Question: Name Jakai Roldage Date 09/15/26 Period 7th
Classwork - Applications of Linear Relationships
1. A coffee truck owner sells three sizes of iced coffee. The small sells for $5.50$, the medium sells for $6.75$, and the large is $7.50$. Last Saturday, the vendor sold $95$ items for
a total of $628$. The vendor sold $13$ more items for $6.75$ than for $7.50$. Write a system of
equations you could use to determine how many iced coffees were sold. Define the
variables.
2. At a school fundraiser, boxes of candy sell for $3.00$, $4.00$, or $5.00$, depending on
the size of the box. Altogether, $90$ boxes were sold, and $345$ was earned from the sales.
Twice as many $3.00$ boxes were sold as $5.00$ boxes. Formulate a system of equations to represent the situation. Be sure to define your
variables.
3. What is the solution to the system of equations represented by this matrix?
$$\begin{bmatrix}4 & 1 & -2 & 5 \\ 8 & -2 & 1 & -19 \\ 4 & 3 & 5 & 10\end{bmatrix}$$
4. Graph the system of linear inequalities. Shade the region that represents the solution
set of the system.
$$y \leq -x + 4$$
$$y \geq \frac{1}{2}x - 10$$
$$x > -1$$
5. Solve the system of equations using substitution. Show work for credit.
$$x + y + z = 2$$
$$2x + z = -1$$
$$x - 3z = -11$$
1. **Problem:** Determine how many small, medium, and large iced coffees were sold given prices and total sales.
2. **Define variables:**
- Let $s$ = number of small coffees sold
- Let $m$ = number of medium coffees sold
- Let $l$ = number of large coffees sold
3. **Write equations from problem:**
- Total items sold: $$s + m + l = 95$$
- Total sales amount: $$5.50s + 6.75m + 7.50l = 628$$
- Medium coffees sold 13 more than large: $$m = l + 13$$
4. **System of equations:**
$$\begin{cases}
s + m + l = 95 \\
5.50s + 6.75m + 7.50l = 628 \\
m = l + 13
\end{cases}$$
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1. **Problem:** Formulate system for candy box sales with given prices, total boxes, total sales, and relation between quantities.
2. **Define variables:**
- Let $x$ = number of $3.00$ boxes
- Let $y$ = number of $4.00$ boxes
- Let $z$ = number of $5.00$ boxes
3. **Write equations:**
- Total boxes sold: $$x + y + z = 90$$
- Total sales: $$3x + 4y + 5z = 345$$
- Twice as many $3.00$ boxes as $5.00$ boxes: $$x = 2z$$
4. **System of equations:**
$$\begin{cases}
x + y + z = 90 \\
3x + 4y + 5z = 345 \\
x = 2z
\end{cases}$$
---
1. **Problem:** Solve the system represented by the augmented matrix
$$\begin{bmatrix}4 & 1 & -2 & 5 \\ 8 & -2 & 1 & -19 \\ 4 & 3 & 5 & 10\end{bmatrix}$$
2. **Write system of equations:**
$$\begin{cases}
4x + y - 2z = 5 \\
8x - 2y + z = -19 \\
4x + 3y + 5z = 10
\end{cases}$$
3. **Solve by elimination:**
- Multiply first equation by 2 and subtract second:
$$2(4x + y - 2z) = 8x + 2y - 4z = 10$$
$$8x - 2y + z = -19$$
Subtracting:
$$(8x + 2y - 4z) - (8x - 2y + z) = 10 - (-19)$$
$$4y - 5z = 29$$
4. **Use first and third equations:**
Subtract first from third:
$$(4x + 3y + 5z) - (4x + y - 2z) = 10 - 5$$
$$2y + 7z = 5$$
5. **Solve system for $y$ and $z$:**
$$\begin{cases}
4y - 5z = 29 \\
2y + 7z = 5
\end{cases}$$
Multiply second by 2:
$$4y + 14z = 10$$
Subtract first:
$$(4y + 14z) - (4y - 5z) = 10 - 29$$
$$19z = -19$$
$$z = -1$$
6. **Find $y$:**
$$2y + 7(-1) = 5$$
$$2y - 7 = 5$$
$$2y = 12$$
$$y = 6$$
7. **Find $x$ using first equation:**
$$4x + 6 - 2(-1) = 5$$
$$4x + 6 + 2 = 5$$
$$4x + 8 = 5$$
$$4x = -3$$
$$x = -\frac{3}{4}$$
**Solution:**
$$\boxed{\left(-\frac{3}{4}, 6, -1\right)}$$
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1. **Problem:** Graph the system of inequalities:
$$y \leq -x + 4$$
$$y \geq \frac{1}{2}x - 10$$
$$x > -1$$
2. **Explanation:**
- The first inequality is a line with slope $-1$ and y-intercept $4$, shading below.
- The second inequality is a line with slope $\frac{1}{2}$ and y-intercept $-10$, shading above.
- The third inequality restricts $x$ to values greater than $-1$.
3. **Graph:**
- Plot lines $y = -x + 4$ and $y = \frac{1}{2}x - 10$.
- Shade below the first line and above the second.
- Shade only to the right of the vertical line $x = -1$.
- The solution region is the intersection of these shaded areas.
---
1. **Problem:** Solve the system using substitution:
$$\begin{cases}
x + y + z = 2 \\
2x + z = -1 \\
x - 3z = -11
\end{cases}$$
2. **From second equation:**
$$2x + z = -1 \implies z = -1 - 2x$$
3. **Substitute $z$ into third equation:**
$$x - 3(-1 - 2x) = -11$$
$$x + 3 + 6x = -11$$
$$7x + 3 = -11$$
$$7x = -14$$
$$x = -2$$
4. **Find $z$:**
$$z = -1 - 2(-2) = -1 + 4 = 3$$
5. **Find $y$ using first equation:**
$$x + y + z = 2$$
$$-2 + y + 3 = 2$$
$$y + 1 = 2$$
$$y = 1$$
**Solution:**
$$\boxed{(x, y, z) = (-2, 1, 3)}$$