Solving Linear Equations
- Page ID
- 29779
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Solving Linear Equations in One Variable
Linear equations in one variable involve two expressions separated by an \(=\) sign where at least one side of the equation contains the variable \(x\).
Example 1: \(0.4(60-c)=10\)
Example 2: \(\frac{c-16}{18}=0.5\)
Example 3: \(7=\frac{x-6}{\frac{9}{\sqrt{36}}}\)
Example 4: \(0.7(x-1)=28\) Practice on your own. (Solution: \(x=41\))
To solve linear equations perform the following steps. (Note: sometimes, some of the steps may not be necessary.)
Solution to Example 1: \(0.4(60-c)=10\)
Step 1: Eliminate parentheses by multiplying the number outside the parentheses by each number inside the parentheses.
\(0.4(60-c)=10\)
\(0.4\times 60-0.4c = 10\)
\(24-4c = 10\)
Step 2: Eliminate any denominators (if there are any) in the expression containing the variable by multiplying each side of the equation by the denominator.
Note: This is not necessary in this example because \(24-0.4c\) does not contain a denominator.
Step 3: Isolate the variable on one side of the \(=\) sign and the number on the opposite side by adding the opposite to each side of the equation.
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Subtract 24 from each side to get \(-0.4c\) alone on the left side |
\(24-0.4c = 10\) |
Step 4: Isolate the variable by dividing each side of the equation by the number in front of the variable.
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Divide each side by \(-0.4\) |
\(\frac{-0.4c}{-0.4} = \frac{-14}{-0.4}\) |
\(c=35\) answer |
Solution to Example 2: Solve: \(\frac{c-16}{18}=0.5\)
Step 1: Not necessary (there are no parentheses)
Step 2: Multiply each side by 18 to eliminate the denominator:
\(\frac{18}{1}\cdot \frac{c-16}{18} = 0.5 \cdot 18, c - 18 = 9\)
Step 3: Isolate 
by adding 16 to each side
\(c - 16 = 9\)
\(+16\) \(+16\)
___________________
\(c = 25\) answer
Step 4: Not necessary because no number is in front of \(c\).
Solution to Example 3: Solve: \(7=\frac{x-6}{\frac{9}{\sqrt{36}}}\)
Step 1: Not necessary since there are no parentheses.
Step 2: Multiply each side by \(\frac{9}{\sqrt{36}}\) to eliminate the fraction.
\(7\cdot \frac{9}{\sqrt{36}}\cdot \frac{x-6}{\frac{9}{\sqrt{36}}}\)
\(7\cdot \frac{9}{6} = x - 6\)
\(7\cdot 1.5 = x - 6\)
\(10.5 = x - 6\)
Step 3: Add 6 to each side to isolate \(x\).
\(10.5 = x-6\)
\(+6\) = \(+6\)
\(16.5 = x\) answer
Step 4: Not necessary – no number is in front of \(x\).