Equation Lab #9 – Systems of Equations: The Elimination Method (2)

Yesterday, I introduced the concept of solving systems of equations, as the first method discussed is the Elimination Method. To solve a system of equations, you have to solve for one variable at a time. And to do that, you’ll have to eliminate one of the variables, which can be done by combining two equations, either by addition and by subtraction. Once you found one variable, you can use it to find the other, and it takes one equation do to that.

What I didn’t cover in yesterday’s lesson is setting up an equation for elimination. This is what today’s lesson is about. You previously know how to solve for $x$ and $y$ when $x+y=8$ and $x-y=2$. Today, you’ll learn how to solve for $x$ and $y$ when $2x+3y=19$ and $3x-2y=9$.

The Elimination Method – Part 2:

Like yesterday, you’ll have to combine two equations to eliminate the variable and solve for one variable at a time. But today, the setup starts.

Let’s start with these two equations: $4l+3r=100$ and $l+6r=130$. Since all coefficients are positive, we’ll have to subtract the equations. But we can’t do that just yet. If we subtract $4l+3r=100$ from $l+6r=130$, you will get $-3l+3r=30$, which did not eliminate the variable. To set $l$ up for elimination, you’ll have to make sure the coefficients of $l$ in both equations are the same. The easy way is to multiply the second equation by $4$. You will get $4l$, but $6r$ becomes $24r$ since $6\cdot4=24$, and $130$ becomes $520$ since $130\cdot4=520$. So our two equations are $4l+3r=100$ and $4l+24r=520$. We can now apply the elimination method.

For best practice, we must subtract $4l+3r=100$ from $4l+24r=520$. $4l-4l=0$, but $24r-3r=21$, and $520-100=420$. This equation is now $21r=420$. With no constant and one coefficient, we can solve by division. Dividing both sides by $21$, you get $r=20$.

Now that we know that $r=20$, let’s find $l$. Using the second equation, we should get $l+6(20)=130$, which becomes $l+120=130$. Subtracting $120$ from both sides, we get $l=10$. So when $4l+3r=100$ and $l+6r=130$, we get that $l=10$ and $r=20$.

 1. $4l+3r=100$
     $l+6r=130$
 2. $4l+3r=100$
    $4(l+6r=130)$
 3. $4l+3r=100$
    $4l+24r=520$
 4. $(4l-4l)+(24r-3r)=520-100$
 5. $21r=420$
 6. $r=20$
 7. $l+6(20)=130$
 8. $l+120=130$
 9. $l=10$
10. $l=10, r=20$

Let’s check our work to see if both $l=10$ and $r=20$ are correct. In the first equation, $l=10$ and $r=20$ means that $4l+3r=100$ is the same as $4(10)+3(20)=100$. $4\cdot 10+3\cdot 20$ is $40+60$, which equals $100$. In the other equation, we should get $(10)+6(20)=130$. $10+6\cdot 20$ is $10+120$, which equals $130$. Since both equations are true, $l=10$ and $r=20$ are the correct solutions.

When setting up equations for elimination, you may not just have to set up one equation, but rather two equations. You don’t know what I mean? Let’s try another example. Our two equations are $8c+5v=21$ and $5c-2v=8$. Here, it looks like $v$ will become the variable to eliminate, but neither equation is ready. Since the coefficient for $v$ in $8c+5v=21$ is $5$, we must multiply $5c-2v=8$ by $5$. The equation is $25c-10v=40$. Since the coefficient for $v$ in $5c-2v=8$ is $2$, we must multiply $8c+5v=21$ by $2$. The equation is $16c+10v=42$. Now that the coefficients for $v$ in both equations match, we can combine the two equations.

Adding the two equations together, we get that $25c+16c=41c$ and $40+42=82$, while the $v$ gets knocked out of the equation. The combo equation is $41c=82$. Dividing both sides by $41$, the equation is $c=2$. We now have a value for $c$.

Let’s plug it into the original equation. We’ll try $8c+5v=21$, as $c=2$ turns it into $8(2)+5v=21$. The equation is $16+5v=21$. Subtracting $16$ from both sides, we get $5v=5$. Dividing both sides by $5$, we get $v=1$. So when $8c+5v=21$ and $5c-2v=8$, we get that $c=2$ and $v=1$.

 1. $8c+5v=21$
    $5c-2v=8$
 2. $2(8c+5v=21)$
    $5(5c-2v=8)$
 3. $16c+10v=42$
    $25c-10v=40$
 4. $(16c+25c)+(10v-10v)=42+40$
 5. $41c=82$
 6. $c=41$
 7. $8(2)+5v=21$
 8. $16+5v=21$
 9. $5v=5$
10. $v=1$
11. $c=2, v=1$

Let’s check our work to see if both $c=2$ and $v=1$ are correct. For the first equation, $8c+5v=21$ becomes $8(2)+5(1)=21$. $8\cdot 2=16$, and $5\cdot 1=5$. Add them up, we get $21$. for the second equation, $5c-2v=8$ becomes $5(2)-2(1)=8$. $5\cdot2=10$, and $2\cdot 1=2$. Subtract them, we get $8$. Since both equations are true, $c=2$ and $v=1$ are the correct solutions.

So that’s pretty much it for the elimination method. Now that you know how to solve using the elimination method and how to set up equations. But there’s one more thing to go over, besides the daily quiz.

Word Problem #8:

At a street fair, one person is selling plasma balls and lava lamps. The total stock sold was 44 items. Each plasma ball costs $35, and each lava lamp costs $16. The total revenue from selling the plasma balls and lava lamps is $1,008. How many plasma balls have been sold, and how many lava lamps have been sold?

Let’s define our two variables. We’ll use $p$ for the number of plasma balls sold and $l$ for the number of lava lamps sold. The first equation reflects the total number of items sold, which is 44. So our first equation is $p+l=44$. The second equation reflects the total revenue. Since each plasma ball costs $35 and each lava lamp costs $16, we’ll use $35p$ for plasma balls and $16l$ for lava lamps. Since the sum is $1,008, the equation is $35p+16l=1008$.

Let’s solve our two equations. You can see that the system isn’t ready. The guilty part is the equation $p+l=44$. Since both $p$ and $l$ have the same coefficients, any coefficient will work. But if we want to use that as the equation to solve from, let’s try solving for $l$ first. Aside to slapping a coefficient of $35$ on both variables, the only multiplication we really have to do is multiply $44$ by $35$. The product is $1540$. We now have $35p+35l=1540$.

Subtracting $35p+16l=1008$ from $35p+35l=1540$, the $p$ variable is eliminated, but $35l-16l=19l$, while $1540-1008=532$. With the equation being $19l=532$, all we need to do is divide both sides by $19$. The result is $l=28$.

Now let’s solve for $p$. Choosing the easier equation, $p+l=44$ is the same as $p+28=44$. The only step to do is to subtract $28$ from both sides. Our result is $p=16$. So when $l=28$, $p=16$.

 1. Define $p$ as the number of plasma
    balls sold, and define $l$ as the
    number of lava lamps sold.
 2. Each plasma ball costs $35, and
    each lava lamp costs $16, so the
    variables of the second equation
    are $35p$ and $16l$.
 3. The total sold is 44 items. But
    the total revenue is $1,008.
 4.   $p+l=44$
    $35p+16l=1008$
 5. $35(p+l=44)$
    $35p+16l=1008$
 6. $35p+35l=1540$
    $35p+16l=1008$
 7. $(35p-35p)+(35l-16l)=1008-1540$
 8. $19l=532$
 9. $l=28$
10. $p+(28)=44$
11. $p+28=44$
12. $p=16$
13. $p=16, l=28$

16 plasma balls and 28 lava lamps have been sold.

Checking our work, $16+28=44$, which matches the first equation. $35\cdot 16=560$, and $16\cdot28=448$. $560+448=1008$. Our answer is correct.

Quiz #9:

Now that the lesson is over, let’s see if you can complete this quiz. Like always, there are eight questions, but there are four systems of equations to solve for. Whether you put down the missing value or the variable with the missing value is fine for as long as you use the correct variable and correct answer. For instance, when the question is $b+3=6$, you can use $3$ or $b=3$ as your answer, but not $a=3$.