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Section 4.3 Addition/Elimination

Objective: Solve systems of equations using the addition/elimination method.

When solving systems we have found that graphing is very limited when solving equations. We then considered a second method known as substituion. This is probably the most used idea in solving systems in various areas of algebra. How- ever, substitution can get ugly if we don’t have a lone variable. This leads us to our second method for solving systems of equations. This method is known as either Elimination or Addition. We will set up the process in the following examples, then define the five step process we can use to solve by elimination.

Example 4.3.1.

\begin{align*} \begin{matrix} 3x-4y=8 \\ 5x+4y=-24 \end{matrix} \amp \quad \text{ Notice opposites in front of $y$'s. Add columns. } \\ \amp\quad \text{ } \\ 8x\qquad =-16 \amp\quad \text{ Solve for $x$, divide by } 8 \\ \frac{8x}{8}=\frac{-16}{8} \amp\quad \text{ } \\ x=-2 \amp\quad \text{ We have our } x \\ 5(-2)+4y=-24 \amp\quad \text{ Plug into either original equation, simplify } \\ -10+4y=-24 \amp\quad \text{ Add $10$ to both sides } \\ +10\qquad \quad +10 \amp\quad \text{ } \\ 4y=-14 \amp\quad \text{ Divide by } 4 \\ \frac{4y}{4}=\frac{-14}{4} \amp\quad \text{ } \\ y=-\frac{7}{2} \amp\quad \text{ Now we have our } y \\ (x,y)=\left(-2,\frac{-7}{2}\right) \amp\quad \text{ Our Solution } \end{align*}

In the previous example one variable had opposites in front of it, \(- 4y\) and \(4y\text{.}\) Adding these together eliminated the y completely. This allowed us to solve for the \(x\text{.}\) This is the idea behind the addition method. However, generally we won’t have opposites in front of one of the variables. In this case we will manipulate the equations to get the opposites we want by multiplying one or both equations (on both sides!). This is shown in the next example.

Example 4.3.2.

\begin{align*} - 6x + 5y = 22 \amp\quad \text{ We can get opposites in front of $x$, by multiplying the } \\ 2x+3y=2 \amp\quad \text{ second equation by $3$, to get $-6x$ and } +6x \\ \amp\quad \text{ } \\ 3(2x + 3y) = (2)3 \amp\quad \text{ Distribute to get new second equation. } \\ 6x+9y=6 \amp\quad \text{ New second equation } \\ -6x+5y=22 \amp\quad \text{ First equation still the same, add } \\ 14y=28 \amp\quad \text{ Divide both sides by } 14 \\ \frac{14y}{14}=\frac{28}{14} \amp\quad \text{ } \\ y=2 \amp\quad \text{ We have our } y \\ 2x+3(2)=2\amp\quad \text{ Plug into one of the original equations, simplify } \\ 2x+6=2 \amp\quad \text{ Subtract $6$ from both sides } \\ -6\ -6 \amp\quad \text{ } \\ 2x=-4 \amp\quad \text{ Divide both sides by } 2 \\ \frac{ 2x}{2}=\frac{-4}{2} \amp\quad \text{ } \\ x=-2 \amp\quad \text{ We also have our } x \\ (x,y)=(-2,2) \amp\quad \text{Our solution } \end{align*}

When we looked at the \(x\) terms, \(- 6x\) and \(2x\) we decided to multiply the \(2x\) by \(3\) to get the opposites we were looking for. What we are looking for with our opposites is the least common multiple (LCM) of the coefficients. We also could have solved the above problem by looking at the terms with \(y\text{,}\) \(5y\) and \(3y\text{.}\) The LCM of \(3\) and \(5\) is \(15\text{.}\) So we would want to multiply both equations, the \(5y\) by \(3\text{,}\) and the \(3y\) by \(- 5\) to get opposites, \(15y\) and \(- 15y\text{.}\) This illustrates an important point, some problems we will have to multiply both equations by a constant (on both sides) to get the opposites we want.

Example 4.3.3.

\begin{align*} 3x+6y=-9 \amp \quad \text{ We can get opposites in front of $x$, find LCM of $6$ and $9$, } \\ 2x+9y=-26 \amp\quad \text{ The LCM is $18$. We will multiply to get $18y$ and $-18y$ } \\ \amp\quad \text{ } \\ 3(3x+6y)=(-9)3 \amp\quad \text{ Multiply the first equation by $3$, both sides! } \\ 9x+18y=-27 \amp\quad \text{ } \\ \amp\quad \text{ } \\ -2(2x+9y)=(-26)(-2) \amp\quad \text{ Multiply the second equation by $- 2$, both sides! } \\ -4x-18y=52 \amp\quad \text{ } \\ \amp\quad \text{ } \\ \begin{matrix}9x+18y=-27\\-4x-18y=52 \end{matrix} \amp\quad \text{ Add two new equations together } \\ 5x\qquad = 25 \amp\quad \text{ Divide both sides by } 5 \\ x=5 \amp\quad \text{ We have our solution for } x \\ 3(5)+6y=-9 \amp\quad \text{ Plug into either original equation, simplify } \\ 15+6y=-9 \amp\quad \text{ Subtract $15$ from both sides} \\ -15 \quad \quad -15 \amp\quad \text{ } \\ 6y=-24 \amp\quad \text{ Divide both sides by } 6 \\ y=-4 \amp\quad \text{ Now we have our solution for } y \\ (x,y)=(5,-4) \amp\quad \text{ Our solution } \end{align*}

It is important for each problem as we get started that all variables and constants are lined up before we start multiplying and adding equations. This is illustrated in the next example which includes the five steps we will go through to solve a problem using elimination.

Table 4.3.4. Solving a System of Equations by Addition/Elimination
Problem \(\begin{matrix} 2x-5y=-13 \\ -3y+4=-5x\end{matrix} \)
\(\begin{matrix} \end{matrix} \)
1. Line up the variables and constants \(\begin{matrix} \text{Second Equation:}\\ -3y+4=-5x \\+5x-4 \quad +5x-4 \\ 5x-3y=-4 \end{matrix} \)
2. Multiply to get opposites (use LCD) \(\begin{matrix} 2x-5y=-13 \\ 5x-3y=-4\\ \text{First Equation: multiply by } - 5 \\ -5(2x-5y)=(-13)(-5) \\ -10x+25y=65 \\ \\ \text{Second Equation: multiply by } 2 \\ 2(5x-3y)=(-4)2 \\ 10x-6y=-8 \\ \\ -10x+25y=65 \\10x-6y=-8\end{matrix} \)
3. Add \(19y=57 \)
4. Solve \(\begin{matrix} \frac{19y}{19}=\frac{57}{19} \\ y=3\end{matrix} \)
5. Plug into either original and solve \(\begin{matrix} 2x-5(3)=-13 \\ 2x-15=-13 \\ +15 \quad +15 \\ 2x =2 \\ \frac{2x}{2} =\frac{2}{2} \\ x=1\end{matrix} \)
Solution \((x,y)=(1,3) \)

World View Note: The famous mathematical text, The Nine Chapters on the Mathematical Art, which was printed around \(179\) AD in China describes a for- mula very similar to Gaussian elimination which is very similar to the addition method.

Just as with graphing and substution, it is possible to have no solution or infinite solutions with elimination. Just as with substitution, if the variables all disappear from our problem, a true statment will indicate infinite solutions and a false stat- ment will indicate no solution.

Example 4.3.5.

\begin{align*} \begin{matrix} 2x-5y=3 \\ -6x+15y=-9 \end{matrix} \amp \quad \text{ To get opposites in front of $x$, multiply first equation by $3$. } \\ \amp\quad \text{ } \\ 3(2x - 5y) = (3)3\amp\quad \text{ Distribute } \\ 6x-15y=9 \amp\quad \text{ } \\ \amp\quad \text{ } \\ 6x-15y=9 \amp\quad \text{ Add equations together} \\ -6x+15y=-9\amp\quad \text{ } \\ 0=0 \amp\quad \text{ Tru statement } \\ \text{Infinite Solution} \amp\quad \text{ Our Solution } \end{align*}

Example 4.3.6.

\begin{align*} \begin{matrix} 4x-6y=8 \\ 6x-9y=15 \end{matrix} \amp \quad \text{ LCM for $x$'s is $12$. } \\ \amp\quad \text{ } \\ 3(4x - 6y) = (8)3 \amp\quad \text{ Multiply first equation by } 3 \\ 12x-18y=24 \amp\quad \text{ } \\ \amp\quad \text{ } \\ -2(6x-9y)=(15)(-2) \amp\quad \text{ Multiply second equation by } -2 \\ -12x+18y=-30 \amp\quad \text{ } \\ \amp\quad \text{ } \\ 12x-18y=24 \amp\quad \text{ Add both new equations together } \\ -12x+18y=-30 \amp\quad \text{ } \\ 0=-6 \amp\quad \text{ False statement } \\ \text{No Solution} \amp\quad \text{ Our Solution } \end{align*}

We have covered three different methods that can be used to solve a system of two equations with 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 lone variable, and addition works great when the other two methods fail. As each method has its own strengths, it is important you are familiar with all three methods.

Exercises Exercises - Addition/Elimination

Exercise Group.

Solve each system by elimination.

1.

\(\begin{matrix} 4x+2y=0\\ -4x-9y=-28 \end{matrix}\)

2.

\(\begin{matrix} -7x+y=-10\\ -9x-y=-22 \end{matrix}\)

3.

\(\begin{matrix} -9x+5y=-22\\ 9x-5y=13 \end{matrix}\)

4.

\(\begin{matrix} -x-2y=-7\\ x+2y=7 \end{matrix}\)

5.

\(\begin{matrix} -6x+9y=3\\ 6x-9y=-9 \end{matrix}\)

6.

\(\begin{matrix} 5x-5y=-15\\5x-5y=-15 \end{matrix}\)

7.

\(\begin{matrix}4x-6y=-10 \\4x-6y=-14 \end{matrix}\)

8.

\(\begin{matrix} -3x+3y=-12\\-3x+9y=-24 \end{matrix}\)

9.

\(\begin{matrix} -x-5y=28 \\-x+4y=-17 \end{matrix}\)

10.

\(\begin{matrix}-10x-5y=0 \\-10x-10y=-30 \end{matrix}\)

11.

\(\begin{matrix} 2x-y=5\\5x+2y=-28 \end{matrix}\)

12.

\(\begin{matrix} -5x+6y=-17\\x-2y=5 \end{matrix}\)

13.

\(\begin{matrix} 10x+6y=24\\-6x+y=4 \end{matrix}\)

14.

\(\begin{matrix} x+3y=-1\\10x+6y=-10 \end{matrix}\)

15.

\(\begin{matrix} 2x+4y=24\\4x-12y=8 \end{matrix}\)

16.

\(\begin{matrix} -6x+4y=12\\ 12x+6y=18 \end{matrix}\)

17.

\(\begin{matrix} -7x+4y=-4\\ 10x-8y=-8 \end{matrix}\)

18.

\(\begin{matrix} -6x+4y=4\\ -3x-y=26 \end{matrix}\)

19.

\(\begin{matrix} 5x+10y=20\\-6x-5y=-3 \end{matrix}\)

20.

\(\begin{matrix} -9x-5y=-19\\3x-7y=-11 \end{matrix}\)

21.

\(\begin{matrix} -7x-3y=12\\ -6x-5y=20 \end{matrix}\)

22.

\(\begin{matrix} -5x+4y=4\\ -7x-10y=-10 \end{matrix}\)

23.

\(\begin{matrix} 9x-2y=-18\\5x-7y=-10 \end{matrix}\)

24.

\(\begin{matrix} 3x+7y=-8\\4x+6y=-4 \end{matrix}\)

25.

\(\begin{matrix} 9x+6y=-21\\-10x-9y=28 \end{matrix}\)

26.

\(\begin{matrix}-4x-5y=12 \\-10x+6y=30 \end{matrix}\)

27.

\(\begin{matrix} -7x+5y=-8\\ -3x-3y=12 \end{matrix}\)

28.

\(\begin{matrix} 8x+7y=-24\\ 6x+3y=-18 \end{matrix}\)

29.

\(\begin{matrix}6x+3y=-18 \\10x+9y=1 \end{matrix}\)

30.

\(\begin{matrix} -7x+10y=13\\4x+9y=22 \end{matrix}\)

31.

\(\begin{matrix}9y = 7 - x \\-18y+4x=-26 \end{matrix}\)

32.

\(\begin{matrix} 0=-9x-21+12y\\ 1 + \frac{4}{3} y + \frac{7}{3} x = 0 \end{matrix}\)

33.

\(\begin{matrix} 0=9x+5y\\ y = \frac{2}{7} x \end{matrix}\)

34.

\(\begin{matrix} -6-42y=-12x\\ x-\frac{1}{2}-\frac{7}{2}y=0 \end{matrix}\)