Solve 14n/n-3+12n^2/(3-n)^2*15-5n/4

$$\frac { 14 n } { n - 3 } + \frac { 12 n ^ { 2 } } { ( 3 - n ) ^ { 2 } } \cdot \frac { 15 - 5 n } { 4 }$$

Evaluate

$$\frac{n\left(15n-14\right)}{3-n}$$

Short Solution Steps

Multiply \frac{12n^{2}}{\left(3-n\right)^{2}} times \frac{15-5n}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{14n}{n-3}+\frac{12n^{2}\left(15-5n\right)}{\left(3-n\right)^{2}\times 4}$$

Cancel out 4 in both numerator and denominator.

$$\frac{14n}{n-3}+\frac{3\left(-5n+15\right)n^{2}}{\left(-n+3\right)^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of n-3 and \left(-n+3\right)^{2} is \left(n-3\right)^{2}. Multiply \frac{14n}{n-3} times \frac{n-3}{n-3}.

$$\frac{14n\left(n-3\right)}{\left(n-3\right)^{2}}+\frac{3\left(-5n+15\right)n^{2}}{\left(n-3\right)^{2}}$$

Since \frac{14n\left(n-3\right)}{\left(n-3\right)^{2}} and \frac{3\left(-5n+15\right)n^{2}}{\left(n-3\right)^{2}} have the same denominator, add them by adding their numerators.

$$\frac{14n\left(n-3\right)+3\left(-5n+15\right)n^{2}}{\left(n-3\right)^{2}}$$

Do the multiplications in 14n\left(n-3\right)+3\left(-5n+15\right)n^{2}.

$$\frac{14n^{2}-42n-15n^{3}+45n^{2}}{\left(n-3\right)^{2}}$$

Combine like terms in 14n^{2}-42n-15n^{3}+45n^{2}.

$$\frac{59n^{2}-42n-15n^{3}}{\left(n-3\right)^{2}}$$

Expand \left(n-3\right)^{2}.

$$\frac{59n^{2}-42n-15n^{3}}{n^{2}-6n+9}$$

Expand

$$\frac{14n-15n^{2}}{n-3}$$

Short Solution Steps

Multiply \frac{12n^{2}}{\left(3-n\right)^{2}} times \frac{15-5n}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{14n}{n-3}+\frac{12n^{2}\left(15-5n\right)}{\left(3-n\right)^{2}\times 4}$$

Cancel out 4 in both numerator and denominator.

$$\frac{14n}{n-3}+\frac{3\left(-5n+15\right)n^{2}}{\left(-n+3\right)^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of n-3 and \left(-n+3\right)^{2} is \left(n-3\right)^{2}. Multiply \frac{14n}{n-3} times \frac{n-3}{n-3}.

$$\frac{14n\left(n-3\right)}{\left(n-3\right)^{2}}+\frac{3\left(-5n+15\right)n^{2}}{\left(n-3\right)^{2}}$$

Since \frac{14n\left(n-3\right)}{\left(n-3\right)^{2}} and \frac{3\left(-5n+15\right)n^{2}}{\left(n-3\right)^{2}} have the same denominator, add them by adding their numerators.

$$\frac{14n\left(n-3\right)+3\left(-5n+15\right)n^{2}}{\left(n-3\right)^{2}}$$

Do the multiplications in 14n\left(n-3\right)+3\left(-5n+15\right)n^{2}.

$$\frac{14n^{2}-42n-15n^{3}+45n^{2}}{\left(n-3\right)^{2}}$$

Combine like terms in 14n^{2}-42n-15n^{3}+45n^{2}.

$$\frac{59n^{2}-42n-15n^{3}}{\left(n-3\right)^{2}}$$

Expand \left(n-3\right)^{2}.

$$\frac{59n^{2}-42n-15n^{3}}{n^{2}-6n+9}$$