10.1.1: Solving One-Step Equations Using Properties of Equality
- Page ID
- 66918
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- Solve algebraic equations using the multiplication property of equality.
Introduction
Writing and solving equations is an important part of mathematics. Algebraic equations can help you model situations and solve problems in which quantities are unknown. The simplest type of algebraic equation is a linear equation that has just one variable.
Expressions and Equations
An equation is a mathematical statement that two expressions are equal. An equation will always contain an equal sign with an expression on each side. Expressions are made up of terms, and the number of terms in each expression in an equation may vary.
Algebraic equations contain variables, symbols that stand for an unknown quantity. Variables are often represented with letters, like \(\ x\), \(\ y\), or \(\ z\). Sometimes a variable is multiplied by a number. This number is called the coefficient of the variable. For example, the coefficient of \(\ 3x\) is 3.
Using the Addition Property of Equality
An important property of equations is one that states that you can add the same quantity to both sides of an equation and still maintain an equivalent equation. Sometimes people refer to this as keeping the equation “balanced.” If you think of an equation as being like a balance scale, the quantities on each side of the equation are equal, or balanced.
Let’s look at a simple numeric equation, \(\ 3+7=10\), to explore the idea of an equation as being balanced.
The expressions on each side of the equal sign are equal, so you can add the same value to each side and maintain the equality. Let’s see what happens when 5 is added to each side.
\(\ 3+7+5=10+5\)
Since each expression is equal to 15, you can see that adding 5 to each side of the original equation resulted in a true equation. The equation is still “balanced.”
On the other hand, let’s look at what would happen if you added 5 to only one side of the equation.
\(\ \begin{array}{c}
3+7=10 \\
3+7+5=10 \\
15 \neq 10
\end{array}\)
Adding 5 to only one side of the equation resulted in an equation that is false. The equation is no longer “balanced”, and it is no longer a true equation!
For all real numbers \(\ a\), \(\ b\), and \(\ c\): If \(\ a=b\), then \(\ a+c=b+c\).
If two expressions are equal to each other, and you add the same value to both sides of the equation, the equation will remain equal.
When you solve an equation, you find the value of the variable that makes the equation true. In order to solve the equation, you isolate the variable. Isolating the variable means rewriting an equivalent equation in which the variable is on one side of the equation and everything else is on the other side of the equation.
When the equation involves addition or subtraction, use the inverse operation to “undo” the operation in order to isolate the variable. For addition and subtraction, your goal is to change any value being added or subtracted to 0, the additive identity.
Solve \(\ x-6=8\).
Solution
| \(\ x-6=8\) | This equation means that if you begin with some unknown number, \(\ x\), and subtract 6, you will end up with 8. You are trying to figure out the value of the variable \(\ x\). |
| \(\ \begin{array}{rr} x-6&=\ \ 8 \\ \ +6 &\ +6 \\ \hline x+0&= 14 \end{array}\) |
Using the Addition Property of Equality, add 6 to both sides of the equation to isolate the variable. You choose to add 6, as 6 is being subtracted from the variable. Subtracting 6 from both sides leaves you with \(\ x+0=14\). |
\(\ x=14\)
Since subtraction can be written as addition (adding the opposite), the addition property of equality can be used for subtraction as well. So just as you can add the same value to each side of an equation without changing the meaning of the equation, you can subtract the same value from each side of an equation.
Solve \(\ x+7=42\).
Solution
| \(\ x+7=42\) | Since 7 is being added to the variable, subtract 7 to isolate the variable. |
| \(\ \begin{array}{rr} x+7= & 42 \\ \ -7 \ \ \ \ & -7 \\ \hline x+0= & 35 \end{array}\) |
To keep the equation balanced, subtract 7 from both sides of the equation. This gives you \(\ x+0=35\). |
\(\ x=35\)
Solve \(\ 12.5+x=-7.5\).
Solution
| \(\ 12.5+x=-7.5\) | Since 12.5 is being added to the variable, subtract 12.5 to isolate the variable. |
| \(\ \begin{array}{rr} 12.5+x= & -7.5 \\ -12.5 \ \ \ \ \ \ \ \ \ \ \ & -12.5 \\ \hline 0+x= & -20\ \ \ \end{array}\) |
To keep the equation balanced, subtract 12.5 from both sides of the equation. This gives you \(\ 0+x=-20\). |
\(\ x=-20\)
The examples above are sometimes called one-step equations because they require only one step to solve. In these examples, you either added or subtracted a constant from both sides of the equation to isolate the variable and solve the equation.
What would you do to isolate the variable in the equation below, using only one step?
\(\ x+10=65\)
- Add 10 to both sides of the equation.
- Subtract 10 from the left side of the equation only.
- Add 65 to both sides of the equation.
- Subtract 10 from both sides of the equation.
- Answer
-
- Incorrect. Adding 10 to both sides of the equation gives an equivalent equation, \(\ x+20=65+10\), but this step does not get the variable alone on one side of the equation. The correct answer is: Subtract 10 from both sides of the equation.
- Incorrect. Subtracting 10 from the left side will isolate the variable, but subtracting 10 from only one side of the equation does not keep the equation balanced. According to the properties of equality, you must perform the same exact operation to each side of the equation, so you must also subtract 10 from 65 to keep the equation balanced. The correct answer is: Subtract 10 from both sides of the equation.
- Incorrect. This step will not isolate the variable. It will only give an equivalent equation. \(\ x+10+65=65+65\). The correct answer is: Subtract 10 from both sides of the equation.
- Correct. Subtracting 10 from each side of the equation yields an equivalent equation with the variable isolated to give the solution: \(\ x+10-10=65-10\), so \(\ x=55\).
What would you do to isolate the variable in the equation below, using only one step? \(\ x-\frac{1}{4}=\frac{7}{2}\)
- Subtract \(\ \frac{1}{4}\) from both sides of the equation.
- Add \(\ \frac{1}{4}\) to both sides of the equation.
- Subtract \(\ \frac{7}{2}\) from both sides of the equation.
- Add \(\ \frac{7}{2}\) to both sides of the equation.
- Answer
-
- Incorrect. Subtracting \(\ \from both sides of the equation gives the equation,frac{1}{4}\) from both sides of the equation gives the equation, \(\ x-\frac{1}{4}-\frac{1}{4}=\frac{7}{2}-\frac{1}{4}\), which is the same as \(\ x-\frac{1}{4}-\frac{1}{4}=\frac{14}{4}-\frac{1}{4}\) or \(\ x-\frac{1}{2}=\frac{13}{4}\). However, this step does not get the variable alone on one side of the equation. The correct answer is: Add \(\ \frac{1}{4}\) to both sides of the equation.
- Correct. Adding \(\ \frac{1}{4}\) to each side of the equation yields an equivalent equation and isolates the variable: \(\ x-\frac{1}{4}+\frac{1}{4}=\frac{7}{2}+\frac{1}{4}\) and \(\ x=\frac{14}{4}+\frac{1}{4}\), so \(\ x=\frac{15}{4}\).
- Incorrect. Subtracting \(\ \frac{7}{2}\) from both sides will result in the equivalent expression \(\ x-\frac{1}{4}-\frac{7}{2}=\frac{7}{2}-\frac{7}{2}\), which can be rewritten \(\ x-\frac{1}{4}-\frac{14}{4}=0\), or \(\ x-\frac{15}{4}=0\), but this step does not get the variable alone on one side of the equation. The correct answer is: Add \(\ \frac{1}{4}\) to both sides of the equation.
- Incorrect. Adding \(\ \frac{7}{2}\) to both sides will result in the equivalent expression \(\ x-\frac{1}{4}+\frac{7}{2}=\frac{7}{2}+\frac{7}{2}\), which can be rewritten \(\ x-\frac{1}{4}+\frac{14}{4}=\frac{14}{2}\), or \(\ x+\frac{13}{4}=7\), but this step does not get the variable alone on one side of the equation. The correct answer is: Add \(\ \frac{1}{4}\) to both sides of the equation.
With any equation, you can check your solution by substituting the value for the variable in the original equation. In other words, you evaluate the original equation using your solution. If you get a true statement, then your solution is correct.
Solve \(\ x+10=-65\). Check your solution.
Solution
| \(\ x+10=-65\) | Since 10 is being added to the variable, subtract 10 from both sides. | |
| \(\ \begin{aligned} x+10=&-65 \\ \ -10\ \ \ \ &-10 \\ \hline x=&-75 \end{aligned}\) |
Note that subtracting 10 is the same as adding -10. You get \(\ x=-75\). | |
| Check: | \(\ \begin{aligned} x+10 &=-65 \\ -75+10 &=-65 \\ -65 &=-65 \end{aligned}\) |
To check, substitute the solution, -75 for \(\ x\) in the original equation. Simplify. This equation is true, so the solution is correct. |
\(\ x=-75\) is the solution to the equation
\(\ x+10=-65\).
It is always a good idea to check your answer whether it is requested or not.
Using the Multiplication Property of Equality
Just as you can add or subtract the same exact quantity on both sides of an equation, you can also multiply both sides of an equation by the same quantity to write an equivalent equation. Let’s look at a numeric equation, \(\ 5 \cdot 3=15\), to start. If you multiply both sides of this equation by 2, you will still have a true equation.
\(\ \begin{aligned}
5 \cdot 3 &=15 \\
5 \cdot 3 \cdot 2 &=15 \cdot 2 \\
30 &=30
\end{aligned}\)
This characteristic of equations is generalized in the multiplication property of equality.
For all real numbers \(\ a\), \(\ b\), and \(\ c\): If \(\ a=b\), then \(\ a \cdot c=b \cdot c\) (or \(\ a b=a c\)).
If two expressions are equal to each other and you multiply both sides by the same number, the resulting expressions will also be equivalent.
When the equation involves multiplication or division, you can “undo” these operations by using the inverse operation to isolate the variable. When the operation is multiplication or division, your goal is to change the coefficient to 1, the multiplicative identity.
Solve \(\ 3 x=24\). Check your solution.
Solution
| \(\ 3 x=24\) | Divide both sides of the equation by 3 to isolate the variable (have a coefficient of 1). |
| \(\ \begin{array}{l} \frac{3 x}{3}&=\frac{24}{3} \\ x&=8 \end{array}\) |
Dividing by 3 is the same as having multiplied by \(\ \frac{1}{3}\). |
| \(\ \begin{aligned} 3 x &=24 \\ 3 \cdot 8 &=24 \\ 24 &=24 \end{aligned}\) |
Check by substituting your solution, 8, for the variable in the original equation.he solution is correct! |
\(\ x=8\)
You can also multiply the coefficient by the multiplicative inverse (reciprocal) in order to change the coefficient to 1!
Solve \(\ \frac{1}{2} x=8\). Check your solution.
Solution
| \(\ \begin{array}{r} \frac{1}{2} x=8 \\ 2\left(\frac{1}{2} x\right)=2(8) \end{array}\) |
The coefficient of \(\ \frac{1}{2} x\) is \(\ \frac{1}{2}\). Since the multiplicative inverse of \(\ \frac{1}{2}\) is 2, you can multiply both sides of the equation by 2 to get a coefficient of 1 for the variable. |
| \(\ \begin{aligned} \frac{2}{2} x &=16 \\ x &=16 \end{aligned}\) |
Multiply. |
| \(\ \frac{1}{2}(16)=8\) | Check by substituting your solution into the original equation. |
| \(\ \begin{array}{c} \frac{16}{2}=8 \\ 8=8 \end{array}\) |
The solution is correct! |
\(\ x=16\)
Solve \(\ \left(-\frac{1}{4}\right) x=2\). Check your solution.
Solution
| \(\ \left(-\frac{1}{4}\right) x=2\) | The coefficient of the variable is \(\ -\frac{1}{4}\). Multiply both sides by the multiplicative inverse of \(\ -\frac{1}{4}\), which is -4. |
| \(\ (-4)\left(-\frac{1}{4}\right) x=(-4) 2\) | Multiply. |
| \(\ \begin{array}{r} \text { (1) } x=-8 \\ x=-8 \end{array}\) |
Any number multiplied by its multiplicative inverse is equal to 1, so \(\ x=-8\). |
| \(\ \begin{array}{r} \left(-\frac{1}{4}\right)(-8)=2 \\ \frac{8}{4}=2 \\ 2=2 \end{array}\) |
Check by substituting your solution into the original equation. The solution is correct. |
\(\ x=-8\)
Solve \(\ -\frac{7}{2}=\frac{x}{10}\). Check your solution.
Solution
| \(\ \begin{aligned} 10\left(-\frac{7}{2}\right) &=10\left(\frac{x}{10}\right) \\ -\frac{70}{2} &=\frac{10 x}{10} \\ -\frac{70}{2} &=x \\ -35 &=x \end{aligned}\) |
This problem contains two fractions. Multiply both sides by 10 in order to isolate the variable \(\ x\). Then simplify the fractions. |
|
| Check | \(\ \begin{aligned} -\frac{7}{2} &=\frac{x}{10} \\ -\frac{7}{2} &=\frac{-35}{10} \\ -\frac{7}{2} \cdot \frac{5}{5} &=\frac{-35}{10} \\ -\frac{35}{10} &=\frac{-35}{10} \end{aligned}\) |
Check your answer by substituting -35 in for \(\ x\). The solution is correct. |
\(\ x=-35\)
Solve for \(\ x\): \(\ 5 x=-100\)
- \(\ x=20\)
- \(\ x=-20\)
- \(\ x=500\)
- \(\ x=-500\)
- Answer
-
- Incorrect. To isolate the variable, you can divide both sides by 5, because any number divided by itself is 1. \(\ \frac{-100}{5}=-20\). The correct answer is -20.
- Correct. Dividing both sides by 5, you find that \(\ x=-20\). To check this answer, you can substitute -20 in for \(\ x\) in the original equation to get a true statement: \(\ 5(-20)=-100\).
- Incorrect. You probably multiplied both sides by -5. To isolate the variable, remember to divide, not multiply, both sides by 5 because any number divided by itself is 1. The correct answer is -20.
- Incorrect. You probably multiplied both sides by 5. To isolate the variable, remember to divide, not multiply, both sides by 5 because any number divided by itself is 1. The correct answer is -20.
Solve for \(\ y\): \(\ 4.2=7 y\)
- \(\ y=0.6\)
- \(\ y=29.4\)
- \(\ y=1.67\)
- \(\ y=-2.8\)
- Answer
-
- Correct. To isolate the variable \(\ y\) on the right side of the equation, you have to divide both sides of the equation by \(\ \text { 7. } \frac{4.2}{7}=0.6\), so \(\ y=0.6\).
- Incorrect. It looks like you multiplied \(\ 4.2 \cdot 7\) to get an answer of 29.4. Remember that if you multiply the left side by 7, you also have to multiply the right by 7; this does not isolate the variable! The correct answer is \(\ y=0.6\).
- Incorrect. It looks like you divided 7 by 4.2 to get an answer of 1.67. However, dividing both sides by 4.2 does not isolate the variable! The correct answer is \(\ y=0.6\).
- Incorrect. It looks like you subtracted 7 from 4.2 to get an answer of -2.8. Remember that the expression \(\ 7y\) means “7 times \(\ y\).” To isolate the variable, you need to use the inverse of multiplication, not subtraction. The correct answer is \(\ y=0.6\).
Summary
Equations are mathematical statements that combine two expressions of equal value. An algebraic equation can be solved by isolating the variable on one side of the equation using the properties of equality. To check the solution of an algebraic equation, substitute the value of the variable into the original equation.