Question

$\frac{3x-1}{5}-\frac{x}{7}=3$

Answer 1

$\begin{array}{l}\\ \text{We have:}\\ \frac{3x-1}{5}-\frac{x}{7}=3\\ \Rightarrow \frac{7(3x-1)-5\times x}{35}=3\\ \Rightarrow \left(\frac{21x-7-5x}{35}\right)=3\\ \Rightarrow \left(\frac{16x-7}{35}\right)=3\\ \Rightarrow 16x-7=3\times 35\left(\mathrm{T}\text{ransposing 35 to RHS}\right)\\ \Rightarrow 16x-7=105\\ \Rightarrow 16x=105+7\\ \Rightarrow 16x=112\\ \Rightarrow x=\frac{\sout{1}\sout{1}{\sout{2}}^7}{\sout{1}{\sout{6}}^1}\\ \Rightarrow x=7\\ \\ \text{CHECK: Substituting}x\text{=7 in the given equation, we get:}\\ \\ \text{LHS=}\frac{3x-1}{5}-\frac{x}{7}\\ \text{=}\frac{7(3x-1)-5\times x}{35}\\ \text{=}\left(\frac{21x-7-5x}{35}\right)\\ \text{=}\left(\frac{16x-7}{35}\right)\\ \text{=}\left(\frac{16\times 7-7}{35}\right)\\ \text{=}\frac{112-7}{35}\\ \text{=}\frac{\sout{1}\sout{0}{\sout{5}}^3}{\sout{3}{\sout{5}}^1}\\ \text{=3}\\ \\ \text{RHS=3}\\ \text{∴ LHS=RHS}\\ \text{Hence,}x\text{=3 is a solution of the given equation}\text{.}\end{array}$