Equation Lab #13 – Systems of Three Equations

During the last five days, I went over how to solve systems of equations using two different methods. But we primarily covered systems of two equations. Now, we’re going up a level. Instead of two equations, we’ll be solving systems of three equations, with three variables per equation.

Systems of Three Equations:

Solving systems of three equations is the same as solving for systems of two equations, where you focus on one variable at a time. You know how to solve systems like $y=2x+4$ and $x+y=13$. What you haven’t seen yet, are systems like $z=2x+3y$, $x+y=\frac{z}{2}-4$, and $x+y+z=50$. You’ll have to substitute or eliminate two variables before you can get to solve one. And yes, if you know how to solve systems of three equations, you’ll know how to solve systems of four or more equations. But we’re not going to continue going higher up.

Let’s start with the first system: $5b+2c+3s=90$, $c+1=s+b$, and $2s+1=b+c$. This one will require double substitution. But first, we need to set up the equations for substitution. We’re going to put $c$ on the backburner first by plugging in $c$ in both equations. Using the equation $c+1=s+b$, it requires subtracting $1$ from both sides to make it the equation for $c$. So we’ll need to substitute $c$ with $s+b-1$ in both $2s+1=b+c$ and $5b+2c+3s=90$. The first equation becomes $5b+2s+2b-2+3s=90$, and the second equation becomes $2s+1=b+s+b-1$. We’ll have to simplify both equations before going to the next step. This should make the two equations $7b+5s-2=90$ and $2s+1=2b+s-1$.

The next variable to substitute is $s$, and that would require setting up $2s+1=2b+s-1$ for operation. Subtracting $s$ from both sides, you get $s+1=2b-1$. Subtracting $1$ from both sides, you get $s=2b-2$. Now we can plug it into the other equation, which will make it $7b+10b-10-2=90$.

With only $b$ left, we can solve for $b$. First, we combine like terms, which will make the equation $17b-12=90$. Then we add $12$ to both sides, which gives us $17b=102$. Finally, we divide both sides by $17$, giving us $b=6$ as the result. We have found our first variable.

Now let’s solve for $s$. Using the equation $s=2b-2$, $2\cdot6=12$, but $12-2=10$. The result is $s=10$, revealing us the next variable.

The final variable to solve for is $c$. Since $b=6$ and $s=10$, we can plug both values into the equation $c+1=s+b$. The equation you’ll be solving is $c+1=10+6$. Combining like terms, you have $c+1=16$. Subtracting $1$ from both sides, you get $c=15$. So, $b=6$, $c=15$, and $s=10$.

 1. $5b+2c+3s=90$
    $c+1=s+b$
    $2s+1=b+c$
 2. $5b+2c+3s=90$
    $c=s+b-1$
    $2s+1=b+c$
 3. $5b+2(s+b-1)+3s=90$
    $2s+1=b+(s+b-1)$
 4. $5b+2s+2b-2+3s=90$
    $2s+1=b+s+b-1$
 5. $7b+5s-2=90$
    $2s+1=2b+s-1$
 6. $7b+5s-2=90$
    $s+1=2b-1$
 7. $7b+5s-2=90$
    $s=2b-2$
 8. $7b+5(2b-2)-2=90$
 9. $7b+10b-10-2=90$
10. $17b-12=90$
11. $17b=102$
12. $b=6$
13. $s=2(6)-2$
14. $s=12$
15. $s=10$
16. $c+1=(10)+(6)$
17. $c+1=10+6$
18. $c+1=16$
19. $c=15$
20. $b=6, c=15, s=10$

Now let’s check our work. For the third equation, you should get $2(10)+1=(6)+(15)$. $2\cdot 10+1=21$, and $6+15=21$, so the third equation holds. For the second equation, we should have $(15)+1=(10)+(6)$. $15+1=16$, and $10+6=16$, so the second equation holds. Finally, for the first equation, you should get $5(6)+2(15)+3(10)=90$. $5\cdot 6=30$, $2\cdot 15=30$, and $3\cdot 10=30$. $30+30+30=90$, so the third equation holds. As a result, $b=6$, $c=15$, and $s=10$ are the correct solutions.

Now let’s do another system. It is $6k+j+8p=248$, $7k+5j+3p=224$, and $4k+9j+2p=232$. This time, we’ll strictly use the elimination method. The easiest equation to set up for elimination is $6k+j+8p=248$, which you’ll have to multiply by $5$ before working with $7k+5j+3p=224$, and $9$ before working with $4k+9j+2p=232$. $6k+j+8p=248$ times $5$ is $30k+5j+40p=1240$, and $6k+j+8p=248$ times $9$ is $54k+9j+72p=2232$. If we subtract $7k+5j+3p=224$ from $30k+5j+40p=1240$, the equation is $23k+37p=1016$. If we subtract $4k+9j+2p=232$ from $54k+9j+72p=2232$, the equation is $50k+70p=2000$.

Our two equations we got are $23k+37p=1016$ and $50k+70p=2000$. The next step is to set up $k$ for elimination, leaving us with $p$. $50$ times $23k+37p=1016$ is $1150k+1850p=50800$, and $23$ times $50k+70p=2000$ is $1150k+1610p=46000$. Subtract both equations, you should be left with $240p=4800$.

With $240p=4800$ being the single-variable equation, we got through two steps of elimination, all we have to do is divide both sides by $240$. The result is $p=20$.

Now let’s solve for $k$. We’ll use $50k+70p=2000$. $70\cdot 20=1400$, so the equation is $50k+1400=2000$. Subtract $1400$ from both sides, we get $50k=600$. Divide both sides by $50$, we get $12$. The result is $k=12$.

Finally, let’s find $j$. Using the equation $6k+j+8p=248$, $6\cdot 12=72$, and $8\cdot 20=160$. You have $72+j+160=248$. Combining like terms, you get $j+232=248$. Subtracting $232$ from both sides, you get $j=16$. We have our values for $k$, $j$, and $p$, which are $12$, $16$, and $20$, respectively.

 1. $6k+j+8p=248$
    $7k+5j+3p=224$
    $4k+9j+2p=232$
 2. $5(6k+j+8p=248)$
    $7k+5j+3p=224$
 3. $30k+5j+40p=1240$
    $7k+5j+3p=224$
 4. $(30k-7k)+(5j-5j)+(40p-3p)=1240-224$
 5. $23k+37p=1016$
 6. $9(6k+j+8p=248)$
    $4k+9j+2p=232$
 7. $54k+9j+72p=2232$
    $4k+9j+2p=232$
 8. $(54k-4k)+(9j-9j)+(72p-2p)=2232-232$
 9. $50k+70p=2000$
10. $23k+37p=1016$
    $50k+70p=2000$
11. $50(23k+37p=1016)$
    $23(50k+70p=2000)$
12. $1150k+1850p=50800$
    $1150k+1610p=46000$
13. $(1150k-1150k)+(1850p-1610p)=50800-46000$
14. $240p=4800$
15. $p=20$
16. $50k+70(20)=2000$
17. $50k+1400=2000$
18. $50k=600$
19. $k=12$
20. $6(12)+j+8(20)=248$
21. $72+j+160=248$
22. $j+232=248$
23. $j=16$
24. $k=12, j=16, p=20$

Let’s check our work. $6\cdot12+16+8\cdot 20$ becomes $72+16+160$, which results in $248$ as the sum. $7\cdot 12+5\cdot 16+3\cdot 20$ becomes $84+80+60$, which results in $224$ as the sum. And $4\cdot 12+9\cdot16+2\cdot 20$ becomes $48+144+40$, which results in $232$ as the sum. All three equations hold, so $k=12$, $j=16$, and $p=20$ are the correct answers.

Word Problem #12:

A science test contains multiple questions with different weights. Questions included are life science questions, physical science questions, and earth science questions. The total number of questions is 100 questions. Each life science question is worth 4 points, each physical science question is worth 5 points, and each earth science question is worth 3 points, making the science test having a total of 420 points. An interesting fact is that the ratio between life science and physical science questions is 2 to 3. How many life science, physical science, and earth science questions are there.

First, let’s define the variables. We’ll use $b$ for life science questions since “biology” starts with a ‘b’, $p$ for physical science questions since “physics” starts with a ‘p’, and $g$ for earth science questions since geology starts with a ‘g’. One of the equations states the total number of questions, which is 100. Therefore, the first equation is $b+p+g=100$. Since life science questions are worth 4 points, physical science questions are worth 5 points, and earth science questions are worth 3 points, the expression for the second equation is $4b+5p+3g$, which has a sum of 420. The last one is the ratio, where there’s 3 physical science questions for every 2 life science questions. So the equation is $\frac{b}{2}=\frac{p}{3}$. With the equations being $b+p+g=100$, $4b+5p+3g=420$, and $\frac{b}{2}=\frac{p}{3}$, we are ready to solve.

First, let’s cross-multiply the ratio equation. We should have $3b=2p$. Since no equation is ready for substitution yet, we’ll have to multiply the coefficients. For $b+p+g=100$, we can multiply the whole equation by $3$, making it $3b+3p+3g=300$. Since the coefficients for $g$ match in both $3b+3p+3g=300$ and $4b+5p+3g=420$, we can try subtraction. This will eliminate $g$ from the equation and result in the rest being $b+2p=120$. At this point, we can substitute $2p$ with $3b$ since $3b=2p$. The equation is $b+3b=120$. Combining like terms, we get $4b=120$. Dividing both sides by $4$, we get $b=30$.

With the value for $b$ found, let’s find $p$. This means substituting $b$ with $30$ in the equation $b+2p=120$. The equation becomes $30+2p=120$. Subtracting $30$ from both sides, we get $2p=90$. Dividing both sides by $2$, we get $p=45$.

The last variable to solve for is $g$. Since $b=30$ and $p=45$, we can plug them into the equation $b+p+g=100$. The equation is $30+45+g=100$. Combining like terms, we get $g+75=100$. Subtracting $75$ from both sides, we get $g=25$.

 1. Define $b$ as number of life science
    questions, define $p$ as number of
    physical science questions, and
    define $g$ as number of earth
    science questions.
 2. There are a total of $100$ questions,
    and $2$ life science questions for
    every $3$ physical science questions.
 3. There are a total of $420$ points for
    the $4$ points for every life science
    question, $5$ points for every
    physical science question, and $3$
    points for every earth science
    question.
 4. $b+p+g=100$
    $4b+5p+3g=420$
    $\frac{b}{2}=\frac{p}{3}$
 5. $b+p+g=100$
    $4b+5p+3g=420$
    $3b=2p$
 6. $3(b+p+g=100)$
    $4b+5p+3g=420$
    $3b=2p$
 7. $3b+3p+3g=300$
    $4b+5p+3g=420$
    $3b=2p$
 8. $(4b-3b)+(5p-3p)+(3g-3g)=420-300$
    $3b=2p$
 9. $b+2p=120$
    $3b=2p$
10. $b+(3b)=120$
11. $b+3b=120$
12. $4b=120$
13. $b=30$
14. $(30)+2p=120$
15. $30+2p=120$
16. $2p=90$
17. $p=45$
18. $(30)+(45)+g=100$
19. $30+45+g=100$
20. $75+g=100$
20. $g=25$
21. $b=30, p=45, g=25$

There are 30 life science questions, 45 physical science questions, and 25 earth science questions.

Quiz #13:

Now that the lesson is over, let’s see if you can complete this quiz. This time, there are nine questions, but there are three different systems to solve. 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 $a=3$.