Solve y/(y ^ 2 - (y ^ 2 - 1)/(y + 1/(y + 1))) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply $y$ times $\frac{y+1}{y+1}$.

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

Since $\frac{y\left(y+1\right)}{y+1}$ and $\frac{1}{y+1}$ have the same denominator, add them by adding their numerators.

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

Do the multiplications in $y\left(y+1\right)+1$.

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

Divide $y^{2}-1$ by $\frac{y^{2}+y+1}{y+1}$ by multiplying $y^{2}-1$ by the reciprocal of $\frac{y^{2}+y+1}{y+1}$.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply $y^{2}$ times $\frac{y^{2}+y+1}{y^{2}+y+1}$.

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

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

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

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

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

Combine like terms in $y^{4}+y^{3}+y^{2}-y^{3}-y^{2}+y+1$.

$$\frac{y}{\frac{y^{4}+y+1}{y^{2}+y+1}}$$

Divide $y$ by $\frac{y^{4}+y+1}{y^{2}+y+1}$ by multiplying $y$ by the reciprocal of $\frac{y^{4}+y+1}{y^{2}+y+1}$.

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

Use the distributive property to multiply $y$ by $y^{2}+y+1$.

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

Expand

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply $y$ times $\frac{y+1}{y+1}$.

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

Since $\frac{y\left(y+1\right)}{y+1}$ and $\frac{1}{y+1}$ have the same denominator, add them by adding their numerators.

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

Do the multiplications in $y\left(y+1\right)+1$.

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

Divide $y^{2}-1$ by $\frac{y^{2}+y+1}{y+1}$ by multiplying $y^{2}-1$ by the reciprocal of $\frac{y^{2}+y+1}{y+1}$.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply $y^{2}$ times $\frac{y^{2}+y+1}{y^{2}+y+1}$.

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

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

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

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

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

Combine like terms in $y^{4}+y^{3}+y^{2}-y^{3}-y^{2}+y+1$.

$$\frac{y}{\frac{y^{4}+y+1}{y^{2}+y+1}}$$

Divide $y$ by $\frac{y^{4}+y+1}{y^{2}+y+1}$ by multiplying $y$ by the reciprocal of $\frac{y^{4}+y+1}{y^{2}+y+1}$.

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

Use the distributive property to multiply $y$ by $y^{2}+y+1$.

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