13.2: Solving Equations- Multiplication and Division
- Page ID
- 523461
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Just as you can add or subtract the same exact quantity on both sides of an equation, you can also multiply or divide 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.
\(\begin{array}{r}5\cdot 3=15\,\,\,\,\,\,\, \\ 2\cdot5\cdot3=15\cdot2 \\ 30=30\,\,\,\,\,\,\,\end{array}\)
This characteristic of equations is generalized in the multiplication property of equality.
Multiplication Property of Equality
For all real numbers a, b, and c: If a = b, then \(a\cdot{c}=b\cdot{c}\) (or ac = bc).
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 \(3x=24\). When you are done, check your solution.
Solution
Divide both sides of the equation by 3 to isolate the variable (have a coefficient of 1). Dividing by 3 is the same as having multiplied by \(\frac{1}{3}\).
\(\begin{array}{r}\underline{3x}=\underline{24}\\3\,\,\,\,\,\,\,\,\,\,\,\,3\,\\x=8\,\,\,\end{array}\)
Check by substituting your solution, 8, for the variable in the original equation.
\(\begin{array}{r}3x=24 \\ 3\cdot8=24 \\ 24=24\end{array}\)
The solution is correct!
You can also multiply the coefficient by the multiplicative inverse (reciprocal) in order to change the coefficient to 1.
Solve \(\frac{1}{2} x = 8\)
Solution
Multiply both sides by 2 to get x by itself:
\((2) \frac{1}{2} x = 8 (2)\)
On the left side the 2's cancel leaving x by itself. On the right side (8)(2) = 16
x = 16
In the next example, we will solve a one-step equation using the multiplication property of equality. You will see that the variable is part of a fraction in the given equation, and using the multiplication property of equality allows us to remove the variable from the fraction. Remember that fractions imply division, so you can think of this as the variable k is being divided by 10. To “undo” the division, you can use multiplication to isolate k. Lastly, note that there is a negative term in the equation, so it will be important to think about the sign of each term as you work through the problem. Stop after each step you take to make sure all the terms have the correct sign.
Solve \(\dfrac{7}{2}=\dfrac{k}{10}\) for k.
Solution
We want to isolate the k, which is being divided by 10. The first thing we should do is multiply both sides by 10.
\(\dfrac{10\cdot7}{2}=\dfrac{k\cdot10}{10}\)
On the right side, the 10's divide out. Solve the left side to get the value for k.
\(\dfrac{10\cdot7}{2}=k\)
\(k = 35\)
A common equation in an introductory chemistry course is \(d = \dfrac{m}{V}\). Rearrange this equation to solve for m.
- Answer
-
"Solve for m" means to get the variable m by itself. To do so we need to move V to the other side of the equation using the opposite of the current operation. In the equation above m is divided by V so we will multiply both sides by V.
\(V\cdot d = \dfrac{m\cdot V}{V}\)
On the right side the V's cancel out leaving
\(V\cdot d = m\)
If you had values for the density and volume you could plug them into the answer to solve for the mass of a substance.
Students often find it more difficult to solve for a value in the denominator. Rearrange the density equation from the previous question to solve for V.
\(d = \dfrac{m}{V}\)
Tip: You must ensure that the variable you are solving for is in the numerator, not the denominator, so start by multiplying both sides by V. Then rearrange the equation to get V by itself.
- Answer
-
Since the requested variable is in the denominator, start by multiplying both sides by V to get it into the numerator.
\(V\cdot d = \dfrac{m\cdot V}{V}\)
Now that V is in the numerator we can proceed as usual to get it by itself. V is being multiplied by d. The opposite of multiplying by d is dividing by d, so divide both sides by d in order to get V by itself.
\(\dfrac{V\cdot d}{d} = \dfrac{m}{d}\)
\(V = \dfrac{m}{d}\)
- Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution
- Graphic: Equation, term, expression. Provided by: Lumen Learning. 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