Question

If $\frac{a}{b}=\frac{2}{3}$ then find the values of the following expressions.

1. $\frac{4a+3b}{3b}$

2.$\frac{5a^2+2b^2}{5a^2-2b^2}$

3.$\frac{a^3+b^3}{b^3}$

4.$\frac{7b-4a}{7b+4a}$

Answer 1

$\frac{a}{b}=\frac{2}{3}$

Let $a=2k$, $b=3k$

(A) $\frac{4a+3b}{3b}=\frac{4\left(2k\right)+3\left(3k\right)}{3\left(3k\right)}$

$=\frac{8k+9k}{9k}=\frac{17k}{9k}=\frac{17}{9}$

(B) $\frac{5a^2+2b^2}{5a^2-2b^2}$

$=\frac{5(2k{)}^2+2(3k{)}^2}{5(2k{)}^2-2(3k{)}^2}=\frac{5\left(4k^2\right)+2\left(9k^2\right)}{5\left(4k^2\right)-2\left(9k^2\right)}$

$=\frac{20k^2+18k^2}{20k^2-18k^2}=\frac{38k^2}{2k^2}=19$

(C) $\frac{a^3+b^3}{b^3}={\left(\frac{a}{b}\right)}^3+1$

$=1+\frac{8}{27}=\frac{27+8}{27}=\frac{35}{27}$

(D) $\frac{7b-4a}{7b+4a}=\frac{7\left(3k\right)-4\left(2k\right)}{7\left(3k\right)+4\left(2k\right)}$

$=\frac{21k-8k}{21k+8k}=\frac{13k}{29k}=\frac{13}{29}$.