Solve A+beta^2-3x=5+x3

$$A+ { \beta }^{ 2 } -3x = 5+x3$$

Solve for A

$$A=3x+x_{3}-\beta ^{2}+5$$

Solution Steps

Subtract \beta ^{2} from both sides.

$$A-3x=5+x_{3}-\beta ^{2}$$

Add 3x to both sides.

$$A=5+x_{3}-\beta ^{2}+3x$$

Solve for x

$$x=\frac{-x_{3}+\beta ^{2}+A-5}{3}$$

Steps for Solving Linear Equation

Subtract A from both sides.

$$\beta ^{2}-3x=5+x_{3}-A$$

Subtract \beta ^{2} from both sides.

$$-3x=5+x_{3}-A-\beta ^{2}$$

The equation is in standard form.

$$-3x=x_{3}-\beta ^{2}-A+5$$

Divide both sides by -3.

$$\frac{-3x}{-3}=\frac{x_{3}-\beta ^{2}-A+5}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x=\frac{x_{3}-\beta ^{2}-A+5}{-3}$$

Divide 5+x_{3}-A-\beta ^{2} by -3.

$$x=\frac{-x_{3}+\beta ^{2}+A-5}{3}$$