Question
Solve: $\frac{x+2}{3}-\frac{x+1}{5}=\frac{x-3}{4}-1$
✓
Answer
$ \frac{x+2}{3}-\frac{x+1}{5}=\frac{x-3}{4}-1 \ldots(\text { L.C.M}$. of $3$ and $5=15)$
$ \Rightarrow \frac{5(\mathrm{x}+2)-3(\mathrm{x}+1)}{15}=\frac{\mathrm{x}-3-4}{4}$
$ \Rightarrow \frac{5 x+10-3 x-3}{15}=\frac{x-7}{4}$
$ \Rightarrow \frac{2 x+7}{15}=\frac{x-7}{4}$
$ \Rightarrow 4(2 x+7)=15(x-7)$
$ \Rightarrow 8 x+28=15 x-105$
$ \Rightarrow 8 \mathrm{x}-15 \mathrm{x}=-105-28$
$ \Rightarrow-7 \mathrm{x}=-133$
$ x=\frac{-133}{-7}$
$ \therefore \mathrm{x}=19$
$ \Rightarrow \frac{5(\mathrm{x}+2)-3(\mathrm{x}+1)}{15}=\frac{\mathrm{x}-3-4}{4}$
$ \Rightarrow \frac{5 x+10-3 x-3}{15}=\frac{x-7}{4}$
$ \Rightarrow \frac{2 x+7}{15}=\frac{x-7}{4}$
$ \Rightarrow 4(2 x+7)=15(x-7)$
$ \Rightarrow 8 x+28=15 x-105$
$ \Rightarrow 8 \mathrm{x}-15 \mathrm{x}=-105-28$
$ \Rightarrow-7 \mathrm{x}=-133$
$ x=\frac{-133}{-7}$
$ \therefore \mathrm{x}=19$
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