Equation Lab #11 – Systems of Equations: The Substitution Method (2)

Yesterday’s lesson covered the Substitution Method. As you seen, you can plug in variables with expressions to turn one two-variable equation into a one-variable equation, whether you’re putting expression against expression or expression into expression. What I haven’t covered, are the additional steps.

The Substitution Method – Part 2:

So here are some questions to ask. If $y=x+3$, then what is $5x+2y$? If $3y=9x-21$, then what is $4x+4y$? Not like yesterday’s lesson, you can’t easily plug in the equations. One of them will require an additional step of distribution, while the other requires setting the equation up for substitution. Just like the elimination method, you’ll have to set up an equation for substitution.

Let’s start with these two equations: $a=4p-20$ and $3a+2p=24$. Just like yesterday’s lesson, we can already plug $a=4p-20$ into $3a+2p=24$. But, since $a$ is scaled by $3$, you’ll have to multiply $4p-20$ by $3$. The result should by $12p-60$. The new equation is $12p+2p-60=24$. Combining like terms, you get $14p-60=24$. Adding $60$ to both sides, you get $14p=84$. Finally, we divide both sides by $14$, giving you $p=6$.

With $p$ found, let’s plug that into $a=4p-20$. $4\cdot 6=24$, but $24-20=4$. The result is $a=4$. So when $a=4p-20$ and $3a+2p=24$, we get that $a=4$ and $p=6$.

 1. $a=4p-20$
    $3a+2p=24$
 2. $3(4p-20)+2p=24$
 3. $12p-60+2p=24$
 4. $14p-60=24$
 5. $14p=84$
 6. $p=6$
 7. $a=4(6)-20$
 8. $a=24-20$
 9. $a=4$
10. $a=4, p=6$

Let’s check our work. We already seen the first equation, so how do they fit the second equation? $3\cdot 4=12$, and $2\cdot 6=12$. $12+12=24$, so $a=4$ and $p=6$ are the correct solutions.

Now let’s get more complicated. The equations are $4t=12c+8$ and $7c+3t=54$. We’ll start solving for $c$, so the $t$ variable needs to be substituted. The problem is that $t$ has a coefficient of $4$ in the first equation, but $3$ in the second equation. We can fix the first equation by dividing both sides by $4$. The equation to plug in is $t=3c+2$.

For the next step, we multiply $3c+2$ by $3$. The expression becomes $9c+6$, so the equation is $7c+9c+6=54$. Combining like terms, we get $16c+6=54$. Subtracting $6$ from both sides, we get $16c=48$. Dividing both sides by $16$, we get $c=3$.

Now let’s plug $c=3$ into $4t=12c+8$. $12\cdot 3=36$, but $36+8=44$. So, $4t=44$, but that’s not enough. We must divide both sides by $4$. This means $t=11$. So when $4t=12c+8$ and $7c+3t=54$, we get that $c=3$ and $t=11$.

 1. $4t=12c+8$
    $7c+3t=54$
 2. $(4t=12c+8)/4$
    $7c+3t=54$
 3. $t=3c+2$
    $7c+3t=54$
 4. $7c+3(3c+2)=54$
 5. $7c+9c+6=54$
 6. $16c+6=54$
 7. $16c=48$
 8. $c=3$
 9. $4t=12(3)+8$
10. $4t=36+8$
11. $4t=44$
12. $t=11$
13. $c=3, t=11$

Let’s plug in our values to see if they are correct. $7\cdot 3=21$, and $3\cdot 11=33$. $21+33=54$, proving the second equation true. So $c=3$ and $t=11$ are the correct solutions.

With more mastery on the substitution method, let’s do a word problem.

Word Problem #10:

NostalgiaTech is making miniature TVs and jukeboxes for the table, which can be functioned as actual video players and music players, respectively. The number of jukeboxes sold is 7 less than 3 times as many TVs sold. Each TV costs $85, and each jukebox costs $55, with the total revenue being $5,365. How many TVs and jukeboxes were sold?

Like always, we define our variables as the first step. We’ll use $t$ for retro TVs and $j$ for jukeboxes. The second equation reflects the total revenue, which is $5,365. Since each TV costs $85 and each jukebox costs $55, the equation is $85t+55j=5365$. As for the first equation, we are given a formula of how many jukeboxes are sold. NostalgiaTech sold three times as many jukeboxes as they sold TVs. Actually, they sold seven less than that. So the other equation is $j=3t-7$. Let’s solve.

We don’t need to set up for substitution this time, but $3t-7$ won’t come as $3t-7$ when we plug $j=3t-7$ into $85t+55j=5365$. We’ll have to multiply it by $55$ first. The product is $165t-385$, so the equation to solve is $85t+165t-385=5365$. Combining like terms, we get $250t-385=5365$. Adding $385$ to both sides, we get $250t=5750$. Divide both sides by $250$, we get $t=23$.

Now let’s plug in $t=23$ into $j=3t-7$. $23\cdot 3=69$, but $69-7=62$. The result is $j=62$.

 1. Define $t$ as number of TVs sold, and
    define $j$ as number of jukeboxes sold.
 2. There are $7$ less jukeboxes sold than
    $3$ times as many TVs sold.
 3. The cost of each TV is $85. The cost
    of each jukebox is $55. The total
    revenue is $5,365.
 4. $j=3t-7$
    $85t+55j=5365$
 5. $85t+55(3t-7)=5365$
 6. $85t+165t-385=5365$
 7. $250t-385=5365$
 8. $250t=5750$
 9. $t=23$
10. $j=3(23)-7$
11. $j=69-7$
12. $j=62$
13. $t=23, j=62$

23 TVs and 62 jukeboxes have been sold.

Plugging these values into the other equation, $23\cdot 85=1955$, and $62\cdot 55=3410$. Their sum is $5365$, so our answers are correct.

Quiz #11:

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 $a=3$.