CEMC's Open Courseware - Lesson 14: Equations with Multiple Variables

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If $x=4$ and $y=9$, then\[\begin{align*} P &= 2x + 2y \\ &= 2(4) + 2(9) \\ &= 8 + 18 \\ &= 26 \end{align*}\]Therefore, the value of $P$ is $26$.
1. If you have $4$ nickels and $7$ quarters then $n=4$ and $q=7$.
  We can substitute this information into the equation.\[\begin{align*} V &= 5n+25q \\ &= 5(4) + 25(7) \\ &= 20 + 175 \\ &= 195 \end{align*}\]Therefore, the total value of the nickels and quarters is $195$ cents or $$1.95$.
2. To find a solution to the equation $200 = 5n+25q$, we can assign an integer to $q$ and then solve for $n$.
  For example, if $q=1$ then\[\begin{align*} 200 &= 5n+25q \\ 200 &= 5n + 25(1) \\ 200 &= 5n + 25 \\ 200 \class{hl2}{-25} &= 5n+25 \class{hl2}{-25} \\ 175 &= 5n \\ \dfrac{175}{\class{hl2}{5}} &= \dfrac{5n}{\class{hl2}{5}} \\ 35 &= n \end{align*}\]Therefore, one possible solution to this equation is $n=35$ and $q=1$, which translates to $35$ nickels and $1$ quarter that amount to $200$ cents.
3. To show that there is more than one solution to the equation $200 = 5n+25q$, we must find a second solution that is different than the one found in part b). We can do this by assigning a different integer to $q$ and then solving for $n$.
  For example, if $q=2$ then\[\begin{align*} 200 &= 5n+25q \\ 200 &= 5n + 25(2) \\ 200 &= 5n + 50 \\ 200 \class{hl2}{-50} &= 5n+50 \class{hl2}{-50} \\ 150 &= 5n \\ \dfrac{150}{\class{hl2}{5}} &= \dfrac{5n}{\class{hl2}{5}} \\ 30 &= n \end{align*}\]Therefore, a second solution to this equation is $n=30$ and $q=2$, which translates to $30$ nickels and $2$ quarters that amount to $200$ cents. Since we have already found two different solutions, we have shown that there is more than one solution to this equation.
1. $T=48a + 30s + 24c$
2. If a group spent $$150$ on admissions, then we can substitute $T=150$ into the equation from part a).\[150 = 48a+30s+24c\]We must find values for $a$, $s$, and $c$ that make this equation true.
  If $a=1$, $s=1$, and $c=1$ then $48a+30s+24c=48(1)+30(1)+24(1)=102$.
  We note that $102 \neq 150$, in fact $102 \lt 150$. This tells us that there must have been more tickets purchased. Since the cost should be $$150$, we need to increase the current value $102$ by $48$. To do this, we can either purchase $1$ more adult ticket or $2$ children's tickets.
  Alternatively, you might notice that $5$ students could have attended since $30(5)=150$.
  Therefore, either
  - $2$ adults, $1$ student and $1$ child;
  - $1$ adult, $1$ student and $3$ children, or;
  - $5$ students.
  went to the amusement park.
1. Answers may vary.
  We know that $100 = 70 + 30$.
  Since $70=2 \times 35$ and $30=3 \times 10$, the numbers $70$ and $30$ meet the description.
2. Using our values from part a), since we know that $70=2 \times 35$ and $30=3 \times 10$, we can rewrite the equation as\[\begin{align*} 100 &= 70 + 30 \\ 100 &= 2(35) + 3(10) \end{align*}\]Therefore, a solution to the equation $100=2x+3y$ is $x=35, y=10$.
Let the number of adult tickets sold be $a$. Since the price for each adult ticket is $$12$, then the revenue from all adult tickets sold (in dollars) is $12\times a $ or $12a$.
We know that the number of child tickets sold is equal to the number of adult tickets sold, we can let the number of child tickets sold be $a$, and the total revenue from all $$6$ child tickets be $6a$ (in dollars).
In dollars, the combined revenue of all adult tickets and child tickets is $12a+6a=18a$.
Let $s$ represent the number of senior tickets sold. Since the cost of a senior's ticket is $$10$, the revenue from all senior tickets sold (in dollars) is $10s$.
Therefore, the equation $18a + 10s = 1100$ represents the relationship between the number of tickets sold and the total revenue.
We also know that $120$ tickets were sold in total. Since $a$ adult tickets were sold, $a$ child tickets were sold and $s$ senior tickets were sold, we have that $2a+s=120$. We need to find values for $a$ and $s$ that satisfy both equations.
Through trial and error:
If $a=10$ and $s=100$, then
- $2a+s=2(10)+100=120$ ✔
- $18a+10s=18(10)+10(100)=1180$ ✘
If $a=20$ and $s=80$, then
- $2a+s=2(20)+80=120$ ✔
- $18a+10s=18(20)+10(80)=1160$ ✘
If $a=50$ and $s=20$, then
- $2a+s=2(50)+20=120$ ✔
- $18a+10s=18(50)+10(20)=1100$ ✔
Therefore, $20$ senior tickets were sold.
1. Let $b$ represent the cost of one soccer ball and let $j$ represent the cost of one soccer jersey.
  Then the equation $b+j=100$ represents this problem.
2. Since $b$ and $j$ must be integer values, we know that $b$ could be any number from the sequence \[1,~2,~3,~4,\ldots,~99\]For example, there are $99$ solutions in total, so we do not list them all, but we can describe them as $(b, 100-b)$ where $b$ is any positive integer from $1$ to $99$.
  - If $b=1$ then $j=100-1=99$.
  - If $b=2$ then $j=100-2=98$.
    $\qquad \qquad \qquad \vdots$
  - If $b=99$ then $j=100-99=1$.
The relationship between the number of vertices, faces, and edges in a dodecahedron must satisfy Euler's formula.\[\begin{align*} e &= v+f-2 \\ e &= v+12-2 \\ e&=v+10 \end{align*}\]

But in this example, we still have $2$ unknown values, $v$ and $e$. Each of the $12$ faces of the dodecahedron is pentagonal. Since a pentagon has $5$ sides, we can deduce that there are a total of $12 \times 5=60$ sides.

Looking at the following diagram, we notice that each edge is shared by exactly $2$ faces, which means that it takes $2$ sides to make $1$ edge.

Source: Dodecahedron - eestingnef/iStock/Getty Images
This tells us that in counting the sides, we have double counted the number of edges. Therefore, the number of edges is equal to $60 \div 2 = 30$.
We can substitute this value into the equation.\[\begin{align*} e&=v+10 \\ 30 &= v+10 \end{align*}\]We can now solve the equation for $v$.\[\begin{align*} 30 &= v+10 \\ 30 \class{hl2}{-10} &= v+10 \class{hl2}{-10} \\ 20 &= v \end{align*}\]Therefore, the dodecahedron has $30$ edges and $20$ vertices.
You could solve this problem using trial and error. Instead of showing this solution, we will highlight how algebra can be used to find a sequence of moves that results in $1$ for this game.
This game is essentially asking whether or not there are integers $x$ and $y$ that satisfy the equation\[5x+17y=1\]Try testing different values for $x$ and $y$. If $x=3$ and $y=-1$ then\[\begin{align*} 5x+17y &= 5(3) + 17(-1) \\ &= 15 + (-17) \\ &= -2 \end{align*}\]This is not a solution since $-2 \neq 1$.
You can continue to try testing different values for $x$ and $y$. To avoid testing randomly, notice that $17(2)=34$. Since $35-34=1$, to get our equation to show this, we might try substituting $x=7, y=-2$.\[\begin{align*} 5x+17y &= 5(7) + 17(-2) \\ &= 35 + (-34) \\ &= 1 \end{align*}\]Therefore, it is possible to eventually get to the number $1$ since $x=7,~y=-2$ is a solution to the equation. This means that one way to complete the game is to add $5$ seven times, then subtract $17$ twice.