4.3: System of Equations - The Addition Method
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We first found that graphing isn’t a sophisticated way for generally solving systems. We then considered a second method known as substitution. The substitution method is often used for solving systems in various areas of algebra. However, substitution can get quite involved, especially if there are fractions because this only allows more room for error. Hence, we need an even more sophisticated way for solving systems in general. We call this method the addition method, also called the elimination method. We will build the concept in the following examples, then define a four-step process we can use to solve by elimination.
The Addition Method
Solve the system by addition (elimination).
\[\left\{\begin{array}{l}3x-4y=8 \\ 5x+4y=-24\end{array}\right.\nonumber\]
Solution
We solve the system by addition because we do just that- add. We want to add the two equations together to obtain an equation of one variable. Hence, we cannot just add right away; we need to make sure that when we add, we will eliminate one of the variables. Looking at the \(y\) variable terms, we can see that the coefficients of \(y\) are the same but opposite signs. We can foresee that when we add these two equations, the \(y\) variable terms will cancel:
\[\begin{array}{l} \quad 3x-4y=8 \\ \underline{+5x+4y=-24} \\ \qquad\quad 8x=-16 \end{array}\qquad \text{ Add the equations}\nonumber\]
Notice the \(y\) variable terms canceled and we are left with one equation in one variable. This is always the goal. Now, we can easily solve as usual:
\[\begin{array}{rl}8x=-16 &\text{Multiply by the reciprocal of }8 \\ x=-2&x\text{-coordinate of the solution}\end{array}\nonumber\]
Since \(x = −2\), then we can plug-n-chug \(x = −2\) into one of the equations to obtain \(y\):
\[\begin{array}{rl}3x-4y=8&\text{Plug-n-chug }x=-2 \\ 3\color{blue}{(-2)}\color{black}{}-4y=8&\text{Evaluate} \\ -6-4y=8&\text{Isolate the variable term} \\ -4y=14&\text{Multiply by the reciprocal of }-4 \\ y=-\dfrac{14}{4}&\text{Reduce the fraction} \\ y=-\dfrac{7}{2}&y\text{-coordinate of the solution}\end{array}\nonumber\]
The solution to the system is the ordered-pair \(\left(-2,-\dfrac{7}{2}\right)\). Furthermore, if we were to graph these two lines, we know they would intersect at \(\left(-2,-\dfrac{7}{2}\right)\). Also, we know this system is a consistent system with an independent solution.
The Addition Method with Multiplication
In Example 4.3.1 , one of the variable terms had the same coefficient, but opposite signs, and adding these together eliminated the variable terms completely, which allowed us to solve for the other variable. This is the idea behind the addition method. However, generally, we aren’t given variable terms that have the same coefficient with opposite signs. So, we will manipulate the equations by multiplying one or both equations by the LCM of the coefficients for one of the variables. We want to work smarter, not harder, so we should be clever in which variable we choose.
Solve the system by addition (elimination).
\[\left\{\begin{array}{l}-6x+5y=22 \\ 2x+3y=2\end{array}\right.\nonumber\]
Solution
Since none of the variable terms have the same coefficient with opposite signs, we need to choose a variable and rewrite the equations so we can cancel the variable. Recall, the goal is to obtain an equation in one variable after adding. Looking at the \(x\) variable terms, we can see that the coefficients of \(x\) have opposite signs. So let’s choose to eliminate \(x\) and we multiply the second equation by 3 to obtain the \(\text{LCM}(2, 6) = 6\):
\[\begin{array}{rl}\color{blue}{3}\color{black}{}\cdot (2x+3y)=(2)\cdot\color{blue}{3}\color{black}{}&\text{Distribute} \\ 6x+9y=6\end{array}\nonumber\]
Notice the \(x\) variable terms have the same coefficients with opposite signs. Now we can add and eliminate \(x\):
\[\begin{array}{l} -6x+5y=22 \\ \underline{+\quad 6x+9y=6}\:\: \\ \qquad\quad 14y=28 \end{array}\qquad \text{ Add the equations}\nonumber\]
Notice the \(x\) variable terms canceled and we are left with one equation in one variable. This is always the goal. Now, we can easily solve as usual:
\[\begin{array}{rl}14y=28&\text{Multiply by the reciprocal of }14 \\ y=2&y\text{-coordinate of the solution}\end{array}\nonumber\]
Since \(y = 2\), then we can plug-n-chug \(y = 2\) into one of the equations to obtain \(x\):
\[\begin{array}{rl}2x+3y=2&\text{Plug-n-chug }y=2 \\ 2x+3\color{blue}{(2)}\color{black}{}=2&\text{Evaluate} \\ 2x+6=2&\text{Isolate the variable term} \\ 2x=-4&\text{Multiply by the reciprocal of }2 \\ x=-2&x\text{-coordinate of the solution}\end{array}\nonumber\]
The solution to the system is the ordered-pair \((−2, 2)\). Furthermore, if we were to graph these two lines, we know they would intersect at \((−2, 2)\). Also, we know this system is a consistent system with an independent solution.
Multiplying Two Equations
Solve the system by addition (elimination).
\[\left\{\begin{array}{l}3x+6y=-9 \\ 2x+9y=-26\end{array}\right.\nonumber\]
Solution
Since none of the variable terms have the same coefficient with opposite signs, we need to choose a variable and rewrite the equations so we can cancel the variable. Recall, the goal is to obtain an equation in one variable after adding. Looking at the \(x\) and \(y\) variable terms, we can see that none of the coefficients are the same or with opposite signs. So we can choose any variable to eliminate. Let’s choose to eliminate \(y\) and we multiply both equations by a factor to obtain the \(\text{LCM}(9, 6) = 18\) with opposite signs:
\[\begin{aligned}\color{blue}{-3}\color{black}{}\cdot (3x+6y)&=(-9)\cdot\color{blue}{-3}\color{black}{} \\ \color{blue}{2}\color{black}{}\cdot (2x+9y)&=(-26)\cdot\color{blue}{2}\color{black}{}\end{aligned}\]
Notice the \(y\) variable terms have the same coefficients with opposite signs
\[\begin{array}{r}-9x-18y=27 \\ 4x+18y=-52\end{array}\nonumber\]
Now we can add and eliminate \(y\):
\[\begin{array}{l} \quad -9x-18y=27 \\ \underline{+\quad 4x+18y=-52}\:\: \\ \qquad\quad -5x=-25 \end{array}\qquad \text{ Add the equations}\nonumber\]
Notice the \(y\) variable terms canceled and we are left with one equation in one variable. This is always the goal. Now, we can easily solve as usual:
\[\begin{array}{rl}-5x=-25&\text{Multiply by the reciprocal of }-5 \\ x=5&x\text{-coordinate of the solution}\end{array}\nonumber\]
Since \(x = 5\), then we can plug-n-chug \(x = 5\) into one of the equations to obtain \(y\):
\[\begin{array}{rl}2x+9y=-26 &\text{Plug-n-chug }x=5 \\ 2\color{blue}{(5)}\color{black}{}+9y=-26&\text{Evaluate} \\ 10+9y=-26&\text{Isolate the variable term} \\ 9y=-36&\text{Multiply by the reciprocal of }9 \\ y=-4&y\text{-coordinate of the solution}\end{array}\nonumber\]
The solution to the system is the ordered-pair \((5, −4)\). Furthermore, if we were to graph these two lines, we know they would intersect at \((5, −4)\). Also, we know this system is a consistent system with an independent solution.
Given a system of two linear equations in two variables, we can use the following steps to solve by addition (elimination).
- Step 1. Choose a variable to eliminate. (Choose the variable with the variable terms with opposite signs, same coefficient, or both, if possible.)
- Step 2. Multiply one or both equations so that the coefficients of this variable are the LCM of the coefficients with opposite signs.
- Step 3. Add the equations together, then solve.
- Step 4. Substitute the value into one of the original equations to find the remaining variable.
It is common practice to write your answer as an ordered-pair of the form \((x, y)\) since this is the point of intersection. Be sure to verify the solution.
Solve the system by addition (elimination):
\[\left\{\begin{array}{l}2x-5y=-13 \\ 5x-3y=-4\end{array}\right.\nonumber\]
Solution
Since none of the variable terms have the same coefficient, opposite signs, or both, we need to choose a variable and rewrite the equations so we can cancel the variable. We can choose any variable to eliminate.
Step 1. Let’s choose to eliminate \(x\) and we multiply both equations by a factor to obtain the \(\text{LCM}(2, 5) = 10\) with opposite signs.
Step 2. We can mulitply the first equation by a factor of \(5\) and the second equation by a factor of \(−2\) so that we obtain variable terms with the same coefficient, \(10\), with opposite signs: \[\begin{aligned}\color{blue}{5}\color{black}{}\cdot (2x-5y)&=(-13)\cdot\color{blue}{5}\color{black}{} \\ \color{blue}{-2}\color{black}{}\cdot (5x-3y)&=(-4)\cdot\color{blue}{-2}\color{black}{}\end{aligned}\] Notice the \(x\) variable terms have the same coefficients with opposite signs: \[\begin{aligned}10x-25y&=-65 \\ -10x+6y&=8\end{aligned}\]
Step 3. Now we can add and eliminate \(x\): \[\begin{array}{l} \quad 10x-25y=-65 \\ \underline{+\quad -10x+6y=8}\:\: \\ \qquad\quad -19y=-57 \end{array}\qquad \text{ Add the equations}\nonumber\] Notice the \(x\) variable terms canceled and we are left with one equation in one variable. This is always the goal. Now, we can easily solve as usual: \[\begin{array}{rl}-19y=-57&\text{Multiply by the reciprocal of }-19 \\ y=3&y\text{-coordinate of the solution}\end{array}\nonumber\]
Step 4. Since \(y = 3\), then we can plug-n-chug \(y = 3\) into one of the equations to obtain \(x\): \[\begin{array}{rl}5x-3y=-4&\text{Plug-n-chug }y=3 \\ 5x-3\color{blue}{(3)}\color{black}{}=-4&\text{Evaluate} \\ 5x-9=-4&\text{Isolate the variable term} \\ 5x=5&\text{Multiply by the reciprocal of }5 \\ x=1&x\text{-coordinate of the solution}\end{array}\nonumber\] The solution to the system is the ordered-pair \((1, 3)\). Furthermore, if we were to graph these two lines, we know they would intersect at \((1, 3)\). Also, we know this system is a consistent system with an independent solution.
The famous mathematical text, The Nine Chapters on the Mathematical Art, which was printed around 179 AD in China, describes a process very similar to Gaussian elimination which is very similar to the addition method.
Addition: Special Cases
Just as with graphing and substitution, it is possible to have no solution or infinite solutions with elimination. Just as with substitution, if the variables eliminate, a true statement will indicate infinitely many solutions and a false statement will indicate there is no solution.
Solve the system by addition (elimination):
\[\left\{\begin{array}{l}2x-5y=3 \\ -6x+15y=-9\end{array}\right.\nonumber\]
Solution
Since none of the variable terms have the same coefficient, but both have opposite signs, we can choose any variable to eliminate.
Step 1. Let’s choose to eliminate \(y\) and we multiply both equations by a factor to obtain the \(\text{LCM}(5, 15) = 15\) with opposite signs.
Step 2. We can mulitply the first equation by a factor of \(3\) and leave the second equation alone so that we obtain variable terms with the same coefficient, \(15\), with opposite signs: \[\begin{array}{rl}\color{blue}{3}\color{black}{}\cdot (2x-5y)=(3)\cdot\color{blue}{3}\color{black}{}&\text{Distribute} \\ 6x-15y=9\end{array}\nonumber\] Notice the \(y\) variable terms have the same coefficients with opposite signs \[\begin{array}{l} 6x-5y=9 \\ -6x+15y=-9\end{array}\nonumber\]
Step 3. Now we can add and eliminate \(x\): \[\begin{array}{l} \qquad\quad 6x-15y=9 \\ \underline{+\quad -6x+15y=-9}\:\: \\ \qquad\qquad\qquad\quad 0=0 \end{array}\qquad \text{ Add the equations}\nonumber\] Since all the variables cancel and we are left with a statement without variables, we ask,“ Is this statement true?” \[\begin{array}{rl}0\stackrel{?}{=}0&\text{Is this true?} \\ 0=0&\checkmark\text{ True}\end{array}\nonumber\] Since this statement is true, then there are infinitely many solutions to this system. Furthermore, if we were to graph these two lines, we know they would be the same line. Hence, this system is a consistent system with a dependent solution.
Solve the system by addition (elimination):
\[\left\{\begin{array}{l}4x-6y=8 \\ 6x-9y=15\end{array}\right.\nonumber\]
Solution
Since none of the variable terms have the same coefficient, opposite signs, or both, we need to choose a variable and rewrite the equations so we can cancel the variable. We can choose any variable to eliminate.
Step 1. Let’s choose to eliminate \(x\) and we multiply both equations by a factor to obtain the \(\text{LCM}(4, 6) = 12\) with opposite signs.
Step 2. We can multiply the first equation by a factor of \(3\) and the second equation by \(−2\) so that we obtain variable terms with the same coefficient, \(12\), with opposite signs: \[\begin{aligned}\color{blue}{3}\color{black}{}\cdot (4x-6y)&=(8)\cdot\color{blue}{3}\color{black}{} \\ \color{blue}{-2}\color{black}{}\cdot (6x-9y)&=(15)\cdot\color{blue}{-2}\end{aligned}\] Notice the \(y\) variable terms have the same coefficients with opposite signs: \[\begin{array}{l}12x-18y=24 \\ -12x+18y=-20\end{array}\nonumber\]
Step 3. Now we can add and eliminate \(x\): \[\begin{array}{l} \qquad\quad 12x-18y=24 \\ \underline{+\quad -12x+18y=-20}\:\: \\ \qquad\qquad\qquad\quad 0=4 \end{array}\qquad \text{ Add the equations}\nonumber\] Since all the variables cancel and we are left with a statement without variables, we ask,“ Is this statement true?” \[\begin{array}{rl}0\stackrel{?}{=}4&\text{Is this true?} \\ 0\neq 4&X\text{ False} \end{array}\nonumber\] Since this statement is false, then there is no solution to this system. Furthermore, if we were to graph these two lines, we know they would be parallel. Hence, this system is an inconsistent system.
We discussed three different methods that can be used to solve a system of two equations in two variables. While all three can be used to solve any system, graphing works great for small integer solutions. Substitution works great when we have a given variable term with a coefficient of one, and addition works great for all other cases. As each method has its own strengths, it is important to be familiar with all three methods. Next, we use these methods to solve application problems.
System of Equations: The Addition Method Homework
Solve each system by addition (elimination). Determine if each system is consistent, independent or dependent, or inconsistent.
\(\begin{array}{l}4x+2y=0 \\ -4x-9y=-28 \end{array}\)
\(\begin{array}{l}-9x+5y=-22 \\ 9x-5y=13 \end{array}\)
\(\begin{array}{l}-6x+9y=3 \\ 6x-9y=-9 \end{array}\)
\(\begin{array}{l}4x-6y=-10 \\ 4x-6y=-14 \end{array}\)
\(\begin{array}{l}-x-5y=28 \\ -x+4y=-17 \end{array}\)
\(\begin{array}{l}2x-y=5 \\ 5x+2y=-28 \end{array}\)
\(\begin{array}{l}10x+6y=24 \\ -6x+y=4 \end{array}\)
\(\begin{array}{l}2x+4y=24 \\ 4x-12y=8 \end{array}\)
\(\begin{array}{l}-7x+4y=-4 \\ 10x-8y=-8 \end{array}\)
\(\begin{array}{l}5x+10y=20 \\ -6x-5y=-3 \end{array}\)
\(\begin{array}{l}-7x-3y=12 \\ -6x-5y=20 \end{array}\)
\(\begin{array}{l}9x-2y=-18 \\ 5x-7y=-10 \end{array}\)
\(\begin{array}{l}9x+6y=-21 \\ -10x-9y=28 \end{array}\)
\(\begin{array}{l}-7x+5y=-8 \\ -3x-3y=12 \end{array}\)
\(\begin{array}{l}-8x-8y=-8 \\ 10x+9y=1 \end{array}\)
\(\begin{array}{l}9y=7-x \\ -18y+4x=-26 \end{array}\)
\(\begin{array}{l}0=9x+5y \\ y=\dfrac{2}{7}x \end{array}\)
\(\begin{array}{l}-7x+y=-10 \\ -9x-y=-22 \end{array}\)
\(\begin{array}{l}-x-2y=-7 \\ x+2y=7 \end{array}\)
\(\begin{array}{l}5x-5y=-15 \\ 5x-5y=-15 \end{array}\)
\(\begin{array}{l}-3x+3y=-12 \\ -3x+9y=-24 \end{array}\)
\(\begin{array}{l}-10x-5y=0 \\ -10x-10y=-30 \end{array}\)
\(\begin{array}{l}-5x+6y=-17 \\ x-2y=5 \end{array}\)
\(\begin{array}{l}x+3y=-1 \\ 10x+6y=-10 \end{array}\)
\(\begin{array}{l}-6x+4y=12 \\ 12x+6y=18 \end{array}\)
\(\begin{array}{l}-6x+4y=4 \\ -3x-y=26 \end{array}\)
\(\begin{array}{l}-9x-5y=-19 \\ 3x-7y=-11 \end{array}\)
\(\begin{array}{l}-5x+4y=4 \\ -7x-10y=-10 \end{array}\)
\(\begin{array}{l}3x+7y=-8 \\ 4x+6y=-4 \end{array}\)
\(\begin{array}{l}-4x-5y=12 \\ -10x+6y=30 \end{array}\)
\(\begin{array}{l}8x+7y=-24 \\ 6x+3y=-18 \end{array}\)
\(\begin{array}{l}-7x+10y=13 \\ 4x+9y=22 \end{array}\)
\(\begin{array}{l}0=-9x-21+12y \\ 1+\dfrac{4}{3}y+\dfrac{7}{3}x=0 \end{array}\)
\(\begin{array}{l}-6-42y=-12x \\ x-\dfrac{1}{2}-\dfrac{7}{2}y=0 \end{array}\)