No The number was 1÷4 not just 4 pls resolve

My apologies for the oversight. Let's resolve the equation with the correct value. The original equation is: $$ \frac{3}{2}(2x-4) = \frac{1}{4}(5x-3) $$ Step 1: Distribute the numbers on both sides of the equation. $$ \left(\frac{3}{2} \times 2x\right) - \left(\frac{3}{2} \times 4\right) = \left(\frac{1}{4} \times 5x\right) - \left(\frac{1}{4} \times 3\right) $$ $$ 3x - 6 = \frac{5x}{4} - \frac{3}{4} $$ Step 2: Eliminate the denominators by multiplying the entire equation by the least common multiple (LCM) of the denominators. The LCM of $1$ and $4$ is $4$. Multiply every term by $4$: $$ 4(3x) - 4(6) = 4\left(\frac{5x}{4}\right) - 4\left(\frac{3}{4}\right) $$ $$ 12x - 24 = 5x - 3 $$ Step 3: Collect all terms with $x$ on one side and constant terms on the other side. Subtract $5x$ from both sides: $$ 12x - 5x - 24 = -3 $$ $$ 7x - 24 = -3 $$ Add $24$ to both sides: $$ 7x = -3 + 24 $$ $$ 7x = 21 $$ Step 4: Solve for $x$. Divide both sides by $7$: $$ x = \frac{21}{7} $$ $$ x = 3 $$ The final answer is $\boxed{3}$. That's 2 down. 3 left today — send the next one.