Solve -y_0^2/2a+frac{u_{1}*v_{0}}{a}=omega

$$- \frac { y _ { 0 } ^ { 2 } } { 2 a } + \frac { u _ { 1 } \cdot v _ { 0 } } { a } = \omega$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{u_{1}v_{0}}{y_{0}^{2}+\omega }\text{, }&v_{0}\neq 0\text{ and }u_{1}\neq 0\text{ and }y_{0}\neq -i\sqrt{\omega }\text{ and }y_{0}\neq i\sqrt{\omega }\text{ and }\omega \neq -y_{0}^{2}\\a\neq 0\text{, }&\left(v_{0}=0\text{ or }u_{1}=0\right)\text{ and }\omega =-y_{0}^{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2a, the least common multiple of 2a,a.

$$-y_{0}^{2}\times 2a+2u_{1}v_{0}=\omega \times 2a$$

Subtract \omega \times 2a from both sides.

$$-y_{0}^{2}\times 2a+2u_{1}v_{0}-\omega \times 2a=0$$

Multiply -1 and 2 to get -2.

$$-2y_{0}^{2}a+2u_{1}v_{0}-\omega \times 2a=0$$

Multiply -1 and 2 to get -2.

$$-2y_{0}^{2}a+2u_{1}v_{0}-2\omega a=0$$

Subtract 2u_{1}v_{0} from both sides. Anything subtracted from zero gives its negation.

$$-2y_{0}^{2}a-2\omega a=-2u_{1}v_{0}$$

Combine all terms containing a.

$$\left(-2y_{0}^{2}-2\omega \right)a=-2u_{1}v_{0}$$

Divide both sides by -2y_{0}^{2}-2\omega .

$$\frac{\left(-2y_{0}^{2}-2\omega \right)a}{-2y_{0}^{2}-2\omega }=-\frac{2u_{1}v_{0}}{-2y_{0}^{2}-2\omega }$$

Dividing by -2y_{0}^{2}-2\omega undoes the multiplication by -2y_{0}^{2}-2\omega .

$$a=-\frac{2u_{1}v_{0}}{-2y_{0}^{2}-2\omega }$$

Divide -2u_{1}v_{0} by -2y_{0}^{2}-2\omega .

$$a=\frac{u_{1}v_{0}}{y_{0}^{2}+\omega }$$

Variable a cannot be equal to 0.

$$a=\frac{u_{1}v_{0}}{y_{0}^{2}+\omega }\text{, }a\neq 0$$

Solve for u_1 (complex solution)

$$\left\{\begin{matrix}u_{1}=\frac{a\left(y_{0}^{2}+\omega \right)}{v_{0}}\text{, }&v_{0}\neq 0\text{ and }a\neq 0\\u_{1}\in \mathrm{C}\text{, }&\left(y_{0}=-i\sqrt{\omega }\text{ or }y_{0}=i\sqrt{\omega }\right)\text{ and }v_{0}=0\text{ and }a\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2a, the least common multiple of 2a,a.

$$-y_{0}^{2}\times 2a+2u_{1}v_{0}=\omega \times 2a$$

Add y_{0}^{2}\times 2a to both sides.

$$2u_{1}v_{0}=\omega \times 2a+y_{0}^{2}\times 2a$$

The equation is in standard form.

$$2v_{0}u_{1}=2ay_{0}^{2}+2a\omega$$

Divide both sides by 2v_{0}.

$$\frac{2v_{0}u_{1}}{2v_{0}}=\frac{2a\left(y_{0}^{2}+\omega \right)}{2v_{0}}$$

Dividing by 2v_{0} undoes the multiplication by 2v_{0}.

$$u_{1}=\frac{2a\left(y_{0}^{2}+\omega \right)}{2v_{0}}$$

Divide 2a\left(\omega +y_{0}^{2}\right) by 2v_{0}.

$$u_{1}=\frac{a\left(y_{0}^{2}+\omega \right)}{v_{0}}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{u_{1}v_{0}}{y_{0}^{2}+\omega }\text{, }&v_{0}\neq 0\text{ and }u_{1}\neq 0\text{ and }\left(|y_{0}|\neq \sqrt{-\omega }\text{ or }\omega >0\right)\text{ and }\omega \neq -y_{0}^{2}\\a\neq 0\text{, }&\left(v_{0}=0\text{ or }u_{1}=0\right)\text{ and }\omega =-y_{0}^{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2a, the least common multiple of 2a,a.

$$-y_{0}^{2}\times 2a+2u_{1}v_{0}=\omega \times 2a$$

Subtract \omega \times 2a from both sides.

$$-y_{0}^{2}\times 2a+2u_{1}v_{0}-\omega \times 2a=0$$

Multiply -1 and 2 to get -2.

$$-2y_{0}^{2}a+2u_{1}v_{0}-\omega \times 2a=0$$

Multiply -1 and 2 to get -2.

$$-2y_{0}^{2}a+2u_{1}v_{0}-2\omega a=0$$

Subtract 2u_{1}v_{0} from both sides. Anything subtracted from zero gives its negation.

$$-2y_{0}^{2}a-2\omega a=-2u_{1}v_{0}$$

Combine all terms containing a.

$$\left(-2y_{0}^{2}-2\omega \right)a=-2u_{1}v_{0}$$

Divide both sides by -2y_{0}^{2}-2\omega .

$$\frac{\left(-2y_{0}^{2}-2\omega \right)a}{-2y_{0}^{2}-2\omega }=-\frac{2u_{1}v_{0}}{-2y_{0}^{2}-2\omega }$$

Dividing by -2y_{0}^{2}-2\omega undoes the multiplication by -2y_{0}^{2}-2\omega .

$$a=-\frac{2u_{1}v_{0}}{-2y_{0}^{2}-2\omega }$$

Divide -2u_{1}v_{0} by -2y_{0}^{2}-2\omega .

$$a=\frac{u_{1}v_{0}}{y_{0}^{2}+\omega }$$

Variable a cannot be equal to 0.

$$a=\frac{u_{1}v_{0}}{y_{0}^{2}+\omega }\text{, }a\neq 0$$

Solve for u_1

$$\left\{\begin{matrix}u_{1}=\frac{a\left(y_{0}^{2}+\omega \right)}{v_{0}}\text{, }&v_{0}\neq 0\text{ and }a\neq 0\\u_{1}\in \mathrm{R}\text{, }&v_{0}=0\text{ and }\omega =-y_{0}^{2}\text{ and }a\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2a, the least common multiple of 2a,a.

$$-y_{0}^{2}\times 2a+2u_{1}v_{0}=\omega \times 2a$$

Add y_{0}^{2}\times 2a to both sides.

$$2u_{1}v_{0}=\omega \times 2a+y_{0}^{2}\times 2a$$

The equation is in standard form.

$$2v_{0}u_{1}=2ay_{0}^{2}+2a\omega$$

Divide both sides by 2v_{0}.

$$\frac{2v_{0}u_{1}}{2v_{0}}=\frac{2a\left(y_{0}^{2}+\omega \right)}{2v_{0}}$$

Dividing by 2v_{0} undoes the multiplication by 2v_{0}.

$$u_{1}=\frac{2a\left(y_{0}^{2}+\omega \right)}{2v_{0}}$$

Divide 2a\left(\omega +y_{0}^{2}\right) by 2v_{0}.

$$u_{1}=\frac{a\left(y_{0}^{2}+\omega \right)}{v_{0}}$$