Chapter 5 ยท Expressions

Adding and Subtracting Linear Expressions

Add and subtract linear expressions by grouping like terms. Learn why subtracting an expression changes the sign of every term in it.

13 min read

So far you have simplified one expression at a time. In this lesson you will add and subtract whole expressions.

A linear expression is made of terms, the parts that are added together. Here each term is either a number times a variable, such as 3x3x, or a plain number, such as 44. In 3x+43x + 4, the letter xx is the variable (it stands for a number), 33 is the coefficient (the number multiplying the variable), and 44 is a constant (a term with no variable).

Any number can be the value of xx in these expressions, because they only multiply xx by a number and then add or subtract.

Think of a linear expression as a rule: choose a value of xx, and the rule gives you exactly one number.

To add two linear expressions means to add the two numbers they give for the same value of xx. To subtract means to take the second number away from the first. Your goal is to write the result as one simpler expression that is equivalent to it. Two expressions are equivalent only when they are equal for every value of the variable.

The table adds 3x+43x + 4 and 5xโˆ’75x - 7 for four values of xx. Its last column shows one simpler expression, 8xโˆ’38x - 3.

Adding the values of two expressions
Value of x3x + 45x - 7Sum of the two values8x - 3
โˆ’1-1
11
โˆ’12-12
1+(โˆ’12)=โˆ’111 + (-12) = -11
โˆ’11-11
00
44
โˆ’7-7
4+(โˆ’7)=โˆ’34 + (-7) = -3
โˆ’3-3
0.50.5
5.55.5
โˆ’4.5-4.5
5.5+(โˆ’4.5)=15.5 + (-4.5) = 1
11
22
1010
33
10+3=1310 + 3 = 13
1313

In every row, the sum of the two values equals the value of 8xโˆ’38x - 3. One matching row would not be enough: 2x+92x + 9 also gives 1313 when x=2x = 2, but it gives 99, not โˆ’3-3, when x=0x = 0. Even four matching rows say nothing about all the other values of xx, so you need a reason that works for every value.

That reason comes from two properties of addition that are true for all numbers. You can add numbers in any order (the commutative property) and group them in any way (the associative property), and the sum stays the same. Whatever number xx is, every term is just a number, so both properties apply. They are properties of addition, not subtraction, so you first rewrite a subtraction such as 5xโˆ’75x - 7 as the addition 5x+(โˆ’7)5x + (-7), which has the same value for every xx.

Then you can move like terms (terms with the same variable part, such as 3x3x and 5x5x, or two constants) next to each other and combine them, as you did in the last lesson. That step is also true for every value of xx.

Worked example

Adding two linear expressions

Find the sum (3x+4)+(5xโˆ’7)(3x + 4) + (5x - 7), where xx stands for any number. Write it as one simpler expression.

  1. Rewrite subtracting 7 as adding negative 7, so that every term is being added. Subtracting 7 from a number always gives the same result as adding negative 7 to it, so the value does not change for any value of x. The terms are now 3x, 4, 5x and negative 7.

    (3x+4)+(5xโˆ’7)=(3x+4)+(5x+(โˆ’7))(3x + 4) + (5x - 7) = (3x + 4) + (5x + (-7))
  2. Only addition is left. For all numbers, addition allows any order and any grouping, so remove the parentheses and move the like terms next to each other.

    3x+4+5x+(โˆ’7)=(3x+5x)+(4+(โˆ’7))3x + 4 + 5x + (-7) = (3x + 5x) + (4 + (-7))
  3. Combine the like terms. For the x terms, add the coefficients: 3 + 5 = 8, so 3x + 5x = 8x. For the constants, 4 plus negative 7 is negative 3.

    (3x+5x)+(4+(โˆ’7))=(3+5)x+(4+(โˆ’7))=8x+(โˆ’3)(3x + 5x) + (4 + (-7)) = (3 + 5)x + (4 + (-7)) = 8x + (-3)
  4. Adding negative 3 gives the same result as subtracting 3, so write the answer in the shorter form. It matches the last column of the table.

    8x+(โˆ’3)=8xโˆ’38x + (-3) = 8x - 3

Answer

The sum (3x+4)+(5xโˆ’7)(3x + 4) + (5x - 7) is equivalent to 8xโˆ’38x - 3. The two expressions are equal for every value of xx, because each step used a property that is true for all numbers.

Subtraction needs one more idea. The opposite of a number is the number you add to it to get 00. The opposite of 55 is โˆ’5-5, and the opposite of โˆ’5-5 is 55. Subtracting a number always gives the same result as adding its opposite: 9โˆ’5=9+(โˆ’5)=49 - 5 = 9 + (-5) = 4 and 9โˆ’(โˆ’5)=9+5=149 - (-5) = 9 + 5 = 14.

For any value of xx, an expression such as 3xโˆ’53x - 5 gives one number, so subtracting that expression means adding its opposite. To find the opposite of an expression, take the opposite of each term: the opposite of 3xโˆ’53x - 5 is โˆ’3x+5-3x + 5. This works because each term and its opposite add to 00, so the two whole expressions add to 00 for every value of xx:

(3xโˆ’5)+(โˆ’3x+5)=(3x+(โˆ’3x))+(โˆ’5+5)=0+0=0(3x - 5) + (-3x + 5) = (3x + (-3x)) + (-5 + 5) = 0 + 0 = 0

Key concept

Add like terms; to subtract, add the opposite

To add linear expressions, rewrite each subtraction as adding a negative term, move like terms together and combine them.

To subtract a linear expression, add its opposite: replace every term of the expression you are subtracting with its opposite, then add in the same way.

Each step uses a property that is true for all numbers. That is why the result is equivalent to what you started with: the two are equal for every value of the variable, not only for the values you test.

Worked example

Subtracting a linear expression

Find the difference (7x+2)โˆ’(3xโˆ’5)(7x + 2) - (3x - 5), where xx stands for any number. Write it as one simpler expression.

  1. Subtracting an expression means adding its opposite. Take the opposite of each term in the second expression: the opposite of 3x is negative 3x, and the opposite of negative 5 is positive 5.

    (7x+2)โˆ’(3xโˆ’5)=(7x+2)+(โˆ’3x+5)(7x + 2) - (3x - 5) = (7x + 2) + (-3x + 5)
  2. Only addition is left, so remove the parentheses and move like terms together. Addition allows any order and any grouping.

    7x+2+(โˆ’3x)+5=(7x+(โˆ’3x))+(2+5)7x + 2 + (-3x) + 5 = (7x + (-3x)) + (2 + 5)
  3. Combine like terms. For the x terms, add the coefficients: 7 plus negative 3 is 4. For the constants, 2 plus 5 is 7.

    (7x+(โˆ’3x))+(2+5)=(7+(โˆ’3))x+(2+5)=4x+7(7x + (-3x)) + (2 + 5) = (7 + (-3))x + (2 + 5) = 4x + 7
  4. Check with x = 1. The first expression gives 7 + 2 = 9, and the second gives 3 minus 5, which is negative 2. The original difference is 9 minus negative 2, which is 9 + 2 = 11. The answer gives 4 + 7 = 11. The values agree. One value can catch a mistake but cannot prove equivalence. Steps 1 to 3 do that, because each one is true for every value of x.

    (7โ‹…1+2)โˆ’(3โ‹…1โˆ’5)=9โˆ’(โˆ’2)=11and4โ‹…1+7=11(7 \cdot 1 + 2) - (3 \cdot 1 - 5) = 9 - (-2) = 11 \quad \text{and} \quad 4 \cdot 1 + 7 = 11

Answer

The difference (7x+2)โˆ’(3xโˆ’5)(7x + 2) - (3x - 5) is equivalent to 4x+74x + 7, so the two expressions are equal for every value of xx.

Worked example

Subtracting with fraction coefficients

Find the difference (12xโˆ’3)โˆ’(34xโˆ’5)\left(\frac{1}{2}x - 3\right) - \left(\frac{3}{4}x - 5\right), where xx stands for any number. Write it as one simpler expression.

  1. Add the opposite of the expression being subtracted. Every term changes to its opposite: three-fourths x becomes negative three-fourths x, and negative 5 becomes positive 5.

    (12xโˆ’3)โˆ’(34xโˆ’5)=(12xโˆ’3)+(โˆ’34x+5)\left(\frac{1}{2}x - 3\right) - \left(\frac{3}{4}x - 5\right) = \left(\frac{1}{2}x - 3\right) + \left(-\frac{3}{4}x + 5\right)
  2. Write subtracting 3 as adding negative 3, then group like terms. Addition allows any order and any grouping.

    12x+(โˆ’3)+(โˆ’34x)+5=(12x+(โˆ’34x))+(โˆ’3+5)\frac{1}{2}x + (-3) + \left(-\frac{3}{4}x\right) + 5 = \left(\frac{1}{2}x + \left(-\frac{3}{4}x\right)\right) + (-3 + 5)
  3. Combine the x terms by adding the coefficients. Rewrite one-half as two-fourths so that both fractions have the denominator 4. Two-fourths plus negative three-fourths is negative one-fourth.

    12x+(โˆ’34x)=24x+(โˆ’34x)=(24+(โˆ’34))x=โˆ’14x\frac{1}{2}x + \left(-\frac{3}{4}x\right) = \frac{2}{4}x + \left(-\frac{3}{4}x\right) = \left(\frac{2}{4} + \left(-\frac{3}{4}\right)\right)x = -\frac{1}{4}x
  4. Combine the constants: negative 3 plus 5 is 2. Write the two results together.

    โˆ’14x+(โˆ’3+5)=โˆ’14x+2-\frac{1}{4}x + (-3 + 5) = -\frac{1}{4}x + 2
  5. Check with x = 4. Half of 4 is 2 and three-fourths of 4 is 3. The first expression gives 2 minus 3, which is negative 1. The second gives 3 minus 5, which is negative 2. Their difference is negative 1 plus 2, which is 1. The answer gives negative 1 plus 2, also 1. A match at one value is not proof; the steps are, since each is true for every value of x.

    (2โˆ’3)โˆ’(3โˆ’5)=โˆ’1โˆ’(โˆ’2)=1andโˆ’14โ‹…4+2=โˆ’1+2=1(2 - 3) - (3 - 5) = -1 - (-2) = 1 \quad \text{and} \quad -\frac{1}{4} \cdot 4 + 2 = -1 + 2 = 1

Answer

The difference is equivalent to โˆ’14x+2-\frac{1}{4}x + 2. It equals the original expression for every value of xx.

Adding or subtracting linear expressions means adding or subtracting the numbers they give for each value of the variable. To add, move like terms together and combine them. To subtract, add the opposite, which means taking the opposite of every term in the expression you subtract. Each step uses a property that is true for all numbers, so your answer is equivalent to the original: equal for every value of the variable. A check with one value can expose a mistake, but it cannot prove equivalence.

Next, you will expand linear expressions with the distributive property and factor them.