Solve 8y+y^2+16/15y^2+3y:16-y^2/25y^2-1=

$$\frac { 8 y + y ^ { 2 } + 16 } { 15 y ^ { 2 } + 3 y } : \frac { 16 - y ^ { 2 } } { 25 y ^ { 2 } - 1 } =$$

Evaluate

$$-\frac{\left(5y-1\right)\left(y+4\right)}{3y\left(y-4\right)}$$

Short Solution Steps

Divide \frac{8y+y^{2}+16}{15y^{2}+3y} by \frac{16-y^{2}}{25y^{2}-1} by multiplying \frac{8y+y^{2}+16}{15y^{2}+3y} by the reciprocal of \frac{16-y^{2}}{25y^{2}-1}.

$$\frac{\left(8y+y^{2}+16\right)\left(25y^{2}-1\right)}{\left(15y^{2}+3y\right)\left(16-y^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(5y-1\right)\left(5y+1\right)\left(y+4\right)^{2}}{3y\left(y-4\right)\left(-y-4\right)\left(5y+1\right)}$$

Cancel out 5y+1 in both numerator and denominator.

$$\frac{\left(5y-1\right)\left(y+4\right)^{2}}{3y\left(y-4\right)\left(-y-4\right)}$$

Expand the expression.

$$\frac{5y^{3}+39y^{2}+72y-16}{-3y^{3}+48y}$$

Expand

$$\frac{5y^{2}+19y-4}{3y\left(4-y\right)}$$

Short Solution Steps

Divide \frac{8y+y^{2}+16}{15y^{2}+3y} by \frac{16-y^{2}}{25y^{2}-1} by multiplying \frac{8y+y^{2}+16}{15y^{2}+3y} by the reciprocal of \frac{16-y^{2}}{25y^{2}-1}.

$$\frac{\left(8y+y^{2}+16\right)\left(25y^{2}-1\right)}{\left(15y^{2}+3y\right)\left(16-y^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(5y-1\right)\left(5y+1\right)\left(y+4\right)^{2}}{3y\left(y-4\right)\left(-y-4\right)\left(5y+1\right)}$$

Cancel out 5y+1 in both numerator and denominator.

$$\frac{\left(5y-1\right)\left(y+4\right)^{2}}{3y\left(y-4\right)\left(-y-4\right)}$$

Expand the expression.

$$\frac{5y^{3}+39y^{2}+72y-16}{-3y^{3}+48y}$$