Solve a^3-2/4a^2y+1/2-a/4y+1/2a^2y

$$\frac { a ^ { 3 } - 2 } { 4 a ^ { 2 } y } + \frac { 1 } { 2 } - \frac { a } { 4 y } + \frac { 1 } { 2 a ^ { 2 } y }$$

Evaluate

$$\frac{1}{2}=0.5$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4a^{2}y and 2 is 4ya^{2}. Multiply \frac{1}{2} times \frac{2ya^{2}}{2ya^{2}}.

$$\frac{a^{3}-2}{4ya^{2}}+\frac{2ya^{2}}{4ya^{2}}-\frac{a}{4y}+\frac{1}{2a^{2}y}$$

Since \frac{a^{3}-2}{4ya^{2}} and \frac{2ya^{2}}{4ya^{2}} have the same denominator, add them by adding their numerators.

$$\frac{a^{3}-2+2ya^{2}}{4ya^{2}}-\frac{a}{4y}+\frac{1}{2a^{2}y}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4ya^{2} and 4y is 4ya^{2}. Multiply \frac{a}{4y} times \frac{a^{2}}{a^{2}}.

$$\frac{a^{3}-2+2ya^{2}}{4ya^{2}}-\frac{aa^{2}}{4ya^{2}}+\frac{1}{2a^{2}y}$$

Since \frac{a^{3}-2+2ya^{2}}{4ya^{2}} and \frac{aa^{2}}{4ya^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{a^{3}-2+2ya^{2}-aa^{2}}{4ya^{2}}+\frac{1}{2a^{2}y}$$

Do the multiplications in a^{3}-2+2ya^{2}-aa^{2}.

$$\frac{a^{3}-2+2ya^{2}-a^{3}}{4ya^{2}}+\frac{1}{2a^{2}y}$$

Combine like terms in a^{3}-2+2ya^{2}-a^{3}.

$$\frac{-2+2ya^{2}}{4ya^{2}}+\frac{1}{2a^{2}y}$$

Factor the expressions that are not already factored in \frac{-2+2ya^{2}}{4ya^{2}}.

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

Cancel out 2 in both numerator and denominator.

$$\frac{ya^{2}-1}{2ya^{2}}+\frac{1}{2a^{2}y}$$

Since \frac{ya^{2}-1}{2ya^{2}} and \frac{1}{2a^{2}y} have the same denominator, add them by adding their numerators.

$$\frac{ya^{2}-1+1}{2ya^{2}}$$

Combine like terms in ya^{2}-1+1.

$$\frac{ya^{2}}{2ya^{2}}$$

Cancel out ya^{2} in both numerator and denominator.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$