Question

show all work and circle your answers.

1. simplify: \\( \frac { t + 1 } { t + 3 } \cdot \frac { 5 t + 15 } { 4 t ^ { 2 } - 4 } \\)
2. simplify: \\( \frac { \frac { x } { 8 } - \frac { 8 } { x } } { \frac { 1 } { 8 } + \frac { 1 } { x } } \\)

Explanation:

Problem 1:

Step1: Factor expressions

Factor \(5t + 15=5(t + 3)\) and \(4t^{2}-4 = 4(t^{2}-1)=4(t - 1)(t + 1)\)
The expression becomes \(\frac{t + 1}{t+3}\cdot\frac{5(t + 3)}{4(t - 1)(t + 1)}\)

Step2: Cancel common factors

Cancel \((t + 1)\) and \((t + 3)\) terms.
\(\frac{\cancel{t + 1}}{\cancel{t+3}}\cdot\frac{5\cancel{(t + 3)}}{4(t - 1)\cancel{(t + 1)}}=\frac{5}{4(t - 1)}\)

Problem 2:

Step1: Simplify numerator and denominator separately

For the numerator: \(\frac{x}{8}-\frac{8}{x}=\frac{x^{2}-64}{8x}=\frac{(x - 8)(x + 8)}{8x}\)
For the denominator: \(\frac{1}{8}+\frac{1}{x}=\frac{x + 8}{8x}\)

Step2: Divide the simplified numerator by denominator

\(\frac{\frac{(x - 8)(x + 8)}{8x}}{\frac{x + 8}{8x}}=\frac{(x - 8)(x + 8)}{8x}\cdot\frac{8x}{x + 8}\)

Step3: Cancel common factors

Cancel \((x + 8)\) and \(8x\) terms.
\(\frac{(x - 8)\cancel{(x + 8)}}{\cancel{8x}}\cdot\frac{\cancel{8x}}{\cancel{(x + 8)}}=x - 8\)

Answer:

1. \(\frac{5}{4(t - 1)}\)
2. \(x - 8\)