Solve y^2+1/2y+8-2(y-1)/3y+12

$$\frac { y ^ { 2 } + 1 } { 2 y + 8 } - \frac { 2 ( y - 1 ) } { 3 y + 12 }$$

Evaluate

$$\frac{3y^{2}-4y+7}{6\left(y+4\right)}$$

Short Solution Steps

Factor 2y+8. Factor 3y+12.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(y+4\right) and 3\left(y+4\right) is 6\left(y+4\right). Multiply \frac{y^{2}+1}{2\left(y+4\right)} times \frac{3}{3}. Multiply \frac{2\left(y-1\right)}{3\left(y+4\right)} times \frac{2}{2}.

$$\frac{3\left(y^{2}+1\right)}{6\left(y+4\right)}-\frac{2\times 2\left(y-1\right)}{6\left(y+4\right)}$$

Since \frac{3\left(y^{2}+1\right)}{6\left(y+4\right)} and \frac{2\times 2\left(y-1\right)}{6\left(y+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3\left(y^{2}+1\right)-2\times 2\left(y-1\right)}{6\left(y+4\right)}$$

Do the multiplications in 3\left(y^{2}+1\right)-2\times 2\left(y-1\right).

$$\frac{3y^{2}+3-4y+4}{6\left(y+4\right)}$$

Combine like terms in 3y^{2}+3-4y+4.

$$\frac{3y^{2}+7-4y}{6\left(y+4\right)}$$

Expand 6\left(y+4\right).

$$\frac{3y^{2}+7-4y}{6y+24}$$

Expand

$$\frac{3y^{2}-4y+7}{6\left(y+4\right)}$$

Short Solution Steps

Factor 2y+8. Factor 3y+12.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(y+4\right) and 3\left(y+4\right) is 6\left(y+4\right). Multiply \frac{y^{2}+1}{2\left(y+4\right)} times \frac{3}{3}. Multiply \frac{2\left(y-1\right)}{3\left(y+4\right)} times \frac{2}{2}.

$$\frac{3\left(y^{2}+1\right)}{6\left(y+4\right)}-\frac{2\times 2\left(y-1\right)}{6\left(y+4\right)}$$

Since \frac{3\left(y^{2}+1\right)}{6\left(y+4\right)} and \frac{2\times 2\left(y-1\right)}{6\left(y+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3\left(y^{2}+1\right)-2\times 2\left(y-1\right)}{6\left(y+4\right)}$$

Do the multiplications in 3\left(y^{2}+1\right)-2\times 2\left(y-1\right).

$$\frac{3y^{2}+3-4y+4}{6\left(y+4\right)}$$

Combine like terms in 3y^{2}+3-4y+4.

$$\frac{3y^{2}+7-4y}{6\left(y+4\right)}$$

Expand 6\left(y+4\right).

$$\frac{3y^{2}+7-4y}{6y+24}$$