13.1: Solving Equations- Addition and Subtraction
- Page ID
- 523437
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Learning Objectives
- Solve algebraic equations using the addition property of equality
- Solve algebraic equations using the multiplication property of equality
Terminology
First, let’s define some important terminology:
- variables: variables are symbols that stand for an unknown quantity, they are often represented with letters, like x, y, or z.
- coefficient: Sometimes a variable is multiplied by a number. This number is called the coefficient of the variable. For example, in 3x the coefficient is 3.
- expression: groups of terms connected by addition and subtraction. \(2x^2-5\) is an expression
- equation: 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. Think of an equal sign as meaning “the same as.” Some examples of equations are \(\frac{3}{4}r = v^{3} - r\), and \(2(6-d) + f(3 +k) = \frac{1}{4}d\)
The following figure shows how coefficients, variables, terms, and expressions all come together to make equations. In the equation \(x\), a coefficient is \(10x\), an expression is \(2x-3^2\).
Addition and Subtraction
An important property of equations is 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.
\(\begin{array}{r}3+7=10\\3+7+5=10\\15\neq 10\end{array}\)
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!
Addition Property of Equality
For all real numbers a, b, and c: If \(a+c=b+c\).
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.
Solve algebraic equations using the addition property of equality
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.
In the following simulation, you can adjust the quantity being added or subtracted to each side of an equation to see how important it is to perform the same operation on both sides of an equation when you are solving.
Solve \(x-6=8\).
Solution
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.
Using the Addition Property of Equality, add 6 to both sides of the equation to isolate the variable. You choose to add 6 because 6 is being subtracted from the variable.
\(\displaystyle \begin{array}{r}x-6\,\,\,=\,\,\,\,8\\\,\,\,\,\,\,\,\underline{+\,6\,\,\,\,\,\,\,\,+6}\\\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,=\, 14\end{array}\)
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 \(x+10=-65\). Check your solution.
Solution
\(x+10=-65\)
Since 10 is being added to the variable, subtract 10 from both sides. Note that subtracting 10 is the same as adding \(–10\).
\(\displaystyle \begin{array}{r}x+10\,\,=\,\,\,\,-65\\\,\,\,\,\,\underline{-10\,\,\,\,\,\,\,\,\,\,-10}\\\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,=\,\,\,-75\end{array}\)
To check, substitute the solution, \(–75\) for x in the original equation, then simplify.
\(\displaystyle \begin{array}{r}\,\,\,\,\,x+10\,\,\,=-65\\-75+\,10\,\,\,=-65\\\,\,\,\,\,\,\,\,\,\,\,-65\,\,\,=-65\end{array}\)
This equation is true, so the solution is correct.
Solve \(x+5=27\).
- Answer
-
This equation means that if you begin with some unknown number, x, and add 5, you will end up with 27. You are trying to figure out the value of the variable x.
Using the Addition Property of Equality, subtract 5 from both sides of the equation to isolate the variable. You choose to subtract 5, as 5 is being added from the variable.
\(\displaystyle \begin{array}{r}x+5\,\,=\,\,27\\\,\,\,\,\,\,\,\underline{-5\,\,\,\,\,\,\,\,-5}\\\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,=\, 22\end{array}\)
Solve \(x-4=-32\). Check your solution.
- Answer
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\(x-4=-32\)
Since 4 is being subtracted from the variable, add 4 to both sides.
\(\displaystyle \begin{array}{r}x-4\,\,=\,\,\,\,-32\\\,\,\,\,\,\underline{+4\,\,\,\,\,\,\,\,\,\,+4}\\\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,=\,\,\,-28\end{array}\)
Check:
To check, substitute the solution, \(–28\) for x in the original equation, then simplify.
\(\displaystyle \begin{array}{r}\,\,\,\,\,x-4\,\,\,=-32\\-28-\,4\,\,\,=-32\\\,\,\,\,\,\,\,\,\,\,\,-32\,\,\,=-32\end{array}\)
This equation is true, so the solution is correct. It is always a good idea to check your answer whether you are requested to or not.
Solve \({12.5}+{ t }= {-7.5}\).
- Answer
-
To solve this equation you need to remember how to add or subtract decimal numbers. You also need to remember that when you subtract a number from a negative number, your result will be negative.
Using the Addition Property of Equality, subtract 12.5 from both sides of the equation to isolate the variable, t. You choose to subtract 12.5 because 12.5 is being added to the variable, t.
\(\displaystyle \begin{array}{r}{12.5}+{t}\,\,\,=\,\,\,\,{-7.5}\\\,\,\,\,\,\,\,\underline{-12.5\,\,\,\,\,\,\,\,-12.5}\\\,\,\,\,\,\,\,\,\,\,\,\,\,t\,\,=\, -20\end{array}\)
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.
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.
Writing and solving algebraic equations is an important part of mathematics. Equations can be used to describe economic, cultural, physical, and biological processes. They help business people make decisions, and help doctors and scientists find ways to heal and help people. Without mathematical equations we would not have the physical infrastructure that rely on every day for transportation and clean water.
Equations can help you model situations and solve problems in which quantities are unknown (like how long Joan should wait before she drives home). The simplest type of algebraic equation is a linear equation that has just one variable.
When you follow the steps to solve an equation, you try to isolate the variable. The variable is a quantity we don’t know yet. You have a solution when you get the equation x = some value.
In chemistry you may use the equation \(T_K = T_{°C} + 273.15\). If \(T_K = 293.65\), what is the value of \(T_{°C}\)?
- Answer
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\(20.5 °C\)
Another equation you may use in chemistry is 14 = pH + pOH. Rearrange this equation to solve for pOH.
- Answer
-
When asked to solve for pOH, that means to get it by itself. You need to move the pH away from pOH by doing the opposite of the current operation. In the given equation pH is being added to pOH. The opposite of addition is subtraction. Remember to do the same thing to both sides of the equation; subtract pH from both sides.
\(14 - pH = pH + pOH - pH\)
The pH's on the right cancel leaving
\(14 - pH = pOH\)
If you were given a value for pH you could plug it into the answer above to solve for pOH.
- Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution
- Graphic: Equation, term, expression. Provided by: Lumen Learning. License: CC BY: Attribution
- Solve One-Step Equations Using Addition and Subtraction (Whole Numbers). Authored by: James Sousa (Mathispower4u.com). Located at: https://youtu.be/VsWrFKFerSY. License: CC BY: Attribution
- Solving One Step Equations Using Addition and Subtraction (Integers). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: https://youtu.be/D3T8eCT5U_w. License: CC BY: Attribution
- Solving One Step Equations Using Addition and Subtraction (Decimals). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: https://youtu.be/D8wKGlxf6bM. License: CC BY: Attribution
- Solving One Step Equations Using Addition and Subtraction (Fractions). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: https://youtu.be/O7SPM7Cs8Ds. License: CC BY: Attribution
- Unit 10: Solving Equations and Inequalities, First Edition Developmental Math: An Open Program . Provided by: Monterey Institute of Technology and Education. Located at: nrocnetwork.org/resources/downloads/nroc-math-open-textbook-units-1-12-pdf-and-word-formats/%20. License: CC BY: Attribution