Question

1. reinforce solve each equation for the indicated variable.

a. solve for x:
\\( \frac { 2 } { x - 4 } = \frac { 1 } { x } \\)
b. solve for t:
\\( \frac { 1 } { t - 3 } + \frac { 1 } { t + 3 } = \frac { 6 } { t ^ { 2 } - 9 } \\)

Explanation:

Part a

Step1: Cross - multiply

Cross - multiply the equation \(\frac{2}{x - 4}=\frac{1}{x}\) to get \(2x=x - 4\).

Step2: Solve for \(x\)

Subtract \(x\) from both sides: \(2x-x=x - 4-x\). So \(x=-4\).

Step1: Factor the denominator

Note that \(t^{2}-9=(t - 3)(t + 3)\). The left - hand side of the equation \(\frac{1}{t - 3}+\frac{1}{t + 3}=\frac{6}{t^{2}-9}\) has a common denominator of \((t - 3)(t + 3)\).

Step2: Combine the left - hand side

\(\frac{t + 3+(t - 3)}{(t - 3)(t + 3)}=\frac{6}{(t - 3)(t + 3)}\). Simplify the numerator on the left: \(\frac{t + 3+t - 3}{(t - 3)(t + 3)}=\frac{2t}{(t - 3)(t + 3)}\).

Step3: Solve for \(t\)

Since \(\frac{2t}{(t - 3)(t + 3)}=\frac{6}{(t - 3)(t + 3)}\) (for \(t
eq3\) and \(t
eq - 3\)), then \(2t=6\). Divide both sides by 2: \(t = 3\). But when \(t = 3\), the original denominators \(t-3\) and \(t^{2}-9\) become 0.

Answer:

\(x = - 4\)

Part b