Solve (y+q)^2-(y-q)^2=-q(q-7y)

$$( y + q ) ^ { 2 } - ( y - q ) ^ { 2 } = - q ( q - 7 y )$$

Solve for y (complex solution)

$$\left\{\begin{matrix}\\y=\frac{q}{3}\text{, }&\text{unconditionally}\\y\in \mathrm{C}\text{, }&q=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+q\right)^{2}.

$$y^{2}+2yq+q^{2}-\left(y-q\right)^{2}=\left(-q\right)\left(q-7y\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-q\right)^{2}.

$$y^{2}+2yq+q^{2}-\left(y^{2}-2yq+q^{2}\right)=\left(-q\right)\left(q-7y\right)$$

To find the opposite of y^{2}-2yq+q^{2}, find the opposite of each term.

$$y^{2}+2yq+q^{2}-y^{2}+2yq-q^{2}=\left(-q\right)\left(q-7y\right)$$

Combine y^{2} and -y^{2} to get 0.

$$2yq+q^{2}+2yq-q^{2}=\left(-q\right)\left(q-7y\right)$$

Combine 2yq and 2yq to get 4yq.

$$4yq+q^{2}-q^{2}=\left(-q\right)\left(q-7y\right)$$

Combine q^{2} and -q^{2} to get 0.

$$4yq=\left(-q\right)\left(q-7y\right)$$

Use the distributive property to multiply -q by q-7y.

$$4yq=\left(-q\right)q-7\left(-q\right)y$$

Multiply -7 and -1 to get 7.

$$4yq=\left(-q\right)q+7qy$$

Subtract 7qy from both sides.

$$4yq-7qy=\left(-q\right)q$$

Combine 4yq and -7qy to get -3yq.

$$-3yq=\left(-q\right)q$$

Multiply q and q to get q^{2}.

$$-3yq=-q^{2}$$

The equation is in standard form.

$$\left(-3q\right)y=-q^{2}$$

Divide both sides by -3q.

$$\frac{\left(-3q\right)y}{-3q}=-\frac{q^{2}}{-3q}$$

Dividing by -3q undoes the multiplication by -3q.

$$y=-\frac{q^{2}}{-3q}$$

Divide -q^{2} by -3q.

$$y=\frac{q}{3}$$

Solve for y

$$\left\{\begin{matrix}\\y=\frac{q}{3}\text{, }&\text{unconditionally}\\y\in \mathrm{R}\text{, }&q=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+q\right)^{2}.

$$y^{2}+2yq+q^{2}-\left(y-q\right)^{2}=\left(-q\right)\left(q-7y\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-q\right)^{2}.

$$y^{2}+2yq+q^{2}-\left(y^{2}-2yq+q^{2}\right)=\left(-q\right)\left(q-7y\right)$$

To find the opposite of y^{2}-2yq+q^{2}, find the opposite of each term.

$$y^{2}+2yq+q^{2}-y^{2}+2yq-q^{2}=\left(-q\right)\left(q-7y\right)$$

Combine y^{2} and -y^{2} to get 0.

$$2yq+q^{2}+2yq-q^{2}=\left(-q\right)\left(q-7y\right)$$

Combine 2yq and 2yq to get 4yq.

$$4yq+q^{2}-q^{2}=\left(-q\right)\left(q-7y\right)$$

Combine q^{2} and -q^{2} to get 0.

$$4yq=\left(-q\right)\left(q-7y\right)$$

Use the distributive property to multiply -q by q-7y.

$$4yq=\left(-q\right)q-7\left(-q\right)y$$

Multiply -7 and -1 to get 7.

$$4yq=\left(-q\right)q+7qy$$

Subtract 7qy from both sides.

$$4yq-7qy=\left(-q\right)q$$

Combine 4yq and -7qy to get -3yq.

$$-3yq=\left(-q\right)q$$

Multiply q and q to get q^{2}.

$$-3yq=-q^{2}$$

The equation is in standard form.

$$\left(-3q\right)y=-q^{2}$$

Divide both sides by -3q.

$$\frac{\left(-3q\right)y}{-3q}=-\frac{q^{2}}{-3q}$$

Dividing by -3q undoes the multiplication by -3q.

$$y=-\frac{q^{2}}{-3q}$$

Divide -q^{2} by -3q.

$$y=\frac{q}{3}$$

Solve for q

$$q=3y$$

$$q=0$$