Solve [(3y - 5y) ^ 2 * (2x ^ 2 * y ^ 3)] / ((- 1/2 * x * y ^ 2) ^ 2) - [20y - (- 8y)] | 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

$$[(3y-5y)^{2}(2x^{2}y^{3})]:(-\frac{1}{2}xy^{2})^{2}-[20y-(-8y)]$$

Evaluate

$4y$

Solution Steps

Combine $3y$ and $-5y$ to get $-2y$.

$$\frac{\left(-2y\right)^{2}\times 2x^{2}y^{3}}{\left(-\frac{1}{2}xy^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

Expand $\left(-2y\right)^{2}$.

$$\frac{\left(-2\right)^{2}y^{2}\times 2x^{2}y^{3}}{\left(-\frac{1}{2}xy^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

Calculate $-2$ to the power of $2$ and get $4$.

$$\frac{4y^{2}\times 2x^{2}y^{3}}{\left(-\frac{1}{2}xy^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

Multiply $4$ and $2$ to get $8$.

$$\frac{8y^{2}x^{2}y^{3}}{\left(-\frac{1}{2}xy^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

To multiply powers of the same base, add their exponents. Add $2$ and $3$ to get $5$.

$$\frac{8y^{5}x^{2}}{\left(-\frac{1}{2}xy^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

Expand $\left(-\frac{1}{2}xy^{2}\right)^{2}$.

$$\frac{8y^{5}x^{2}}{\left(-\frac{1}{2}\right)^{2}x^{2}\left(y^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

To raise a power to another power, multiply the exponents. Multiply $2$ and $2$ to get $4$.

$$\frac{8y^{5}x^{2}}{\left(-\frac{1}{2}\right)^{2}x^{2}y^{4}}-\left(20y-\left(-8y\right)\right)$$

Calculate $-\frac{1}{2}$ to the power of $2$ and get $\frac{1}{4}$.

$$\frac{8y^{5}x^{2}}{\frac{1}{4}x^{2}y^{4}}-\left(20y-\left(-8y\right)\right)$$

Cancel out $x^{2}y^{4}$ in both numerator and denominator.

$$\frac{8y}{\frac{1}{4}}-\left(20y-\left(-8y\right)\right)$$

Divide $8y$ by $\frac{1}{4}$ by multiplying $8y$ by the reciprocal of $\frac{1}{4}$.

$$8y\times 4-\left(20y-\left(-8y\right)\right)$$

The opposite of $-8y$ is $8y$.

$$8y\times 4-\left(20y+8y\right)$$

Combine $20y$ and $8y$ to get $28y$.

$$8y\times 4-28y$$

Multiply $8$ and $4$ to get $32$.

$$32y-28y$$

Combine $32y$ and $-28y$ to get $4y$.

$$4y$$

Expand

$4y$

Solution Steps

Combine $3y$ and $-5y$ to get $-2y$.

$$\frac{\left(-2y\right)^{2}\times 2x^{2}y^{3}}{\left(-\frac{1}{2}xy^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

Expand $\left(-2y\right)^{2}$.

$$\frac{\left(-2\right)^{2}y^{2}\times 2x^{2}y^{3}}{\left(-\frac{1}{2}xy^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

Calculate $-2$ to the power of $2$ and get $4$.

$$\frac{4y^{2}\times 2x^{2}y^{3}}{\left(-\frac{1}{2}xy^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

Multiply $4$ and $2$ to get $8$.

$$\frac{8y^{2}x^{2}y^{3}}{\left(-\frac{1}{2}xy^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

To multiply powers of the same base, add their exponents. Add $2$ and $3$ to get $5$.

$$\frac{8y^{5}x^{2}}{\left(-\frac{1}{2}xy^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

Expand $\left(-\frac{1}{2}xy^{2}\right)^{2}$.

$$\frac{8y^{5}x^{2}}{\left(-\frac{1}{2}\right)^{2}x^{2}\left(y^{2}\right)^{2}}-\left(20y-\left(-8y\right)\right)$$

To raise a power to another power, multiply the exponents. Multiply $2$ and $2$ to get $4$.

$$\frac{8y^{5}x^{2}}{\left(-\frac{1}{2}\right)^{2}x^{2}y^{4}}-\left(20y-\left(-8y\right)\right)$$

Calculate $-\frac{1}{2}$ to the power of $2$ and get $\frac{1}{4}$.

$$\frac{8y^{5}x^{2}}{\frac{1}{4}x^{2}y^{4}}-\left(20y-\left(-8y\right)\right)$$

Cancel out $x^{2}y^{4}$ in both numerator and denominator.

$$\frac{8y}{\frac{1}{4}}-\left(20y-\left(-8y\right)\right)$$

Divide $8y$ by $\frac{1}{4}$ by multiplying $8y$ by the reciprocal of $\frac{1}{4}$.

$$8y\times 4-\left(20y-\left(-8y\right)\right)$$

The opposite of $-8y$ is $8y$.

$$8y\times 4-\left(20y+8y\right)$$

Combine $20y$ and $8y$ to get $28y$.

$$8y\times 4-28y$$

Multiply $8$ and $4$ to get $32$.

$$32y-28y$$

Combine $32y$ and $-28y$ to get $4y$.

$$4y$$