5.01 Review: One-step equations
Checking a solution
Whenever we have an equation, we want to find the solution. And in order to find the solution, we want to find a value that makes the equation true.
When testing which values could be the solution to an equation, we often want to use substitution to check if those values make the equation true.
We can substitute possible solutions into an equation by replacing the variable in the equation with a numeric value. After substituting the possible solution, we can evaluate both sides of the equation to check if the value we substituted is a solution.
- If both sides of the equation evaluate to give the same value, we have found the solution.
- If not, we need to try a different substitution.
Worked example
Question 1
Determine if $x=9$x=9 is a solution for the equation $\frac{63}{x}=7$63x=7.
Think: We can check if the substitution is a solution by replacing the variable $x$x in the equation with the numeric value $9$9. If this equation is true, $x=9$x=9 is a solution for the equation.
Do: We can substitute in our proposed solution by replacing $x$x in the equation with $9$9. This gives us:
| $\frac{63}{x}$63x | $=$= | $7$7 |
Write the original equation. |
| $\frac{63}{9}$639 | ? | $7$7 |
Substitute $9$9 for $x$x. |
| $7$7 | $=$= | $7$7 |
The solution checks, as this is true. |
Therefore, this equation is true and the substitution $x=9$x=9 is a solution to the equation.
Reflect: After substituting the value into the equation, we compared the value of each side of the equation to see if the substitution made the equation true or not.
Solving equations with addition or subtraction
One way to solve an equation is to use inverse operations along with the properties of equality. An inverse operation is an operation that "undoes" another operation. A property of equality is an operation that produces a new equation with the same solution as the original.
Addition and subtraction are inverse operations. For example, adding two to a number is the opposite of subtracting two. As we saw in modeling balanced equations, we can also add or subtract the same amount to both sides of an equation, and it will remain true.
Adding or subtracting the same amount to both sides keeps the equations balanced.
Addition property of equality: Adding the same number to each side of an equation produces an equivalent equation.
Example:
|
If |
$x-2$x−2 | $=$= | $7$7 |
|
Then |
$x-2+2$x−2+2 | $=$= | $7+2$7+2 |
Subtraction property of equality: Subtracting the same number to each side of an equation produces an equivalent equation.
Example:
|
If |
$x+5$x+5 | $=$= | $7$7 |
|
|
Then |
$x+5-5$x+5−5 | $=$= | $7-5$7−5 |
|
Let's apply our knowledge of inverses and the addition and subtraction properties of equality to solve some equations.
Worked examples
Question 2
Solve for $x$x in the equation $x+8=15$x+8=15, showing all of your work algebraically.
Think: The number $8$8 is being added to the variable $x$x. In order to undo the operation of adding $8$8, we should subtract $8$8 from both sides of the equation.
Do:
| $x+8$x+8 | $=$= | $15$15 |
Write the original equation. |
| $x+8-8$x+8−8 | $=$= | $15-8$15−8 |
Subtract $8$8 from each side. |
| $x$x | $=$= | $7$7 |
Simplify by doing the subtraction |
Reflect: We can check our answer by substituting it back into the original equation.
| $x+8$x+8 | $=$= | $15$15 |
Write the original equation. |
| $7+8$7+8 | ? | $15$15 |
Substitute $7$7 for $x$x. |
| $15$15 | $=$= | $15$15 |
The solution checks. |
Question 3
Solve for $m$m in the equation $m-6=8$m−6=8, showing all of your work algebraically.
Think: The number $6$6 is being subtracted from the variable $m$m. In order to undo the operation of subtracting $6$6, we should add $6$6 to both sides of the equation.
Do:
| $m-6$m−6 | $=$= | $8$8 |
Write the original equation. |
| $m-6+6$m−6+6 | $=$= | $8+6$8+6 |
Add $6$6 to each side. |
| $m$m | $=$= | $14$14 |
Simplify by doing the addition. |
Reflect: We can check our answer by substituting it back into the original equation.
| $m-6$m−6 | $=$= | $8$8 |
Write the original equation. |
| $14-6$14−6 | ? | $8$8 |
Substitute $14$14 for $m$m. |
| $8$8 | $=$= | $8$8 |
The solution checks. |
Practice questions
Question 4
Solve: $21=x+13$21=x+13
Question 5
Solve: $x-8=-20$x−8=−20
Solving equations with multiplication or division
Multiplication and division are also inverse operations. For example, multiplying a number by two is the opposite of dividing it by two. As we saw in modeling balanced equations, we can also multiply or divide the same nonzero amount to both sides of an equation, and it will remain true.
Multiplying or dividing both sides by the same nonzero amount keeps the equations balanced.
Multiplication property of equality: Multiplying each side of an equation by the same nonzero number produces an equivalent equation.
Example:
|
If |
$\frac{x}{12}$x12 | $=$= | $4$4 |
|
Then |
$\frac{x}{12}\times12$x12×12 | $=$= | $4\times12$4×12 |
Division properties of equality: Dividing each side of an equation by the same nonzero number produces an equivalent equation.
Example:
|
If |
$6x$6x | $=$= | $12$12 |
|
Then |
$\frac{6x}{6}$6x6 | $=$= | $\frac{12}{6}$126 |
Let's apply our knowledge of inverses and the multiplication and division properties of equality to solve some equations.
Worked examples
Question 6
Solve for $x$x in the equation $3x=15$3x=15, showing all of your work algebraically.
Think: The number $3$3 is being multiplied by the variable $x$x. In order to undo the operation of multiplying by $3$3, we should divide each side of the equation by $3$3.
Do:
| $3x$3x | $=$= | $15$15 |
Write the original equation. |
| $\frac{3x}{3}$3x3 | $=$= | $\frac{15}{3}$153 |
Divide each side by $3$3. |
| $x$x | $=$= | $5$5 |
Simplify. |
Reflect: We can check our answer by substituting it back into the original equation.
| $3x$3x | $=$= | $15$15 |
Write the original equation. |
| $3\times5$3×5 | ? | $15$15 |
Substitute $5$5 for $x$x. |
| $15$15 | $=$= | $15$15 |
The solution checks. |
Question 7
Solve for $w$w in the equation $8=\frac{w}{2}$8=w2, showing all of your work algebraically.
Think: Notice that the variable is on the right side, but this does not change anything about how we solve it. The variable $w$w is being divided by $2$2. In order to undo the operation of dividing by $2$2, we should multiply each side of the equation by $2$2.
Do:
| $8$8 | $=$= | $\frac{w}{2}$w2 |
Write the original equation. |
| $8\times2$8×2 | $=$= | $\frac{w}{2}\times2$w2×2 |
Multiply each side by $2$2. |
| $16$16 | $=$= | $w$w |
Simplify. |
Reflect: We can check our answer by substituting it back into the original equation.
| $8$8 | $=$= | $\frac{w}{2}$w2 |
Write the original equation. |
| $8$8 | ? | $\frac{16}{2}$162 |
Substitute $16$16 for $w$w. |
| $8$8 | $=$= | $8$8 |
The solution checks. |
Practice questions
Question 8
Solve: $5x=45$5x=45
Question 9
Solve: $\frac{x}{8}=6$x8=6