**Author**: John Redden - Professor of Mathematics, author, father of four, and education technology enthusiast. ( jt.redden@gmail.com )

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[ Elementary Algebra Online Textbook ]

[ Intermediate Algebra Online Textbook ]

[ TrigCheatSheet.com ]

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**Solving General Linear Equations in Two Steps**

Steps for solving general linear equations of the form $ ax + b = c$.

$$ 2x + 3 = 13 $$

Add $-3$ on both sides of the equation.

$$ 2x = 10 $$

Divide both sides by $2$.

$$ x = 5 $$

Check: $2(5) + 3 = 10 + 3 = 13$ ✔

Solving Factorable Quadratic Equations Part 1

Solving Factorable Quadratic Equations Part 1

$$x^2 + 2x - 8 = 0$$

Factor: here $-2 \cdot 4 = -8$ where $-2 + 4 = 2$.

$$(x - 2)(x + 4) = 0$$

Set each factor equal to zero and solve.

$$x - 2 = 0 \,\,\, \text{ or } \,\,\, x + 4 = 0$$

$$x = 2 \, \,\,\,\,\,\,\,\,\,\, \, x = -4$$

Check: $(2)^2 + 2(2) - 8 = 4 + 4 - 8 = 0$ ✔

Check: $(-4)^2 + 2(-4) - 8 = 16 - 8 - 8 = 0$ ✔

**Solving Rational Equations**

$$ \frac{x}{x+4} + \frac{1}{x+2} = \frac{x+12}{x^2+6x+8}$$

Factor the denominators.

$$ \frac{x}{x+4} + \frac{1}{x+2} = \frac{x+12}{(x+4)(x+2)}$$

Make note of the restrictions: $x \neq -4$ and $x\neq-2$.

Multiply both sides by the LCD = $(x+4)(x+2)$ and simplify.

$$ x(x+2) + 1(x+4) = x+12$$

$$x^2 + 2x + x + 4 = x+12$$

$$x^2 + 3x + 4 = x+12$$

Set to zero and solve.

$$x^2 + 2x - 8 = 0$$

$$(x - 2)(x + 4) = 0$$

$$x + 4 = 0 \,\,\, \text{ or } \,\,\, x - 2 = 0$$

$$x \neq -4 \, \,\,\,\,\,\,\,\,\,\, \, x = 2 \,\, \checkmark $$

One solution because $-4$ is extraneous (it is a restriction).

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