Question

If $a:b=4:7$, find the following

(ii) $\frac{6a-b}{a+3b}$

Answer 1

Given: $\frac{a}{b}=\frac{4}{7}$

$\frac{6a-b}{a+3b}$

Divide both numerator and denominator by $b$,

$=\frac{6\left(\frac{a}{b}\right)-1}{\left(\frac{a}{b}\right)+3}$

On substituting the value $\frac{a}{b}=\frac{4}{7}$,

$=\frac{6\left(\frac{4}{7}\right)-1}{\left(\frac{4}{7}\right)+3}$

$=\frac{6\times 4-7\times 1}{4+3\times 7}$

$=\frac{24-7}{4+21}$

$=\frac{17}{25}$

Hence, the value of $\frac{6a-b}{a+3b}\text{is}\frac{17}{25}$.