Question
Solve the following equations :
$2 y+\frac{5}{3}=\frac{26}{3}-y$.
$2 y+\frac{5}{3}=\frac{26}{3}-y$.
✓
Answer
We have, $\quad \frac{2 x}{3}+1=\frac{7 x}{15}+3$
$\Rightarrow \quad \frac{2 x}{3}-\frac{7 x}{15}=3-1\quad$ [transposing $\frac{7 x}{15}$ to LHS and 1 to RHS]
$\Rightarrow \quad \frac{10 x-7 x}{15}=2 \quad[\because$ LCM of 3 and 15 is 15$]$
$\Rightarrow \quad \frac{3 x}{15}=2$
$\Rightarrow \quad \frac{3 x}{15} \times 5=2 \times 5 \quad$ [multiplying both sides by 5]
$\Rightarrow \quad \frac{15 x}{15}=2 \times 5 \Rightarrow x=2 \times 5$
$\Rightarrow \quad x=10$, which is the required solution.
Check For $x=10$
LHS $=\frac{2(10)}{3}+1=\frac{23}{3}$
RHS $=\frac{7(10)}{15}+3=\frac{23}{3}$
$\therefore \quad$ LHS $=$ RHS $\quad$ [as required]
$\Rightarrow \quad \frac{2 x}{3}-\frac{7 x}{15}=3-1\quad$ [transposing $\frac{7 x}{15}$ to LHS and 1 to RHS]
$\Rightarrow \quad \frac{10 x-7 x}{15}=2 \quad[\because$ LCM of 3 and 15 is 15$]$
$\Rightarrow \quad \frac{3 x}{15}=2$
$\Rightarrow \quad \frac{3 x}{15} \times 5=2 \times 5 \quad$ [multiplying both sides by 5]
$\Rightarrow \quad \frac{15 x}{15}=2 \times 5 \Rightarrow x=2 \times 5$
$\Rightarrow \quad x=10$, which is the required solution.
Check For $x=10$
LHS $=\frac{2(10)}{3}+1=\frac{23}{3}$
RHS $=\frac{7(10)}{15}+3=\frac{23}{3}$
$\therefore \quad$ LHS $=$ RHS $\quad$ [as required]
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