Equations and Inequalities Review

The Algebra I Toolkit, Sharpened

Every technique in Algebra II builds on the equation-solving skills you developed in Algebra I. This unit is not about learning new material from scratch -- it is about making sure those skills are sharp, automatic, and ready for harder problems. If you can solve the equations in this unit without hesitation, you are prepared for what comes next.

Multi-Step Equations: The Full Toolkit

A multi-step equation requires more than one operation to isolate the variable. The general strategy has not changed since Algebra I:

1. Distribute to eliminate parentheses.
2. Combine like terms on each side.
3. Move variable terms to one side using addition or subtraction.
4. Move constants to the other side.
5. Divide (or multiply) to isolate the variable.

Example: Solve

$$3(2x-5)+4=7x-2$$

.

Step 1: Distribute:

$$6x-15+4=7x-2$$

Step 2: Combine like terms on the left:

$$6x-11=7x-2$$

Step 3: Subtract

$$6x$$

from both sides:

$$-11=x-2$$

Step 4: Add 2 to both sides:

$$-9=x$$

Check:

$$3(2(-9)-5)+4=3(-23)+4=-65$$

and

$$7(-9)-2=-65$$

. Both sides match.

Absolute Value Equations

The absolute value of a number is its distance from zero on the number line. Distance is always non-negative, so

$$|x|\ge 0$$

for every real number

$$x$$

.

This distance interpretation is the key to solving absolute value equations. When you write

$$|x|=5$$

, you are asking: what numbers are exactly 5 units from zero? The answer is

$$x=5$$

or

$$x=-5$$

.

General principle: If

$$|A|=c$$

where

$$c>0$$

, then

$$A=c$$

or

$$A=-c$$

.

If

$$c=0$$

, then

$$A=0$$

(one solution). If

$$c‘<‘0$$

, there is no solution -- distance cannot be negative.

Example: Solve

$$|2x+3|=7$$

.

Case 1:

$$2x+3=7$$

, so

$$2x=4$$

, so

$$x=2$$

.

Case 2:

$$2x+3=-7$$

, so

$$2x=-10$$

, so

$$x=-5$$

.

Check both:

$$|2(2)+3|=|7|=7$$

and

$$|2(-5)+3|=|-7|=7$$

. Both valid.

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❓

Concept Check

Solve |4x - 1| = -3. How many solutions are there?

▸

There are no solutions. Absolute value is always non-negative, so |4x - 1| can never equal -3. Any equation |expression| = negative number has no solution.

Absolute Value Inequalities

Absolute value inequalities fall into two patterns that correspond to distance reasoning on the number line.

Less than (distance is LESS than c):

$$|A|‘<‘c$$

means

$$-c‘<‘A‘<‘c$$

. The solutions are between

$$-c$$

and

$$c$$

-- a single interval.

Greater than (distance is MORE than c):

$$|A|>c$$

means

$$A‘<‘-c$$

or

$$A>c$$

. The solutions are outside the interval -- two separate rays.

Example: Solve

$$|x-4|\le 3$$

.

This means the distance from

$$x$$

to 4 is at most 3.

Rewrite as a compound inequality:

$$-3\le x-4\le 3$$

Add 4 to all parts:

$$1\le x\le 7$$

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Example: Solve

$$|3x+2|>5$$

.

Case 1:

$$3x+2>5$$

, so

$$3x>3$$

, so

$$x>1$$

.

Case 2:

$$3x+2‘<‘-5$$

, so

$$3x‘<‘-7$$

, so

$$x‘<‘-\frac{7}{3}$$

.

Solution:

$$x‘<‘-\frac{7}{3}$$

or

$$x>1$$

.

Literal Equations

A literal equation is an equation with multiple variables where you solve for one variable in terms of the others. Formulas from science, geometry, and finance are all literal equations.

The technique is identical to solving a numerical equation -- isolate the target variable -- except your answer is an expression rather than a number.

Example: Solve

$$V=\pi r^{2}h$$

for

$$h$$

.

Divide both sides by

$$\pi r^2$$

:

$$h=\frac{V}{\pi r^2}$$

Example: Solve

$$ax+by=c$$

for

$$y$$

.

Subtract

$$ax$$

:

$$by=c-ax$$

Divide by

$$b$$

:

$$y=\frac{c-ax}{b}$$

Notice this is the slope-intercept form:

$$y=-\frac{a}{b}x+\frac{c}{b}$$

. The slope is

$$-\frac{a}{b}$$

and the y-intercept is

$$\frac{c}{b}$$

.

❓

Concept Check

The formula for converting Fahrenheit to Celsius is C = (5/9)(F - 32). Solve this equation for F.

▸

Multiply both sides by 9/5: (9/5)C = F - 32. Add 32: F = (9/5)C + 32. This is the formula for converting Celsius back to Fahrenheit.

Looking Ahead

With multi-step equations, absolute value, and literal equations secure, you are ready to move beyond individual equations into the study of functions -- the central organizing concept of Algebra II. In Unit 2, every equation you solve will be understood as a relationship between inputs and outputs, and you will develop tools for combining and manipulating these relationships.