Question

what is the solution to the equation \\(\sqrt{5x - 7} = \sqrt{3x + 5}\\)?
\\(\bigcirc\\) \\(x = 1\\)
\\(\bigcirc\\) \\(x = 6\\)
\\(\bigcirc\\) \\(x = 12\\)
\\(\bigcirc\\) \\(x = 24\\)

Explanation:

Step1: Square both sides

To eliminate the square roots, square both sides of the equation \(\sqrt{5x - 7}=\sqrt{3x + 5}\). This gives \(5x - 7 = 3x + 5\) (since \((\sqrt{a})^2=a\) for \(a\geq0\)).

Step2: Solve for x

Subtract \(3x\) from both sides: \(5x-3x - 7=3x - 3x+ 5\), which simplifies to \(2x - 7 = 5\). Then add 7 to both sides: \(2x-7 + 7=5 + 7\), so \(2x=12\). Divide both sides by 2: \(x = \frac{12}{2}=6\). We should also check if this solution makes the original square roots defined. For \(\sqrt{5x - 7}\), when \(x = 6\), \(5(6)-7=30 - 7 = 23\geq0\). For \(\sqrt{3x + 5}\), when \(x = 6\), \(3(6)+5=18 + 5 = 23\geq0\). So \(x = 6\) is valid.

Answer:

B. \(x = 6\) (assuming the options are labeled as A. \(x = 1\), B. \(x = 6\), C. \(x = 12\), D. \(x = 24\))