Lesson 4: Solving Equations

New Concept

An equation states that two expressions are equal. We'll learn to solve them by using inverse operations to isolate the variable. This algebraic skill is essential for using formulas and writing equations to model and solve real-world problems.

What’s next

This is just the beginning. Next, you'll master inverse operations with practice cards and then apply your skills to solve formulas and word problems.

Property

An equation is a statement that two expressions are equal. It may involve one or more variables. A value of the variable that makes an equation true is called a solution of the equation, and the process of finding this value is called solving the equation.

Examples

• The statement $x + 5 = 12$ is an equation. The value $x=7$ is a solution because $7 + 5 = 12$ is a true statement.
• To check if $y=3$ is a solution to $8y = 24$, we substitute it in: $8(3) = 24$. This is true, so $y=3$ is a solution.
• Is $z=10$ a solution for $z - 4 = 5$? We check: $10 - 4 = 6$. Since $6$ is not equal to $5$, $z=10$ is not a solution.

Explanation

Think of an equation as a perfectly balanced scale. A solution is the specific value for the variable that keeps the scale level. Finding that value is what we call solving the equation.

Property

Multiplication and division are opposite or inverse operations, because each operation undoes the effects of the other.
Addition and subtraction are opposite or inverse operations, because each operation undoes the effects of the other.

Examples

• To undo adding 8, you subtract 8. For example, $x + 8 - 8$ simplifies back to just $x$.
• To undo multiplying by 3, you divide by 3. For example, $\frac{3y}{3}$ simplifies back to just $y$.
• The inverse of subtracting 10 is adding 10, and the inverse of dividing by 5 is multiplying by 5.

Explanation

Inverse operations are pairs of actions that cancel each other out, like locking and unlocking a door. We use them to isolate a variable by undoing whatever operation is being performed on it.

Property

Addition and Subtraction Properties of Equality: If the same quantity is added to or subtracted from both sides of an equation, the solution is unchanged. In symbols, If $a = b$, then $a + c = b + c$ and $a - c = b - c$.
Multiplication and Division Properties of Equality: If both sides of an equation are multiplied or divided by the same nonzero quantity, the solution is unchanged. In symbols, If $a = b$, then $ac = bc$ and $\frac{a}{c} = \frac{b}{c}$, $c \neq 0$.

Examples

• If you have the equation $x - 5 = 10$, you can add 5 to both sides to get $x - 5 + 5 = 10 + 5$, which simplifies to $x = 15$.
• For the equation $4y = 28$, you can divide both sides by 4 to get $\frac{4y}{4} = \frac{28}{4}$, which simplifies to $y = 7$.
• If $z + 9 = 12$, you can subtract 9 from both sides to get $z + 9 - 9 = 12 - 9$, which simplifies to $z = 3$.

Explanation

To keep an equation balanced, you must perform the exact same operation on both sides. Whatever you add, subtract, multiply, or divide on one side, you must do to the other side too.

Property

To solve an equation algebraically:

1. Ask yourself what operation has been performed on the variable.
2. Perform the opposite operation on both sides of the equation in order to isolate the variable.

Examples

• To solve $x + 9 = 20$, 9 is added to $x$. We do the opposite and subtract 9 from both sides: $x + 9 - 9 = 20 - 9$, so $x = 11$.
• To solve $8m = 40$, $m$ is multiplied by 8. We do the opposite and divide both sides by 8: $\frac{8m}{8} = \frac{40}{8}$, so $m = 5$.
• To solve $\frac{w}{4} = 6$, $w$ is divided by 4. We do the opposite and multiply both sides by 4: $4(\frac{w}{4}) = 4(6)$, so $w = 24$.

Explanation

The main goal is to get the variable alone on one side of the equals sign. You do this by identifying the operation connected to the variable and using its inverse to cancel it out.

Property

Steps for solving an applied problem:

1. Identify the unknown quantity and choose a variable to represent it.
2. Find some quantity that can be described in two different ways, and write an equation using the variable to model the problem situation.
3. Solve the equation and answer the question in the problem.

Examples

• Leo had some comics, then bought 5 more. He now has 18 comics. How many did he start with? Let $c$ be his starting comics. The equation is $c + 5 = 18$. Solving gives $c = 13$.
• A pizza was shared equally among 4 friends. Each paid 9 dollars. What was the total cost? Let $p$ be the total cost. The equation is $\frac{p}{4} = 9$. Solving gives $p = 36$ dollars.
• After a discount of 10 dollars, a shirt costs 25 dollars. What was the original price? Let $x$ be the original price. The equation is $x - 10 = 25$. Solving gives $x = 35$ dollars.

Explanation

This three-step process turns a real-world story into a solvable math problem. Define your unknown with a variable, build a balanced equation that describes the situation, and then solve for your variable.