Chapter 6 ยท Equations and inequalities

Checking and Solving One-Step Equations

Test whether a value makes an equation true. Then solve one-step equations, including ones with fractions and decimals, by undoing the operation on both sides.

20 min read

You already know how to evaluate an expression: you replace the variable with a number and compute. In this lesson you use that skill to test equations and then to solve them.

An equation is a statement that two expressions have the same value. You write it with an equals sign between the expressions, as in x+7=12x + 7 = 12. The letter xx is a variable, which means it stands for a number.

Whether x+7=12x + 7 = 12 is true depends on which number xx stands for. A solution of an equation is a value of the variable that makes the equation true.

To decide whether a value is a solution, test it in three moves.

  1. Substitute the value: write the number in place of the variable.
  2. Evaluate each side: compute its value.
  3. Compare the two sides. If they have the same value, the equation is true and the value is a solution. If they have different values, the equation is false and the value is not a solution.

The table tests three values of xx in x+7=12x + 7 = 12.

Three values tested in the same equation
Value testedLeft sideRight sideDo the sides match?Is it a solution?
x=3x = 3
3+7=103 + 7 = 10
1212
No, because 10โ‰ 1210 \neq 12
No
x=5x = 5
5+7=125 + 7 = 12
1212
Yes, because 12=1212 = 12
Yes
x=โˆ’5x = -5
โˆ’5+7=2-5 + 7 = 2
1212
No, because 2โ‰ 122 \neq 12
No

Worked example

Testing two values

The variable yy stands for a number. Is y=โˆ’4y = -4 a solution of โˆ’3y=12-3y = 12? Is y=4y = 4 a solution?

  1. To test -4, substitute it for y in the left side. The expression -3y means -3 times y, so the left side becomes -3 times -4.

    โˆ’3(โˆ’4)-3(-4)
  2. Evaluate the left side. A negative number times a negative number is positive.

    โˆ’3(โˆ’4)=12-3(-4) = 12
  3. Compare with the right side, which is 12. The two sides have the same value, so the equation is true for this value.

    12=1212 = 12
  4. Now test 4. Substitute 4 for y in the left side and evaluate. A negative number times a positive number is negative.

    โˆ’3(4)=โˆ’12-3(4) = -12
  5. Compare with the right side. The two sides have different values, so the equation is false for this value.

    โˆ’12โ‰ 12-12 \neq 12

Answer

The value y=โˆ’4y = -4 is a solution of โˆ’3y=12-3y = 12 because it makes the equation true. The value y=4y = 4 is not a solution because it makes the left side โˆ’12-12, not 1212.

Testing values one at a time is slow, and you may never try the right one. To solve an equation means to find every value of the variable that makes it true. You can reason your way to those values instead of guessing.

The reasoning rests on one fact about equal numbers. Start with a true statement such as 10=1010 = 10. Subtract 3 from both sides and you get 7=77 = 7. Then multiply both sides by โˆ’2-2 and you get โˆ’14=โˆ’14-14 = -14. Each statement is still true, because both sides started as the same number and you did the same thing to each. Change only one side and the statement breaks: 10โˆ’3=1010 - 3 = 10 is false.

Key concept

Do the same thing to both sides

An equation says that its two sides have the same value. When you do the same operation to both sides of a true equation, the two sides still have the same value. You may add the same number to both sides or subtract the same number from both sides. You may also multiply or divide both sides by the same number, as long as that number is not zero. You never divide by zero.

To solve, choose the operation that undoes what the equation does to the variable. Then check your answer: a solution must make the original equation true.

Two operations are inverse operations when each one undoes the other. Adding 9 and subtracting 9 are inverse operations. Multiplying by 4 and dividing by 4 are inverse operations too. Dividing by zero is never an inverse step, because a division such as 5รท05 \div 0 has no answer: no number times 0 equals 5.

A one-step equation applies exactly one operation to the variable. It adds a number to the variable, subtracts a number from the variable, or multiplies or divides the variable by a number that is not zero. One inverse operation, done to both sides, leaves the variable alone on its side.

That step shows what the solution must be. Substituting that value into the original equation confirms that it really is a solution.

Worked example

Undoing addition

The variable xx stands for a number. Solve x+9=4x + 9 = 4.

  1. The left side adds 9 to x. Subtracting 9 undoes adding 9, so subtract 9 from both sides. Doing the same operation to both sides keeps them equal.

    x+9โˆ’9=4โˆ’9x + 9 - 9 = 4 - 9
  2. Simplify the left side. Adding 9 and then subtracting 9 gives back x for every value of x, so x + 9 - 9 and x are equivalent expressions.

    x+9โˆ’9=xx + 9 - 9 = x
  3. Simplify the right side. Starting at 4 and going down 9 lands 5 below zero.

    4โˆ’9=โˆ’54 - 9 = -5
  4. Write the simpler equation. It shows the value that x must have.

    x=โˆ’5x = -5
  5. Check by substituting -5 for x in the original equation. The left side equals 4, which matches the right side, so the equation is true.

    โˆ’5+9=4-5 + 9 = 4

Answer

The solution is x=โˆ’5x = -5, because โˆ’5+9=4-5 + 9 = 4 makes the original equation true.

One inverse step for each operation. Each letter stands for a number.
EquationWhat the equation does to the variableInverse step on both sidesSimpler equationCheck in the original
a+6=โˆ’1a + 6 = -1
Adds 6 to aa
Subtract 6: a+6โˆ’6=โˆ’1โˆ’6a + 6 - 6 = -1 - 6
a=โˆ’7a = -7
โˆ’7+6=โˆ’1-7 + 6 = -1
pโˆ’5=โˆ’2p - 5 = -2
Subtracts 5 from pp
Add 5: pโˆ’5+5=โˆ’2+5p - 5 + 5 = -2 + 5
p=3p = 3
3โˆ’5=โˆ’23 - 5 = -2
โˆ’6n=54-6n = 54
Multiplies nn by โˆ’6-6
Divide by โˆ’6-6, which is not zero: โˆ’6nโˆ’6=54โˆ’6\frac{-6n}{-6} = \frac{54}{-6}
n=โˆ’9n = -9
โˆ’6(โˆ’9)=54-6(-9) = 54
t4=โˆ’3\frac{t}{4} = -3
Divides tt by 4
Multiply by 4: t4โ‹…4=โˆ’3โ‹…4\frac{t}{4} \cdot 4 = -3 \cdot 4
t=โˆ’12t = -12
โˆ’124=โˆ’3\frac{-12}{4} = -3

Fractions and decimals do not change the plan. You still do the inverse operation to both sides, and you still never divide by zero. Only the arithmetic changes.

One fact helps when you divide by a fraction. The reciprocal of a number that is not zero is the number you multiply it by to get 1. The reciprocal of 34\frac{3}{4} is 43\frac{4}{3}, because 34โ‹…43=1\frac{3}{4} \cdot \frac{4}{3} = 1. Dividing by a number that is not zero gives the same result as multiplying by its reciprocal. Zero has no reciprocal, because 0 times any number is 0, never 1.

The number you divide by does not have to be a fraction with 1 on top, such as 14\frac{1}{4}. It can be any fraction, any mixed number such as 1121\frac{1}{2}, or any decimal, positive or negative, as long as it is not zero.

Worked example

Dividing by a fraction

The variable mm stands for a number. Solve 34m=โˆ’9\frac{3}{4}m = -9.

  1. The left side multiplies m by 3/4. Division undoes multiplication, and 3/4 is not zero, so divide both sides by 3/4.

    34mรท34=โˆ’9รท34\frac{3}{4}m \div \frac{3}{4} = -9 \div \frac{3}{4}
  2. Simplify the left side. Multiplying by 3/4 and then dividing by 3/4 gives back m for every value of m.

    34mรท34=m\frac{3}{4}m \div \frac{3}{4} = m
  3. Rewrite the right side. Because 3/4 is not zero, dividing by it gives the same result as multiplying by its reciprocal, 4/3.

    โˆ’9รท34=โˆ’9โ‹…43-9 \div \frac{3}{4} = -9 \cdot \frac{4}{3}
  4. Multiply. A negative number times a positive number is negative. Then 9 times 4 is 36, and 36 divided by 3 is 12.

    โˆ’9โ‹…43=โˆ’363=โˆ’12-9 \cdot \frac{4}{3} = -\frac{36}{3} = -12
  5. Write the simpler equation. It shows the value that m must have.

    m=โˆ’12m = -12
  6. Check by substituting -12 for m in the original equation. The left side equals -9, which matches the right side.

    34(โˆ’12)=โˆ’364=โˆ’9\frac{3}{4}(-12) = -\frac{36}{4} = -9

Answer

The solution is m=โˆ’12m = -12, because 34(โˆ’12)=โˆ’9\frac{3}{4}(-12) = -9 makes the original equation true.

Worked example

Dividing by a negative decimal

The variable kk stands for a number. Solve โˆ’0.4k=2.6-0.4k = 2.6.

  1. The left side multiplies k by -0.4, which is not zero. Divide both sides by -0.4 to undo the multiplication.

    โˆ’0.4kโˆ’0.4=2.6โˆ’0.4\frac{-0.4k}{-0.4} = \frac{2.6}{-0.4}
  2. Simplify the left side to k, because multiplying by -0.4 and then dividing by -0.4 gives back k for every value of k.

    k=2.6โˆ’0.4k = \frac{2.6}{-0.4}
  3. Find the size of the answer by dividing 2.6 by 0.4. Multiplying both numbers by 10 gives a division with the same answer.

    2.6รท0.4=26รท4=6.52.6 \div 0.4 = 26 \div 4 = 6.5
  4. A positive number divided by a negative number is negative, so the answer is negative.

    k=โˆ’6.5k = -6.5
  5. Check in the original equation. A negative number times a negative number is positive, and 0.4 times 6.5 is 2.6.

    โˆ’0.4(โˆ’6.5)=2.6-0.4(-6.5) = 2.6

Answer

The solution is k=โˆ’6.5k = -6.5, because โˆ’0.4(โˆ’6.5)=2.6-0.4(-6.5) = 2.6 makes the original equation true.

The same plan with other fractions and decimals. Each letter stands for a number.
EquationInverse step on both sidesSolution and arithmeticCheck in the original
y+12=13y + \frac{1}{2} = \frac{1}{3}
Subtract 12\frac{1}{2}
y=13โˆ’12=26โˆ’36=โˆ’16y = \frac{1}{3} - \frac{1}{2} = \frac{2}{6} - \frac{3}{6} = -\frac{1}{6}
โˆ’16+12=โˆ’16+36=26=13-\frac{1}{6} + \frac{1}{2} = -\frac{1}{6} + \frac{3}{6} = \frac{2}{6} = \frac{1}{3}
dโˆ’2.75=โˆ’1.5d - 2.75 = -1.5
Add 2.752.75
d=โˆ’1.5+2.75=1.25d = -1.5 + 2.75 = 1.25
1.25โˆ’2.75=โˆ’1.51.25 - 2.75 = -1.5
112โ‹…r=61\frac{1}{2} \cdot r = 6
Divide by 1121\frac{1}{2}, which equals 32\frac{3}{2} and is not zero
r=6รท32=6โ‹…23=123=4r = 6 \div \frac{3}{2} = 6 \cdot \frac{2}{3} = \frac{12}{3} = 4
112โ‹…4=32โ‹…4=122=61\frac{1}{2} \cdot 4 = \frac{3}{2} \cdot 4 = \frac{12}{2} = 6
hรท2.5=โˆ’1.2h \div 2.5 = -1.2
Multiply by 2.52.5
h=โˆ’1.2โ‹…2.5=โˆ’3h = -1.2 \cdot 2.5 = -3
โˆ’3รท2.5=โˆ’1.2-3 \div 2.5 = -1.2

Here is the takeaway. A solution is a value that makes an equation true, and you can test any value by substituting it into the original equation. To solve a one-step equation, do the inverse operation to both sides so that the variable is alone. On both sides, you may add or subtract any number, and you may multiply or divide by any number except zero. The plan is the same for positive and negative numbers, and for fractions and decimals. Finish by checking your answer in the original equation.

Next, you will model and solve equations with algebra tiles and bar diagrams.