4$
Thankful for help. Now I don't get how to write this. d+8 should be the numerator, d-2 the denominator. How do i get that?
Hi
$\displaystyle \frac{d+8}{d-2} = \frac94$
First note that $\displaystyle d \neq 2$ because the denominator must not be equal to 0
Multiply both sides by (d-2)
$\displaystyle \frac{d+8}{d-2} \d-2) = \frac94\d-2)$
$\displaystyle d+8 = \frac94\d-2)$
Multiply both sides by 4
$\displaystyle 4(d+8) = 9(d-2)$
Spoiler:
There is a latex tutorial somewhere. You can type this problem as I have done below and click on someone's text to see what they typed.
\frac{d+8}{d-2} = \frac{9}{4}
Multiply both sides by (d-2) to give
$\displaystyle d+8 = \frac{9d}{4} - \frac{9}{2}$
$\displaystyle \frac{9d-4d}{4} = -\frac{9-16}{2}$
Then multiply by 4/5 to get your answer