8.1 Systems of Two Equations and Two Unknowns
- Page ID
- 1019
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The cost of two cell phone plans can be written as a system of equations based on the number of minutes used and the base monthly rate. As a consumer, it would be useful to know when the two plans cost the same and when is one plan cheaper.
Plan A costs $40 per month plus $0.10 for each minute of talk time.
Plan B costs $25 per month plus $0.50 for each minute of talk time.
Plan B has a lower starting cost, but since it costs more per minute, it may not be the right plan for someone who likes to spend a lot of time on the phone. When do the two plans cost the same amount?
Solving Systems of Equations with Two Unknowns
There are many ways to solve a system that you have learned in the past including substitution and graphical intersection. Here you will focus on solving using elimination because the knowledge and skills used will transfer directly into using matrices.
When solving a system, the first thing to do is to count the number of variables that are missing and the number of equations. The number of variables needs to be the same or fewer than the number of equations. Two equations and two variables can be solved, but one equation with two variables cannot.
Here's the procedure for solving a system using the elimination method:
- Step 1: Write both equations with two variables in standard form, \(A x+B y=C\). This form helps to align the variables.
- Step 2: Determine which variable you want to eliminate.
- Step 3: Scale each equation as necessary by multiplying through by constants.
- Step 4: Add the equations together. This should reduce both the number of equations and the number of variables leaving one equation and one variable.
- Step 5: Solve and substitute to determine the value of the second variable.
Here is a system of two equations and two variables in standard form: \(5 x+12 y=72\) and \(3 x-2 y=18\). Notice that there is an \(x\) column and a \(y\) column on the left hand side and a constant column on the right hand side when you rewrite the equations as shown. Also notice that if you add the system as written no variable will be eliminated.
Equation 1: \(5 x+12 y=72\)
Equation 2: \(3 x-2 y=18\)
Strategically choose to eliminate \(y\) by scaling the second equation by 6 so that the coefficient of \(y\) will match at 12 and -12.
\(\begin{aligned} 5 x+12 y &=72 \\ 18 x-12 y &=108 \end{aligned}\)
Add the two equations:
\(23 x=180\)
\(x=\frac{180}{23}\)
The value for \(x\) could be substituted into either of the original equations and the result could be solved for \(y\); however, since the value is a fraction it will be easier to repeat the elimination process in order to solve for \(x\). This time you will take the first two equations and eliminate \(x\) by making the coefficients of \(x\) to be 15 and \(-15 .\) Scale the first equation by a factor of 3 and scale the second equation by a factor of \(-5 .\)
Equation 1: \(15 x+36 y=216\)
Equation \(2:-15 x+10 y=-90\)
Adding the two equations:
\(0 x+46 y=126\)
\(y=\frac{126}{46}=\frac{63}{23}\)
The point \(\left(\frac{180}{23}, \frac{63}{23}\right)\) is where these two lines intersect.
Examples
Earlier, you were asked about two phone plans.
Plan A costs $40 per month plus $0.10 for each minute of talk time.
Plan B costs $25 per month plus $0.50 for each minute of talk time.
If you want to find out when the two plans cost the same, you can represent each plan with an equation and solve the system of equations. Let \(y\) represent cost and \(x\) represent number of minutes.
\(y=0.10 x+40\)
\(y=0.50 x+25\)
First you put these equations in standard form.
\(x-10 y=-400\)
\(x-2 y=-50\)
Then you scale the second equation by -1 and add the equations together and solve for \(y\).
\(\begin{aligned}-8 y &=-350 \\ y &=43.75 \end{aligned}\)
To solve for \(x\), you can scale the second equation by \(-5,\) add the equations together and solve for \(x\).
\(\begin{aligned}-4 x &=-150 \\ x &=37.5 \end{aligned}\)
The equivalent costs of plan \(\mathrm{A}\) and plan \(\mathrm{B}\) will occur at 37.5 minutes of talk time with a cost of \(\$ 43.75 .\)
Solve the following system of equations:
\(\begin{aligned} 6 x-7 y &=8 \\ 15 x-14 y &=21 \end{aligned}\)
Scaling the first equation by -2 will allow the \(y\) term to be eliminated when the equations are summed.
\(\begin{aligned}-12 x+14 y &=-16 \\ 15 x-14 y &=21 \end{aligned}\)
The sum is:
\(3 x=5\)
\(x=\frac{5}{3}\)
You can substitute \(x\) into the first equation to solve for \(y\).
\(\begin{aligned} 6 \cdot \frac{5}{3}-7 y &=8 \\ 10-7 y &=8 \\ -7 y &=-2 \\ y &=\frac{2}{7} \end{aligned}\)
The point \(\left(\frac{5}{3}, \frac{2}{7}\right)\) is where these two lines intersect.
Solve the following system using elimination:
\(5 x-y=22\)
\(-2 x+7 y=19\)
Start by scaling the first equation by 7 and notice that the \(y\) coefficient will immediately be eliminated when the equations are summed.
\(35 x-7 y=154\)
\(-2 x+7 y=19\)
Start by scaling the first equation by 7 and notice that the \(y\) coefficient will immediately be eliminated when the equations are summed.
\(35 x-7 y=154\)
\(-2 x+7 y=19\)
Add, solve for \(x=\frac{173}{33}\). Instead of substituting, practice eliminating \(x\) by scaling the first equation by 2 and the second equation by 5.
\(10 x-2 y=44\)
\(-10 x+35 y=95\)
Add, solve for \(y\).
Final Answer: \(\left(\frac{173}{33}, \frac{139}{33}\right)\)
Solve the following system of equations:
\(\begin{aligned} 5 \cdot \frac{1}{x}+2 \cdot \frac{1}{y} &=11 \\ \frac{1}{x}+\frac{1}{y} &=4 \end{aligned}\)
The strategy of elimination still applies. You can eliminate the \(\frac{1}{y}\) term if the second equation is scaled by a factor of -2
\(5 \cdot \frac{1}{x}+2 \cdot \frac{1}{y}=11\)
\(-2 \cdot \frac{1}{x}-2 \cdot \frac{1}{y}=-8\)
Add the equations together and solve for \(x\).
\(\begin{aligned}-3 \cdot \frac{1}{x}+0 \cdot \frac{1}{y} &=3 \\ -3 \cdot \frac{1}{x} &=3 \\ \frac{1}{x} &=-1 \\ x &=-1 \end{aligned}\)
Substitute into the second equation and solve for \(y\).
\(\begin{aligned} \frac{1}{-1}+\frac{1}{y} &=4 \\ -1+\frac{1}{y} &=4 \\ \frac{1}{y} &=5 \\ y &=\frac{1}{5} \end{aligned}\)
The point \(\left(-1, \frac{1}{5}\right)\) is the point of intersection between these two curves.
Solve the following system using elimination:
\(\begin{aligned} 11 \cdot \frac{1}{x}-5 \cdot \frac{1}{y} &=-38 \\ 9 \cdot \frac{1}{x}+2 \cdot \frac{1}{y} &=-25 \end{aligned}\)
To eliminate \(\frac{1}{y},\) scale the first equation by 2 and the second equation by \(5 .\)
To eliminate \(\frac{1}{x},\) scale the first equation by -9 and the second equation by 11
Final Answer: \(\left(-\frac{1}{3}, 1\right)\)
Solve each system of equations using the elimination method.
1. \(x+y=-4 ;-x+2 y=13\)
2. \(\frac{3}{2} x-\frac{1}{2} y=\frac{1}{2} ;-4 x+2 y=4\)
3. \(6 x+15 y=1 ; 2 x-y=19\)
4. \(x-\frac{2 y}{3}=\frac{-2}{3} ; 5 x-2 y=10\)
5. \(-9 x-24 y=-243 ; \frac{1}{2} x+y=\frac{21}{2}\)
6. \(5 x+\frac{28}{3} y=\frac{176}{3} ; y+x=10\)
7. \(2 x-3 y=50 ; 7 x+8 y=-10\)
8. \(2 x+3 y=1 ; 2 y=-3 x+14\)
9. \(2 x+\frac{3}{5} y=3 ; \frac{3}{2} x-y=-5\)
10. \(5 x=9-2 y ; 3 y=2 x-3\)
11. How do you know if a system of equations has no solution?
12. If a system of equations has no solution, what does this imply about the relationship of the curves on the graph?
13. Give an example of a system of two equations with two unknowns with an infinite number of solutions. Explain how you know the system has an infinite number of solutions.
14. Solve
\(\begin{aligned} 12 \cdot \frac{1}{x}-18 \cdot \frac{1}{y} &=4 \\ 8 \cdot \frac{1}{x}+9 \cdot \frac{1}{y} &=5 \end{aligned}\)
15. Solve
\(\begin{aligned} 14 \cdot \frac{1}{x}-5 \cdot \frac{1}{y} &=-3 \\ 7 \cdot \frac{1}{x}+2 \cdot \frac{1}{y} &=3 \end{aligned}\)