## Algebra

Author: John Redden - Professor of Mathematics, author, father of four, and education technology enthusiast. Contact on LinkedIn.

Publications:
[ Elementary Algebra Online Textbook ]
[ Intermediate Algebra Online Textbook ]

[ TrigCheatSheet.com ]
help.myalgebrasucks.com ]
this -> study guide
[ Calculus 1 Limits Playlist ]

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[ Dynamic Worksheet - Solving Linear Equations ]

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

Steps for solving factorable quadratic equations of the form $x^2 + bx + c = 0$.
$$x^2 + 2x - 8 = 0$$
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).