Solution. Collect the \(x\) terms on one side and the numbers on the other: \[5x-3x=8+12\quad \implies \quad 2x=20\quad \implies \quad x=10.\]

Check. Substituting back, the left side is \(5(10)-12=38\) and the right is \(3(10)+8=38\). Both agree, so \(x=10\) is correct. Checking costs one line and catches sign errors, which are the usual cause of a wrong answer here.

Solution. (a) The denominators are \(3\) and \(4\), so the LCD is \(12\). Multiply every term — including the \(2\) on the right: \[12\cdot \frac {x}{3}+12\cdot \frac {3x}{4}=12\cdot 2\] \[4x+9x=24\quad \implies \quad 13x=24\quad \implies \quad x=\frac {24}{13}.\]

(b) Denominators \(5\) and \(2\), so the LCD is \(10\): \[10\cdot \frac {4x}{5}-10\cdot \frac {x}{2}=10\cdot 9\] \[8x-5x=90\quad \implies \quad 3x=90\quad \implies \quad x=30.\]

Solution. Cross-multiply: \[(3y-2)(4y+3)=(6y-9)(2y+1).\] Expand both sides carefully: \begin {align*} \text {left} &= 12y^2+9y-8y-6=12y^2+y-6\\ \text {right} &= 12y^2+6y-18y-9=12y^2-12y-9 \end {align*}

The \(12y^2\) appears on both sides and cancels — which is what makes this a linear equation despite its appearance: \[y-6=-12y-9\quad \implies \quad 13y=-3\quad \implies \quad y=-\frac {3}{13}.\]

One caution. Cross-multiplying assumes neither denominator is zero. Here that rules out \(y=-\frac {1}{2}\) and \(y=-\frac {3}{4}\); our answer is neither, so it is valid.