This section covers how to solve linear equations, simple quadratic equations, and linear inequalities. You’ll also learn how to translate verbal descriptions into equations and when to reverse an inequality sign - skills that help you find exact solutions as well as ranges of possible values.

Definitions

Variable

A symbol (often $x$) that represents an unknown number.

Equation

A statement that two expressions are equal, containing a variable to solve for.

Linear equation

An equation of the form $ax+b=c$ where $a\ne 0$, solved by isolating $x$ using inverse operations.

Quadratic equation

An equation in which the variable appears to the second power, such as $x^2=k$ or $(x-a{)}^2=k$. Solutions come from taking square roots.

Inequality

A statement that two expressions are not necessarily equal, connected by one of the symbols $<$, $>$, $\le$, or $\ge$.

Solution

A value of the variable that makes the equation true.

Root

Another term for a solution of an equation, especially used with quadratics.

Solution set

The collection of all values that satisfy an equation or an inequality.

Solving equations and inequalities means isolating the variable or describing its solution range. Three principles run through every method in this chapter:

• Keep balance: whatever you do to one side of $=$, $<$, or $>$, do to the other.
• Undo operations in reverse order: distribute and combine like terms first, then remove constants, then remove coefficients.
• Apply special rules: these come up as each topic is introduced - for example, taking both square roots of a quadratic, or reversing an inequality sign when dividing by a negative number.

Solving linear equations

The goal when solving a linear equation is to isolate the variable on one side. We do this by applying inverse operations - undoing whatever is being done to the variable - in reverse order.

Example: Solve a basic linear equation

$$\begin{array}{rl}3x+5& =20\\ 3x& =15\\ x& =5\end{array}$$

Steps taken:

• Subtract 5 from both sides
• Divide both sides by 3

Answer: $x=5$

Example: Solve a more complex equation

$$\begin{array}{rl}4(x-2)& =2x+6\\ 4x-8& =2x+6\\ 2x-8& =6\\ 2x& =14\\ x& =7\end{array}$$

Steps taken:

• Distribute 4 on the left
• Subtract $2x$ from both sides
• Add 8 to both sides
• Divide both sides by 2

Answer: $x=7$

Solving simple quadratic equations

Quadratic equations often appear when a variable is squared. Unlike linear equations, which have one solution, quadratic equations often have two solutions - because both a positive and a negative number can square to the same value.

Use the square-root method when the variable appears in a single squared expression, such as $x^2=k$ or $(x-a{)}^2=k$. The key idea is that squaring hides sign information:

$$\text{If }{u}^2=k,\text{ then }u=\pm \sqrt{k}$$

This is because both positive and negative numbers square to the same value. For example, $3^2=9$ and $(-3{)}^2=9$, so if $x^2=9$, then $x=3$ or $x=-3$.

Steps to follow:

• Isolate the squared expression on one side of the equation.
• Take the square root of both sides and include the $\pm$.
• Solve the two resulting linear equations.
• If $k<0$, there is no real-number solution - no real number squared gives a negative result.

Common pitfall: When taking the square root of both sides, always write $\pm \sqrt{k}$ - not just $+\sqrt{k}$. Forgetting the negative root is one of the most common errors on this type of problem. For example, if $x^2=49$, the answer is $x=7$ or $x=-7$, not just $x=7$.

One exception: if the problem context makes the negative root impossible - for example, $x$ represents a length or a count - you can drop the negative root and report only the positive value.

Example: Solve $(x-3{)}^2=16$

$$\begin{array}{rl}(x-3{)}^2& =16\\ x-3& =\pm \sqrt{16}\\ x-3& =\pm 4\end{array}$$

• Case 1: $x-3=4\Rightarrow x=7$
• Case 2: $x-3=-4\Rightarrow x=-1$

Answer: $x=7$ or $x=-1$

Translating verbal descriptions

One of the most practical algebra skills is translating a word problem into an equation or expression you can solve. The key is recognizing which words signal which operations:

• Multiplication (“times,” “product of”) → $\times$ or $\cdot$
• Addition (“sum,” “more than”) → $+$
• Subtraction (“difference,” “less than”) → $-$
• Division (“quotient,” “per”) → $÷$ or a fraction bar

Watch out - “less than” reverses order: “A less than B” translates to $B-A$, not $A-B$. For example, “4 less than 7” means $7-4=3$, not $4-7=-3$. Similarly, “2 less than $3x$” means $3x-2$, not $2-3x$. The phrase names what you’re subtracting first, then what you’re subtracting it from.

Example: Translate and solve a multi-step word problem

A store sells notebooks for $\$3$ each and pens for $\$1.50$ each. Keisha buys twice as many pens as notebooks. She spends a total of $\$18$. How many notebooks does she buy?

Solution:

Let $n$ = the number of notebooks. Then the number of pens is $2n$.

Total cost: $3n+1.50(2n)=18$

$$\begin{array}{rl}3n+3n& =18\\ 6n& =18\\ n& =3\end{array}$$

Steps taken:

• Write expressions for each item’s cost in terms of $n$
• Simplify $1.50\times 2n=3n$
• Combine like terms
• Divide both sides by 6

Answer: Keisha buys 3 notebooks (and 6 pens).

Solving linear inequalities

An inequality $ax+b<c$ or $ax+b\ge c$ is solved much like an equation, with one important difference:

• If you multiply or divide both sides by a negative number, reverse the inequality sign.
• Adding or subtracting any number - even a negative one - does not flip the sign. Only multiplication or division by a negative does.

Example: Solve an inequality

$$\begin{array}{rl}2x-5& <7\\ 2x& <12\\ x& <6\end{array}$$

Steps taken:

• Add 5 to both sides
• Divide both sides by 2

Answer: $x<6$

Example: Solve an inequality with negatives

$$\begin{array}{rl}-3x+4& \ge 10\\ -3x& \ge 6\\ x& \le -2\end{array}$$

Steps taken:

• Subtract 4 from both sides
• Divide both sides by $-3$ and reverse the inequality sign

Answer: $x\le -2$

Real-world applications

Use an equation when a relationship is exact; use an inequality when there is a limit or constraint such as a budget or minimum requirement.

Example: Solve an applied inequality

You have $\$20$ to spend on pens. Each pen costs $\$1.50$, and there is a one-time supply fee of $\$2$. What is the maximum number of pens you can buy?

Solution:

Let $p$ = the number of pens. The total cost is $1.50p+2$, which must be no more than $20$:

$$\begin{array}{rl}1.50p+2& \le 20\\ 1.50p& \le 18\\ p& \le 12\end{array}$$

Steps taken:

• Subtract 2 from both sides
• Divide both sides by 1.50

Since $p$ must be a whole number and $p\le 12$, the maximum number of pens you can buy is 12.

Answer: 12 pens