Question

Solve the following equation for $y.$

$\frac{6y-2}{9}+\frac{3y+5}{18}=\frac{1}{3}$

Answer 1

$\frac{6y-2}{9}+\frac{3y+5}{18}=\frac{1}{3}$

Here, HCF of denominators $9,18$and $3$ is $18$

So first of all we make the each denominator same ($18$).

$$\begin{gathered} \Rightarrow \frac{2\times (6y-2)}{2\times 9}+\frac{1\times (3y+5)}{1\times 18}=\frac{1\times 6}{3\times 6} \\ \Rightarrow \frac{12y-4}{18}+\frac{3y+5}{18}=\frac{6}{18} \\ \Rightarrow \frac{12y-4+3y+5}{18}=\frac{6}{18} \\ \Rightarrow \frac{15y+1}{18}=\frac{6}{18} \\ \Rightarrow 15y+1=6 \\ \Rightarrow 15y=6-1 \\ \Rightarrow 15y=5 \\ \Rightarrow y=\frac{5}{15} \end{gathered}$$

Dividing numerator and denominator by common factor $5$we get,

$y=\frac{5÷5}{15÷5}\Rightarrow \frac{1}{3}$

Hence, the value of $y$ is $\frac{1}{3}.$