0
$\begingroup$

Should I distribute the $\frac 45$ first, or find a common denominator between $\frac 45$ and $\frac {-8}9$ and then distribute?

$\endgroup$

3 Answers 3

1
$\begingroup$

neither. multiply both sides by 5/4

$\endgroup$
0
$\begingroup$

$$\frac 45(n-10)=\frac {-8}9$$

$$\frac54\frac 45(n-10)=\frac54\frac {-8}9$$

$$n-10=\frac {-40}{36}$$

$$n-10=\frac {-10}{9}$$

$$n=\frac {-10}{9}+10$$

$$n=\frac {-10+90}{9}=\frac {80}{9}$$

$\endgroup$
0
$\begingroup$

Multiply by $\frac54$ on both sides... Get $n-10=-\frac {10}9$... Now add $10$ to both sides. .. Get $n=10-\frac {10}9=\frac {80}9$...

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .