Solve 1/2t+5-2/25-10t-4/4t^2-25

$$\frac { 1 } { 2 t + 5 } - \frac { 2 } { 25 - 10 t } - \frac { 4 } { 4 t ^ { 2 } - 25 }$$

Evaluate

$$\frac{7}{5\left(2t+5\right)}$$

Short Solution Steps

Factor 25-10t.

$$\frac{1}{2t+5}-\frac{2}{5\left(-2t+5\right)}-\frac{4}{4t^{2}-25}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2t+5 and 5\left(-2t+5\right) is 5\left(-2t+5\right)\left(2t+5\right). Multiply \frac{1}{2t+5} times \frac{5\left(-2t+5\right)}{5\left(-2t+5\right)}. Multiply \frac{2}{5\left(-2t+5\right)} times \frac{2t+5}{2t+5}.

$$\frac{5\left(-2t+5\right)}{5\left(-2t+5\right)\left(2t+5\right)}-\frac{2\left(2t+5\right)}{5\left(-2t+5\right)\left(2t+5\right)}-\frac{4}{4t^{2}-25}$$

Since \frac{5\left(-2t+5\right)}{5\left(-2t+5\right)\left(2t+5\right)} and \frac{2\left(2t+5\right)}{5\left(-2t+5\right)\left(2t+5\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5\left(-2t+5\right)-2\left(2t+5\right)}{5\left(-2t+5\right)\left(2t+5\right)}-\frac{4}{4t^{2}-25}$$

Do the multiplications in 5\left(-2t+5\right)-2\left(2t+5\right).

$$\frac{-10t+25-4t-10}{5\left(-2t+5\right)\left(2t+5\right)}-\frac{4}{4t^{2}-25}$$

Combine like terms in -10t+25-4t-10.

$$\frac{-14t+15}{5\left(-2t+5\right)\left(2t+5\right)}-\frac{4}{4t^{2}-25}$$

Factor 4t^{2}-25.

$$\frac{-14t+15}{5\left(-2t+5\right)\left(2t+5\right)}-\frac{4}{\left(2t-5\right)\left(2t+5\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 5\left(-2t+5\right)\left(2t+5\right) and \left(2t-5\right)\left(2t+5\right) is 5\left(2t-5\right)\left(2t+5\right). Multiply \frac{-14t+15}{5\left(-2t+5\right)\left(2t+5\right)} times \frac{-1}{-1}. Multiply \frac{4}{\left(2t-5\right)\left(2t+5\right)} times \frac{5}{5}.

$$\frac{-\left(-14t+15\right)}{5\left(2t-5\right)\left(2t+5\right)}-\frac{4\times 5}{5\left(2t-5\right)\left(2t+5\right)}$$

Since \frac{-\left(-14t+15\right)}{5\left(2t-5\right)\left(2t+5\right)} and \frac{4\times 5}{5\left(2t-5\right)\left(2t+5\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-\left(-14t+15\right)-4\times 5}{5\left(2t-5\right)\left(2t+5\right)}$$

Do the multiplications in -\left(-14t+15\right)-4\times 5.

$$\frac{14t-15-20}{5\left(2t-5\right)\left(2t+5\right)}$$

Combine like terms in 14t-15-20.

$$\frac{14t-35}{5\left(2t-5\right)\left(2t+5\right)}$$

Factor the expressions that are not already factored in \frac{14t-35}{5\left(2t-5\right)\left(2t+5\right)}.

$$\frac{7\left(2t-5\right)}{5\left(2t-5\right)\left(2t+5\right)}$$

Cancel out 2t-5 in both numerator and denominator.

$$\frac{7}{5\left(2t+5\right)}$$

Expand 5\left(2t+5\right).

$$\frac{7}{10t+25}$$

Factor

$$\frac{7}{5\left(2t+5\right)}$$