Question

how many solutions exist for the given equation?
\\(\frac{1}{2}(x + 12) = 4x - 1\\)
\\(\circ\\) zero
\\(\circ\\) one
\\(\circ\\) two
\\(\circ\\) infinitely many

Explanation:

Step1: Distribute the fraction

Multiply \(\frac{1}{2}\) with each term inside the parentheses: \(\frac{1}{2}(x + 12)=\frac{1}{2}x+6\). So the equation becomes \(\frac{1}{2}x + 6 = 4x-1\).

Step2: Move variable terms to one side

Subtract \(\frac{1}{2}x\) from both sides: \(6 = 4x-\frac{1}{2}x-1\). Combine like terms: \(4x-\frac{1}{2}x=\frac{8}{2}x-\frac{1}{2}x=\frac{7}{2}x\), so the equation is \(6=\frac{7}{2}x - 1\).

Step3: Move constant terms to one side

Add 1 to both sides: \(6 + 1=\frac{7}{2}x\), which simplifies to \(7=\frac{7}{2}x\).

Step4: Solve for x

Multiply both sides by \(\frac{2}{7}\): \(x = 7\times\frac{2}{7}=2\). Since we found a unique value for \(x\), the equation has one solution.

Answer:

one