3.1: Addition, Subtraction, Multiplication, and Division Properties of Equality
The addition, subtraction, multiplication, and division properties of equality allow us to add, subtract, multiply, or divide the same value on both sides of an equation, this guarantees the equation remains true (note, we cannot divide by zero).
Concept: We know this to be a true statement: \(5=5\)
The statement will remain true if we perform the same operation on both sides of the equation.
- Add 4 on both sides of the equation to obtain:\[\begin{align*} 5+4 &=5+4 \\[4pt] 9 &=9 \end{align*}\]
- Subtract 10 on both sides of the original equation to obtain: \[\begin{align*} 5-10&=5-10 \\ -5&=-5 \end{align*}\]
- Multiply by 2 on both sides of the original equation to obtain: \[ \begin{align*} 5\cdot 2&=5\cdot 2 \\ 10&= 10 \end{align*} \]
- Divide by 15 on both sides of the original equation to obtain: \[\begin{align*} \frac{5}{15}&= \frac{5}{15}\\ \frac{1}{3}&=\frac{1}{3} \end{align*}\]
We use the Addition, Subtraction, Multiplication, and Division Properties of Equality to solve equations for a specified variable or unknown.
Process for Solving a Basic Linear Equation in One-Variable
- Isolate the variable by “undoing” the operation on the variable, that is, applying the opposite operation on both sides of the equation using the Properties of Equality
Solve for x: \(x+2=9\)
Solution
Since 2 is being added to x, to isolate the x, we need to “undo” the addition of 2, the opposite of adding 2 is subtracting 2, so using the Subtraction Property of Equality, let’s subtract 2 on both sides of the equation to obtain:
\[x+2-2=9-2\]
\[x=7\]
Solve for x: \(x-7=13\)
Solution
Since 7 is being subtracted from x, to isolate the x, we need to “undo” the subtraction of 7, the opposite of subtracting 7 is adding 7, so using the Addition Property of Equality, let’s add 7 on both sides of the equation to obtain:
\[x-7+7=13+7\]
\[x=20\]
Solve for x: \(3x=12\)
Solution
Since x is being multiplied by 3, to isolate the x, we need to “undo” the multiplication by 3, the opposite of multiplying by 3 is dividing by 3, so using the Division Property of Equality, let’s divide by 3 on both sides of the equation to obtain:
\[\frac{3x}{3}=\frac{12}{3}\]
\[x=\frac{12}{3}=4\]
Solve for x: \(\frac{x}{8}=2\)
Solution
Since x is being divided by 8, to isolate the x, we need to “undo” the division by 8. The opposite of dividing by 8 is multiplying by 8, so using the Multiplication Property of Equality, let’s multiply by 8 on both sides of the equation to obtain:
\[\frac{x}{8}\cdot 8=2\cdot 8\]
\[\frac{x}{8}\cdot \frac{8}{1}=2\cdot 8\]
\[x=16\]
Solve for x: \(\frac{1}{2}x=5\)
Solution
We can approach this problem in a couple of different ways,
Option 1: Read the problem as x is being multiplied by ½, hence we can divide both sides by ½ to isolate the variable, x.
\[\frac{\frac{1}{2}x}{\left(\frac{1}{2}\right)}=\frac{5}{\left(\frac{1}{2}\right)}\]
\[x=\frac{5}{\left(\frac{1}{2}\right)}=\frac{5}{1}\cdot \frac{2}{1}=\frac{10}{1}=10\]
Option 2: Rewrite the problem or think of the problem as being read as x is being divided by 2 since ½ \(x\) is equivalent to \(\frac{x}{2}\), so we can multiply both sides of the equation by 2 to isolate \(x\):
\[\frac{1}{2}x=5\]
\[\frac{x}{2}=5\]
\[\frac{x}{2}\cdot 2=5\cdot 2\]
\[x=10\]
Similarly, if we have a fraction times a variable, let’s say x, then we can multiply both sides of the equation by the reciprocal of the fraction (flip the fraction such that the numerator becomes the denominator, and the denominator becomes the numerator):
Solve for x: \(\frac{2}{3}x=7\)
Solution
\[\frac{2}{3}x=7\]
\[\boldsymbol{\frac{3}{2}}\cdot \frac{2}{3}x=\boldsymbol{\frac{3}{2}}\cdot 7\]
\[x=\frac{3}{2}\cdot \frac{7}{1}=\frac{21}{2}\]
Process for solving a linear equation in one variable with multiple operations
When solving a linear equation with multiple operations, we reverse the order of operations because we are “undoing” the original operations.
Example:
1. Solve for x: \(2x+5=15\)
Order of operations state to perform multiplication, then addition, so when solving, we will reverse this order so we will “undo” the addition first, then we will “undo” the multiplication
\[2x+5=15\nonumber \]
Step 1: “Undo the addition by 5” by subtracting 5 on both sides of the equation
\[2x+5-\boldsymbol{5}=15-\boldsymbol{5} \nonumber \]
\[2x=10\nonumber \]
Step 2: “Undo the multiplication by 2” by dividing by 2 on both sides of the equation
\[\frac{2x}{\boldsymbol{2}}=\frac{10}{\boldsymbol{2}}\nonumber \]
\[x=5 \nonumber \]
We can check our answer by substituting the value for x into the original equation and verifying that the equation is true: \(2\left(5\right)+5=10+5=15 \, ✓ \)
2. Solve for x: \(\frac{2}{3}x+\frac{1}{5}=\frac{2}{7}\)
\[\frac{2}{3}x+\frac{1}{5}=\frac{2}{7}\nonumber \]
Step 1: Subtract \(\frac{1}{5}\) on both sides of the equation
\[\frac{2}{3}x+\frac{1}{5}-\boldsymbol{\frac{1}{5}}=\frac{2}{7}-\boldsymbol{\frac{1}{5}}\nonumber \]
\[\frac{2}{3}x=\frac{2}{7}-\frac{1}{5} \nonumber \]
Step 2: Find an LCD to subtract the fractions on the right side:
\[\frac{2}{3}x=\frac{2}{7}\cdot \frac{5}{5}-\frac{1}{5}\cdot \frac{7}{7}\nonumber \]
\[\frac{2}{3}x=\frac{10}{35}-\frac{7}{35} \nonumber \]
\[\frac{2}{3}x=\frac{3}{35}\nonumber \]
Step 3: Multiply both sides of the equation by the reciprocal of \(\frac{2}{3}\) which would be \(\frac{3}{2}\) :
\[\boldsymbol{\frac{3}{2}}\cdot \frac{2}{3}x=\boldsymbol{\frac{3}{2}}\cdot \frac{3}{35}\nonumber \]
\[x=\frac{9}{70}\nonumber \]
Process for solving linear equations with parenthesis
When an equation contains parentheses, we can clear the parentheses using the distributive property.
Distributive Property: \(a\left(b+c\right)=ab+ac\)
Examples
1. Solve for m: \(5\left(m+3\right)-2\left(7-m\right)=12\)
Step 1: Apply the distributive property to clear the parentheses:
\[5m+5\left(3\right)-2\left(7\right)-2(-m)=12 \nonumber \]
\[5m+15-14+2m=12 \nonumber \]
Step 2: Combine like terms
\[5m+2m+15-14+1=12\nonumber \]
\[7m+1=12 \nonumber \]
Step 3: Isolate the variable by subtracting 1 on both sides, then dividing both sides by 7
\[7m+1-\boldsymbol{1}=12-\boldsymbol{1}\nonumber \]
\[7m=11\nonumber \]
\[\frac{7m}{\boldsymbol{7}}=\frac{11}{\boldsymbol{7}}\nonumber \]
\[m=\frac{11}{7}\nonumber \]
2. Solve for x: \(-7\left(3-x\right)+11=2\left(x-3\right)\)
Step 1: Clear the parentheses by using the distributive property
\[-21+7x+11=2x-6 \nonumber \]
Step 2: Combine like terms
\[-10+7x=2x-6 \nonumber \]
Step 3: Isolate the variable by subtracting \(2x\) on both sides of the equation and adding 10 on both sides of the equation
\[-10+\boldsymbol{10}+7x-\boldsymbol{2x}=2x-\boldsymbol{2x}-6+\boldsymbol{10}\nonumber \]
\[5x=4\nonumber\]
Now, divide both sides by 5:
\[\frac{5x}{\boldsymbol{5}}=\frac{4}{\boldsymbol{5}}\nonumber \]
\[x=\frac{4}{5}\nonumber \]