Solving Addition and Subtraction Equations | Grade 6 Math | LevelKite

Every equation in this lesson hides one number, and your job is to find it. First, the words you need.

A variable is a letter that stands for a number. In an equation, it often stands for a number you do not know yet. An expression is built from numbers, variables and operations, such as $x + 5$ or just $12$ . An equation is a statement that two expressions have the same value, such as $x + 5 = 12$ .

A solution of an equation is a value of the variable that makes the equation true. To solve an equation means to find its solution.

Try two values for $x$ in $x + 5 = 12$ .

If $x$ is $4$ , the left side is $4 + 5 = 9$ . Since $9$ is not $12$ , the number $4$ is not a solution.
If $x$ is $7$ , the left side is $7 + 5 = 12$ . Both sides are $12$ , so $7$ is a solution.

Testing values is slow when the numbers are large, or when they are fractions or decimals. In this lesson you will learn a method that finds the solution directly.

Start with a true statement about numbers: $7 + 5 = 12$ . Each side has the value $12$ .

Subtract $5$ from both sides. The left side becomes $7 + 5 - 5 = 7$ . The right side becomes $12 - 5 = 7$ . The new statement, $7 = 7$ , is still true.

This is no accident. The two sides were the same number, so taking $5$ away from each must leave the same number again.

Now go back to $7 + 5 = 12$ and subtract $5$ from the left side only. You get $7 = 12$ , which is false. Changing the value of one side alone breaks the equality.

So here is the idea behind solving: when two sides are equal, they stay equal if you add the same number to both sides or subtract the same number from both sides. This holds for every number you add or subtract.

Addition and subtraction are inverse operations: each one undoes the other. Add $5$ to a number, then subtract $5$ , and you are back at the number you started with.

In symbols, $x + 5 - 5$ has the same value as $x$ for every number $x$ . Two expressions that have the same value for every value of the variable are called equivalent expressions. In the same way, $x - 5 + 5$ is equivalent to $x$ .

Now put the two ideas together to solve $x + 5 = 12$ . The equation adds $5$ to $x$ . Subtracting $5$ undoes that addition, and subtracting it from both sides keeps the two sides equal:

x + 5 - 5 = 12 - 5

Now simplify each side: rewrite it as a simpler expression with the same value. On the left, $x + 5 - 5$ is equivalent to $x$ , so you can write $x$ in its place without changing that side's value. On the right, $12 - 5 = 7$ . The equation becomes $x = 7$ .

The variable now stands alone on one side. This is called isolating the variable, and it shows you the solution.

Always check your value in the original equation. Replace $x$ with $7$ : the left side is $7 + 5 = 12$ , and the right side is $12$ . Both sides match, so $7$ is a solution. The steps show that a solution has to be $7$ . The check confirms that $7$ really works.

Key concept

Use the inverse operation on both sides

An equation says its two sides have the same value. Apply the same valid operation to both sides and they still have the same value.

In this lesson, every equation has a number added to the variable or subtracted from it. To solve one:

See what the equation does to the variable.
Apply the inverse operation to both sides. Subtract to undo addition. Add to undo subtraction.
Simplify until the variable is alone, then check in the original equation.

Adding or subtracting the same number is valid for every number, so the method works for whole numbers, fractions and decimals. (Division has a limit: you may divide only by a number that is not zero. You will not need division here.)

In the formula, $a$ , $b$ and $c$ stand for any numbers.

\text{If } a = b, \text{ then } a + c = b + c \text{ and } a - c = b - c.

Worked example

Whole numbers: undo a subtraction

Solve $n - 14 = 29$ . The variable $n$ stands for an unknown number.

See what the equation does to n. It subtracts 14 from n. Addition is the inverse of subtraction, so adding 14 will undo it.
$n - 14 = 29$
Add 14 to both sides. Adding the same number to both sides is a valid operation, so the two sides stay equal.
$n - 14 + 14 = 29 + 14$
Simplify each side. On the left, subtracting 14 and then adding 14 leaves any number unchanged, so the left side is equivalent to n. On the right, 29 plus 14 is 43.
$n = 29 + 14 = 43$
Check in the original equation. Replace n with 43 and subtract. The result is 29, which matches the right side, so 43 is a solution.
$43 - 14 = 29$

Answer

The solution of $n - 14 = 29$ is $n = 43$ .

Worked example

Decimals: undo an addition

Solve $d + 2.75 = 6.4$ . The variable $d$ stands for an unknown number.

The equation adds 2.75 to d. Subtraction is the inverse of addition, so subtract 2.75 from both sides. The same valid operation on both sides keeps them equal.
$d + 2.75 - 2.75 = 6.4 - 2.75$
Simplify the left side. Adding 2.75 and then subtracting 2.75 leaves any number unchanged, so the left side is equivalent to d.
$d = 6.4 - 2.75$
Do the subtraction. Write 6.4 as 6.40 so that both numbers show hundredths. Then 640 hundredths minus 275 hundredths is 365 hundredths, which is 3.65.
$d = 6.40 - 2.75 = 3.65$
Check in the original equation. Replace d with 3.65 and add. The sum is 6.40, which is the same number as 6.4, so both sides match and 3.65 is a solution.
$3.65 + 2.75 = 6.40$

Answer

The solution of $d + 2.75 = 6.4$ is $d = 3.65$ .

Worked example

Fractions: undo a subtraction

Solve $y - \frac{1}{4} = \frac{2}{3}$ . The variable $y$ stands for an unknown number.

The equation subtracts one fourth from y, so add one fourth to both sides. The same valid operation on both sides keeps them equal.
$y - \frac{1}{4} + \frac{1}{4} = \frac{2}{3} + \frac{1}{4}$
Simplify the left side. Subtracting one fourth and then adding one fourth leaves any number unchanged, so the left side is equivalent to y.
$y = \frac{2}{3} + \frac{1}{4}$
Add the fractions. Both need the same denominator, so use 12, a multiple of both 3 and 4. Multiplying the top and bottom of two thirds by 4 gives eight twelfths. Multiplying the top and bottom of one fourth by 3 gives three twelfths. Each fraction keeps its value.
$y = \frac{2 \times 4}{3 \times 4} + \frac{1 \times 3}{4 \times 3} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}$
Check in the original equation. Replace y with eleven twelfths, write one fourth as three twelfths, and subtract. Eight twelfths simplifies to two thirds, which matches the right side, so eleven twelfths is a solution.
$\frac{11}{12} - \frac{1}{4} = \frac{11}{12} - \frac{3}{12} = \frac{8}{12} = \frac{2}{3}$

Answer

The solution of $y - \frac{1}{4} = \frac{2}{3}$ is $y = \frac{11}{12}$ .

The same method with different numbers. Each letter stands for an unknown number.

Equation	What the equation does	Do to both sides	Solution	Check
$a + 37 = 90$	Adds $37$ to $a$	Subtract $37$	$a = 90 - 37 = 53$	$53 + 37 = 90$
$b - 45 = 120$	Subtracts $45$ from $b$	Add $45$	$b = 120 + 45 = 165$	$165 - 45 = 120$
$c - 1.8 = 4.5$	Subtracts $1.8$ from $c$	Add $1.8$	$c = 4.5 + 1.8 = 6.3$	$6.3 - 1.8 = 4.5$
$p + \frac{2}{9} = \frac{7}{9}$	Adds $\frac{2}{9}$ to $p$	Subtract $\frac{2}{9}$	$p = \frac{7}{9} - \frac{2}{9} = \frac{5}{9}$	$\frac{5}{9} + \frac{2}{9} = \frac{7}{9}$

Here is the idea to keep. Two equal sides stay equal when you apply the same valid operation to both. When an equation adds a number to the variable or subtracts a number from it, use the inverse operation on both sides: subtract to undo addition, add to undo subtraction. Simplify until the variable is alone, then check your value in the original equation.

The method is the same for whole numbers, fractions and decimals. Only the arithmetic changes.

Next, you will write one-step addition and subtraction equations from words, solve word problems, explain the steps of a solution and spot mistakes in solving.