ORCCA Elimination

Section 4.3 Elimination

Objectives: PCC Course Content and Outcome Guide

permalinkWe learned how to solve a system of linear equations using substitution in Section 4.2. In this section, we will learn a second symbolic method for solving systems of linear equations.

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Figure 4.3.1. Alternative Video Lesson

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Subsection 4.3.1 Solving Systems of Equations by Elimination

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Example 4.3.2.

Alicia has $\$1000$ to give to her two grandchildren for New Year's. She would like to give the older grandchild $\$120$ more than the younger grandchild, because that is the cost of the older grandchild's college textbooks this term. How much money should she give to each grandchild?

To answer this question, we will demonstrate a new technique. You may have a very good way for finding how much money Alicia should give to each grandchild, but right now we will try to see this new method.

Let $A$ be the dollar amount she gives to her older grandchild, and $B$ be the dollar amount she gives to her younger grandchild. (As always, we start solving a word problem like this by defining the variables, including their units.) Since the total she has to give is $\$1000\text{,}$ we can say that $A+B=1000\text{.}$ And since she wants to give $\$120$ more to the older grandchild, we can say that $A-B=120\text{.}$ So we have the system of equations:

\begin{aligned} {\begin{aligned} A + B & = 1000 \\ A - B & = 120 \end{aligned} \end{aligned}

$\begin{align*} \left\{ \begin{aligned} A+B \amp = 1000 \\ A-B \amp = 120 \end{aligned} \right. \end{align*}$

We could solve this system by substitution as we learned in Section 4.2, but there is an easier method. If we add together the left sides from the two equations, it should equal the sum of the right sides:

\begin{aligned} \begin{aligned} A + B \\ + A - B \end{aligned} & = \begin{aligned} 1000 \\ + 120 \end{aligned} \end{aligned}

$\begin{align*} \begin{aligned}A+B\\{}+A-B\end{aligned}\amp=\begin{aligned}1000\\{}+120\end{aligned}\\ \end{align*}$

So we have:

\begin{aligned} 2 A & = 1120 \end{aligned}

$\begin{align*} 2A\amp=1120 \end{align*}$

Note that the variable $B$ is eliminated. This happened because the “ ${}+B$ ” and the “ ${}-B$ ” perfectly cancel each other out when they are added. With only one variable left, it doesn't take much to finish:

\begin{aligned} 2 A & = 1120 \\ A & = 560 \end{aligned}

$\begin{align*} 2A\amp=1120\\ A\amp=560 \end{align*}$

To finish solving this system of equations, we need the value of $B\text{.}$ For now, an easy way to find $B$ is to substitute in our value of $A$ into one of the original equations:

\begin{aligned} A + B & = 1000 \\ 560 + B & = 1000 \\ B & = 440 \end{aligned}

$\begin{align*} A+B\amp=1000\\ \substitute{560}+B\amp=1000\\ B\amp=440 \end{align*}$

To check our work, substitute $A=560$ and $B=440$ into the original equations:

\begin{aligned} A + B & = 1000 & A - B & = 120 \\ 560 + 440 & \overset{?}{=} 1000 & 560 - 440 & \overset{?}{=} 120 \\ 1000 & \overset{✓}{=} 1000 & 120 & \overset{✓}{=} 120 \end{aligned}

$\begin{align*} A+B\amp=1000\amp A-B\amp=120\\ \substitute{560}+\substitute{440}\amp\stackrel{?}{=}1000\amp \substitute{560}-\substitute{440}\amp\stackrel{?}{=}120\\ 1000\amp\stackrel{\checkmark}{=}1000\amp 120\amp\stackrel{\checkmark}{=}120 \end{align*}$

This confirms that our solution is correct. In summary, Alicia should give $\$560$ to her older grandchild, and $\$440$ to her younger grandchild.

This method for solving the system of equations in Example 4.3.2 worked because $B$ and $-B$ add to zero. Once the $B$ -terms were eliminated we were able to solve for $A\text{.}$ This method is called the elimination method. Some people call it the addition method, because we added the corresponding sides from the two equations to eliminate a variable.

permalinkIf neither variable can be immediately eliminated, we can still use this method but it will require that we first adjust one or both of the equations. Let's look at an example where we need to adjust one of the equations.

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Example 4.3.3. Scaling One Equation.

Solve the system of equations using the elimination method.

\begin{aligned} {\begin{aligned} 3 x & - & 4 y & = & 2 \\ 5 x & + & 8 y & = & 18 \end{aligned} \end{aligned}

$\begin{align*} \left\{ \begin{alignedat}{4} 3x\amp {}-{} \amp 4y \amp {}={} \amp 2 \\ 5x\amp {}+{} \amp 8y \amp {}={} \amp 18 \end{alignedat} \right. \end{align*}$

Explanation

To start, we want to see whether it will be easier to eliminate $x$ or $y\text{.}$ We see that the coefficients of $x$ in each equation are $3$ and $5\text{,}$ and the coefficients of $y$ are $-4$ and $8\text{.}$ Because $8$ is a multiple of $4$ and the coefficients already have opposite signs, the $y$ variable will be easier to eliminate.

To eliminate the $y$ terms, we will multiply each side of the first equation by $2$ so that we will have $-8y\text{.}$ We can call this process scaling the first equation by $2\text{.}$

\begin{align*} \amp\left\{ \begin{alignedat}{4} \multiplyleft{2}(3x\amp {}-{} \amp 4y) \amp {}={} \amp \multiplyleft{2}(2) \\ 5x\amp {}+{} \amp 8y \amp {}={} \amp 18 \end{alignedat} \right.\\ \amp\left\{ \begin{alignedat}{4} 6x\amp {}-{} \amp 8y \amp {}={} \amp 4 \\ 5x\amp {}+{} \amp 8y \amp {}={} \amp 18 \end{alignedat} \right. \end{align*}

We now have an equivalent system of equations where the $y$-terms can be eliminated:

\begin{align*} \begin{aligned}6x-8y\\{}+5x+8y\end{aligned}\amp=\begin{aligned}4\\{}+18\end{aligned}\\ \end{align*}

So we have:

\begin{align*} 11x\amp=22\\ x\amp=2 \end{align*}

To solve for $y\text{,}$ we can substitute $2$ for $x$ into either of the original equations or the new one. We use the first original equation, $3x-4y=2\text{:}$

\begin{align*} 3x-4y\amp=2\\ 3(\substitute{2})-4y\amp=2\\ 6-4y\amp=2\\ -4y\amp=-4\\ y\amp=1 \end{align*}

Our solution is $x=2$ and $y=1\text{.}$ We will check this in both of the original equations:

\begin{align*} 5x+8y\amp=18\amp 3x-4y\amp=2\\ 5(\substitute{2})+8(\substitute{1})\amp\stackrel{?}{=}18\amp 3(\substitute{2})-4(\substitute{1})\amp\stackrel{?}{=}2\\ 10+8\amp\stackrel{\checkmark}{=}18\amp 6-4\amp\stackrel{\checkmark}{=}2 \end{align*}

The solution to this system is $(2,1)$ and the solution set is $\{(2,1)\}\text{.}$

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Checkpoint 4.3.4.

permalinkHere's an example where we have to scale both equations.

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Example 4.3.5. Scaling Both Equations.

Solve the system of equations using the elimination method.

\begin{aligned} {\begin{aligned} 2 x & + & 3 y & = & 10 \\ - 3 x & + & 5 y & = & - 15 \end{aligned} \end{aligned}

$\begin{align*} \left\{ \begin{alignedat}{4} 2x \amp {}+{} \amp 3y \amp {}={} \amp 10 \\ -3x \amp {}+{} \amp 5y \amp {}={} \amp {-15} \end{alignedat} \right. \end{align*}$

Explanation

Considering the coefficients of $x$ ($2$ and $-3$) and the coefficients of $y$ ($3$ and $5$) we see that we cannot eliminate the $x$ or the $y$ variable by scaling a single equation. We will need to scale both.

The $x$-terms already have opposite signs, so we choose to eliminate $x\text{.}$ The least common multiple of $2$ and $3$ is $6\text{.}$ We can scale the first equation by $3$ and the second equation by $2$ so that the equations have terms $6x$ and $-6x\text{,}$ which will cancel when added.

\begin{align*} \amp\left\{ \begin{alignedat}{4} \multiplyleft{3}(2x\amp {}+{} \amp 3y) \amp {}={}\amp \multiplyleft{3}(10) \\ \multiplyleft{2}(-3x\amp {}+{} \amp 5y) \amp {}={}\amp \multiplyleft{2}(-15) \end{alignedat} \right.\\ \amp\left\{ \begin{alignedat}{4} 6x\amp {}+{} \amp 9y \amp {}={} \amp 30 \\ -6x\amp {}+{} \amp 10y \amp {}={} \amp {-30} \end{alignedat} \right. \end{align*}

At this point we can add the corresponding sides from the two equations and solve for $y\text{:}$

\begin{align*} \begin{aligned}6x+9y\\{}-6x+10y\end{aligned}\amp=\begin{aligned}30\\{}-30\end{aligned}\\ \end{align*}

So we have:

\begin{align*} 19y\amp=0\\ y\amp=0 \end{align*}

To solve for $x\text{,}$ we'll replace $y$ with $0$ in $2x+3y=10\text{:}$

\begin{align*} 2x+3y\amp=10\\ 2x+3(\substitute{0})\amp=10\\ 2x\amp=10\\ x\amp=5 \end{align*}

We'll check the system using $x=5$ and $y=0$ in each of the original equations:

\begin{align*} 2x+3y\amp=10 \amp -3x+5y\amp=-15\\ 2(\substitute{5})+3(\substitute{0})\amp\stackrel{?}{=}10 \amp -3(\substitute{5})+5(\substitute{0})\amp\stackrel{?}{=}-15\\ 10+0\amp\stackrel{\checkmark}{=}10 \amp -15+0\amp\stackrel{\checkmark}{=}-15 \end{align*}

So the system's solution is $(5,0)$ and the solution set is $\{(5,0)\}\text{.}$

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Checkpoint 4.3.6.

Try a similar exercise.

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Example 4.3.7. Meal Planning.

Javed is on a meal plan and needs to consume $600$ calories and $20$ grams of fat for breakfast. A small avocado contains $300$ calories and $30$ grams of fat. He has bagels that contain $400$ calories and $8$ grams of fat. Write and solve a system of equations to determine how much bagel and avocado would combine to make his target calories and fat.

Explanation

To write this system of equations, we first need to define our variables. Let $A$ be the number of avocados consumed and let $B$ be the number of bagels consumed. Both $A$ and $B$ might be fractions. For our first equation, we will count calories from the avocados and bagels:

\begin{equation*} \left(300\,\tfrac{\text{calories}}{\text{avocado}}\right)(A\,\text{avocados})+\left(400\,\tfrac{\text{calories}}{\text{bagel}}\right)(B\,\text{bagel})=600\,\text{calories} \end{equation*}

Or, without the units:

\begin{equation*} 300A+400B=600 \end{equation*}

Similarly, for our second equation, we will count the grams of fat:

\begin{equation*} \left(30\,\tfrac{\text{g fat}}{\text{avocado}}\right)(A\,\text{avocados})+\left(8\,\tfrac{\text{g fat}}{\text{bagel}}\right)(B\,\text{bagel})=20\,\text{g fat} \end{equation*}

Or, without the units:

\begin{equation*} 30A+8B=20 \end{equation*}

So the system of equations is:

\begin{align*} \left\{ \begin{alignedat}{4} 300A \amp {}+{} \amp 400B \amp {}={} \amp 600 \\ 30A \amp {}+{} \amp 8B \amp {}={} \amp 20 \end{alignedat} \right. \end{align*}

Since none of the coefficients are equal to $1\text{,}$ it will be easier to use the elimination method to solve this system. Looking at the terms $300A$ and $30A\text{,}$ we can eliminate the $A$ variable if we multiply the second equation by $-10$ to get $-300A\text{:}$

\begin{align*} \amp\left\{ \begin{alignedat}{4} 300A \amp {}+{} \amp 400B \amp {}={} 600 \\ \multiplyleft{-10}(30A \amp {}+{} \amp 8B) \amp {}={} \multiplyleft{-10}(20) \end{alignedat} \right.\\ \amp\left\{ \begin{alignedat}{4} 300A \amp {}+{} \amp 400B \amp {}={}\amp 600 \\ -300A \amp {}+{} \amp (-80B) \amp {}={}\amp {-200} \end{alignedat} \right. \end{align*}

When we add the corresponding sides from the two equations together we have:

\begin{align*} \begin{aligned}300A+400B\\{}-300A-80B\end{aligned}\amp=\begin{aligned}600\\{}-200\end{aligned}\\ \end{align*}

So we have:

\begin{align*} 320B\amp=400\\ \divideunder{320B}{320}\amp=\divideunder{400}{320}\\ B\amp=\frac{5}{4} \end{align*}

We now know that Javed should eat $\frac{5}{4}$ bagels (or one and one-quarter bagels). To determine the number of avocados, we will substitute $B$ with $\frac{5}{4}$ in either of our original equations.

\begin{align*} 300A+400B\amp=600\\ 300A+400\left(\substitute{\frac{5}{4}}\right)\amp=600\\ 300A+500\amp=600\\ 300A\amp=100\\ \divideunder{300A}{300}\amp=\divideunder{100}{300}\\ A\amp=\frac{1}{3} \end{align*}

To check this result, try using $B=\frac{5}{4}$ and $A=\frac{1}{3}$ in each of the original equations:

\begin{align*} 300A+400B\amp=600\amp 30A+8B\amp=20\\ 300\left(\substitute{\frac{1}{3}}\right)+400\left(\substitute{\frac{5}{4}}\right)\amp\stackrel{?}{=}600\amp30\left(\substitute{\frac{1}{3}}\right)+8\left(\substitute{\frac{5}{4}}\right)\amp\stackrel{?}{=}20\\ 100+500\amp\stackrel{\checkmark}{=}600\amp10+10\amp\stackrel{\checkmark}{=}20 \end{align*}

In summary, Javed can eat $\frac{5}{4}$ of a bagel (so one and one-quarter bagel) and $\frac{1}{3}$ of an avocado in order to consume exactly $600$ calories and $20$ grams of fat.

permalinkFor summary reference, here is the general procedure.

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Process 4.3.8. Solving Systems of Equations by Elimination.

To solve a system of equations by elimination,

Algebraically manipulate both equations into standard form, unless the variables are already aligned (e.g. if both equations are in slope intercept form already).
Scale one or both of the equations to force one of the variables to have equal but opposite sign in the two equations.
Add the corresponding sides of the two equations together which should have the effect that one variable cancels out entirely.
Solve the resulting equation for the one remaining variable.
Substitute that value into either of the original equations to find the other variable.
Verify your solution.

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Subsection 4.3.2 Solving Special Systems of Equations with Elimination

permalinkRemember the two special cases we encountered when solving by graphing and substitution? Sometimes a system of equations has no solutions at all, and sometimes the solution set is infinite with all of the points on one line satisfying the equations. Let's see what happens when we use the elimination method on each of the special cases.

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Example 4.3.9. A System with Infinitely Many Solutions.

Solve the system of equations using the elimination method.

\begin{aligned} {\begin{aligned} 3 x & + & 4 y & = & 5 \\ 6 x & + & 8 y & = & 10 \end{aligned} \end{aligned}

$\begin{align*} \left\{ \begin{alignedat}{4} 3x \amp {}+{} \amp 4y \amp {}={} \amp 5 \\ 6x \amp {}+{} \amp 8y \amp {}={} \amp 10 \end{alignedat} \right. \end{align*}$

Explanation

To eliminate the $x$-terms, we multiply each term in the first equation by $-2\text{,}$ and we have:

\begin{align*} \amp\left\{ \begin{alignedat}{4} \multiplyleft{-2}(3x \amp {}+{} \amp 4y) \amp {}={} \amp \multiplyleft{-2}5 \\ 6x \amp {}+{} \amp 8y \amp {}={} \amp 10 \end{alignedat} \right.\\ \amp\left\{ \begin{alignedat}{4} -6x \amp {}+{} \amp -8y \amp {}={}\amp -10 \\ 6x \amp {}+{} \amp 8y \amp {}={} \amp 10 \end{alignedat} \right. \end{align*}

We might notice that the equations look very similar. Adding the respective sides of the equation, we have:

\begin{equation*} 0=0 \end{equation*}

Both of the variables have been eliminated. Since the statement $0=0$ is true no matter what $x$ and $y$ are, the solution set is infinite. Specifically, you just need any $(x,y)$ satisfying one of the two equations, since the two equations represent the same line. We can write the solution set as $\{(x,y)\mid 3x+4y=5\}\text{.}$

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Example 4.3.10. A System with No Solution.

Solve the system of equations using the elimination method.

\begin{aligned} {\begin{aligned} 10 x & + & 6 y & = & 9 \\ 25 x & + & 15 y & = & 4 \end{aligned} \end{aligned}

$\begin{align*} \left\{ \begin{alignedat}{4} 10x \amp {}+{} \amp 6y \amp {}={} \amp 9 \\ 25x \amp {}+{} \amp 15y \amp {}={} \amp 4 \end{alignedat} \right. \end{align*}$

Explanation

To eliminate the $x$-terms, we will scale the first equation by $-5$ and the second by $2\text{:}$

\begin{align*} \amp\left\{ \begin{alignedat}{4} \multiplyleft{-5}(10x\amp {}+{} \amp 6y) \amp {}={}\amp \multiplyleft{-5}(9) \\ \multiplyleft{2}(25x\amp {}+{} \amp 15y) \amp {}={}\amp \multiplyleft{2}(4) \end{alignedat} \right.\\ \amp\left\{ \begin{alignedat}{4} -50x\amp {}+{} \amp (-30y) \amp {}={} \amp {-45} \\ 50x\amp {}+{} \amp 30y \amp {}={} \amp 8 \end{alignedat} \right. \end{align*}

Adding the respective sides of the equation, we have:

\begin{equation*} 0=-37 \end{equation*}

Both of the variables have been eliminated. In this case, the statement $0=-37$ is just false, no matter what $x$ and $y$ are. So the system has no solution.

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Subsection 4.3.3 Deciding to Use Substitution versus Elimination

In every example so far from this section, both equations were in standard form, $Ax+By=C\text{.}$ And all of the coefficients were integers . If none of the coefficients are equal to $1$ then it is usually easier to use the elimination method, because otherwise you will probably have some fraction arithmetic to do in the middle of the substitution method. If there is a coefficient of $1\text{,}$ then it is a matter of preference.

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Example 4.3.11.

A college used to have a north campus with $6000$ students and a south campus with $15{,}000$ students. The percentage of students at the north campus who self-identify as LGBTQ was three times the percentage at the south campus. After the merge, $5.5\%$ of students identify as LGBTQ. What percentage of students on each campus identified as LGBTQ before the merge?

Explanation

We will define $N$ as the percentage (as a decimal) of students at the north campus and $S$ as the percentage (as a decimal) of students at the south campus that identified as LGBTQ. Since the percentage of students at the north campus was three times the percentage at the south campus, we have:

\begin{equation*} N=3S \end{equation*}

For our second equation, we will count LGBTQ students at the various campuses. At the north campus, multiply the population, $6000\text{,}$ by the percentage $N$ to get $6000N\text{.}$ This must be the actual number of LGBTQ students. Similarly, the south campus has $15000S$ LGBTQ students, and the combined school has $21000(0.055)=1155\text{.}$ When we combine the two campuses, we have:

\begin{equation*} 6000N+15000S=1155 \end{equation*}

We write the system as:

\begin{equation*} \left\{ \begin{aligned} N\amp=3S \\ 6000N+15000S\amp=1155 \end{aligned} \right. \end{equation*}

Because the first equation is already solved for $N\text{,}$ this is a good time to not use the elimination method. Instead we can substitute $N$ in our second equation with $3S$ and solve for $S\text{:}$

\begin{align*} 6000N+15000S\amp=1155\\ 6000(\substitute{3S})+15000S\amp=1155\\ 18000S+15000S\amp=1155\\ 33000S\amp=1155\\ \divideunder{33000S}{33000}\amp=\divideunder{1155}{33000}\\ S\amp=0.035 \end{align*}

We can determine $N$ using the first equation:

\begin{align*} N\amp=3S\\ N\amp=3(0.035)\\ N\amp=0.105 \end{align*}

Before the merge, $10.5\%$ of the north campus students self-identified as LGBTQ, and $3.5\%$ of the south campus students self-identified as LGBTQ.

permalinkIf you need to solve a system, and one of the equations is not in standard form, substitution may be easier. But you also may find it easier to convert the equations into standard form. Additionally, if the system's coefficients are fractions or decimals, you may take an additional step to scale the equations so that they only have integer coefficients.

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Example 4.3.12.

Solve the system of equations using the method of your choice.

\begin{aligned} {\begin{aligned} - \frac{1}{3} y & = \frac{1}{15} x + \frac{1}{5} \\ \frac{5}{2} x - y & = 6 \end{aligned} \end{aligned}

$\begin{align*} \left\{ \begin{aligned} -\frac{1}{3}y\amp = \frac{1}{15}x + \frac{1}{5} \\ \frac{5}{2}x-y\amp = 6 \end{aligned} \right. \end{align*}$

Explanation

First, we can cancel the fractions by using the least common multiple of the denominators in each equation, similarly to the topic of Section 2.3. We have:

\begin{align*} \amp\left\{ \begin{aligned} \multiplyleft{15}{-\frac{1}{3}y} \amp = \multiplyleft{15}\left(\frac{1}{15}x + \frac{1}{5}\right) \\ \multiplyleft{2}\left(\frac{5}{2}x-y\right) \amp = \multiplyleft{2}(6) \end{aligned} \right.\\ \amp\left\{ \begin{aligned} -5y \amp = x+3 \\ 5x-2y \amp = 12 \end{aligned} \right. \end{align*}

We could put convert the first equation into standard form by subtracting $x$ from both sides, and then use elimination. However, the $x$-variable in the first equation has a coefficient of $1\text{,}$ so the substitution method may be faster. Solving for $x$ in the first equation we have:

\begin{align*} -5y\amp=x+3\\ -5y\subtractright{3}\amp=x+3\subtractright{3}\\ -5y-3\amp=x\\ \end{align*}

Substituting $-5y-3$ for $x$ in the second equation we have:

\begin{align*} 5(\substitute{-5y-3})-2y\amp=12\\ -25y-15-2y\amp=12\\ -27y-15\amp=12\\ -27y\amp=27\\ y\amp=-1 \end{align*}

Using the equation where we isolated $x$ and substituting $-1$ for $y\text{,}$ we have:

\begin{align*} -5(\substitute{-1})-3\amp=x\\ 5-3\amp=x\\ 2\amp=x \end{align*}

The solution is $(2,-1)\text{.}$ Checking the solution is left as an exercise.

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Example 4.3.13.

A penny is made by combining copper and zinc. A chemistry reference source says copper has a density of 9 ^g⁄_cm³ and zinc has a density of 7.1 ^g⁄_cm³. A penny's mass is 2.5 g and its volume is 0.35 cm³. How many cm³ each of copper and zinc go into one penny?

Explanation

Let $c$ be the volume of copper and $z$ be the volume of zinc in one penny, both measured in cm³. Since the total volume is 0.35 cm³, one equation is:

\begin{equation*} \left(c\,\text{cm}^3\right)+\left(z\,\text{cm}^3\right)=0.35\,\text{cm}^3 \end{equation*}

Or without units:

\begin{equation*} c+z=0.35\text{.} \end{equation*}

For the second equation, we will examine the masses of copper and zinc. Since copper has a density of 9 ^g⁄_cm³ and we are using $c$ to represent the volume of copper, the mass of copper is $9c\text{.}$ Similarly, the mass of zinc is $7.1\text{.}$ Since the total mass is 2.5 g, we have the equation:

\begin{equation*} \left(9\,\tfrac{\text{g}}{\text{cm}^3}\right)\left(c\,\text{cm}^3\right)+\left(7.1\,\tfrac{\text{g}}{\text{cm}^3}\right)\left(z\,\text{cm}^3\right)=2.5\,\text{g} \end{equation*}

Or without units:

\begin{equation*} 9c+7.1z=2.5\text{.} \end{equation*}

So we have a system of equations:

\begin{align*} \left\{ \begin{alignedat}{4} c \amp {}+{} \amp z \amp {}={} \amp0.35 \\ 9c \amp {}+{} \amp 7.1z \amp {}={} \amp2.5 \end{alignedat} \right. \end{align*}

Since the coefficient of $c$ (or $z$) in the first equation is $1\text{,}$ we could solve for one of these variables and use substitution to complete the problem. Some decimal arithmetic would be required. Alternatively, we can scale the equations by the right power of $10$ to make all the coefficients integers:

\begin{align*} \amp\left\{ \begin{alignedat}{4} \multiplyleft{100}(c \amp {}+{} \amp z) \amp {}={} \amp\multiplyleft{100}(0.35) \\ \multiplyleft{10}(9c \amp {}+{} \amp 7.1z) \amp {}={} \amp\multiplyleft{10}(2.5) \end{alignedat} \right.\\ \amp\left\{ \begin{alignedat}{4} 100c \amp {}+{} 100z \amp {}={} \amp 35\\ 90c \amp {}+{} 71z \amp {}={} \amp 25 \end{alignedat} \right. \end{align*}

Now to set up elimination, scale each equation again to eliminate $c\text{:}$

\begin{align*} \amp\left\{ \begin{alignedat}{4} \multiplyleft{9}(100c \amp {}+{} \amp 100z) \amp {}={} \amp\multiplyleft{9}(35) \\ \multiplyleft{-10}(90c \amp {}+{} \amp 71z) \amp {}={} \amp\multiplyleft{-10}(25) \end{alignedat} \right.\\ \amp\left\{ \begin{alignedat}{4} 900c \amp {}+{} 900z \amp {}={} \amp 315\\ -900c \amp {}+{} (-710z) \amp {}={} \amp {-250} \end{alignedat} \right. \end{align*}

Adding the corresponding sides from the two equations gives

\begin{equation*} 190z=65\text{,} \end{equation*}

from which we find $z=\frac{65}{190}\approx0.342\text{.}$ So there is about 0.342 cm³ of zinc in a penny.

To solve for $c\text{,}$ we can use one of the original equations:

\begin{align*} c+z\amp=0.35\\ c+\substitute{0.342}\amp\approx0.35\\ c\amp\approx0.008 \end{align*}

Therefore there is about 0.342 cm³ of zinc and 0.008 cm³ of copper in a penny.

To summarize, if a variable is already isolated or has a coefficient of $1\text{,}$ consider using the substitution method. If both equations are in standard form or none of the coefficients are equal to $1\text{,}$ we suggest using the elimination method. Either way, if you have fraction or decimal coefficients, it may help to scale your equations so that only integer coefficients remain.

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Reading Questions 4.3.4 Reading Questions

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1.

What is another name that the “elimination method” goes by?

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2.

To use the elimination method, usually the first step is to at least one equation.

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3.

Describe a good situation to use the substitution method instead of the elimination method for solving a system of two linear equations in two variables.

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Exercises 4.3.5 Exercises

Review and Warmup

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1.

Solve the equation.

$\displaystyle{ {{\frac{3}{4}}-10x}={6} }$

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2.

Solve the equation.

$\displaystyle{ {{\frac{7}{10}}-8r}={6} }$

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3.

Solve the equation.

$\displaystyle{ {{\frac{5}{6}}-{\frac{1}{6}}a}={5} }$

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4.

Solve the equation.

$\displaystyle{ {{\frac{5}{2}}-{\frac{1}{2}}b}={2} }$

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5.

Solve the equation.

$\displaystyle{ {\frac{8A}{9}-9}={-{\frac{89}{9}}} }$

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6.

Solve the equation.

$\displaystyle{ {\frac{4B}{5}-7}={-{\frac{43}{5}}} }$