Algebra worksheet using substitution to solve systems of equations, revealing a joke about an antelope.

Problem Overview:

The task involves solving a series of systems of linear equations using the substitution method. Each solution corresponds to a word that will fill in the blanks of a riddle at the top of the page. Additionally, there is a word problem at the bottom that needs to be solved.

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Step-by-Step Solution:

#### 1. Solve Each System of Equations Using Substitution

##### Problem 1:
\[
\begin{aligned}
& y = 3x \\
& 5x + 2y = 44
\end{aligned}
\]
- Substitute \( y = 3x \) into the second equation:
\[
5x + 2(3x) = 44
\]
\[
5x + 6x = 44
\]
\[
11x = 44
\]
\[
x = 4
\]
- Substitute \( x = 4 \) back into \( y = 3x \):
\[
y = 3(4) = 12
\]
- Solution: \( (4, 12) \)

##### Problem 2:
\[
\begin{aligned}
& x = 5y - 1 \\
& x + 2y = 13
\end{aligned}
\]
- Substitute \( x = 5y - 1 \) into the second equation:
\[
(5y - 1) + 2y = 13
\]
\[
5y + 2y - 1 = 13
\]
\[
7y - 1 = 13
\]
\[
7y = 14
\]
\[
y = 2
\]
- Substitute \( y = 2 \) back into \( x = 5y - 1 \):
\[
x = 5(2) - 1 = 10 - 1 = 9
\]
- Solution: \( (9, 2) \)

##### Problem 3:
\[
\begin{aligned}
& y = 2x + 7 \\
& 3x - y = -9
\end{aligned}
\]
- Substitute \( y = 2x + 7 \) into the second equation:
\[
3x - (2x + 7) = -9
\]
\[
3x - 2x - 7 = -9
\]
\[
x - 7 = -9
\]
\[
x = -2
\]
- Substitute \( x = -2 \) back into \( y = 2x + 7 \):
\[
y = 2(-2) + 7 = -4 + 7 = 3
\]
- Solution: \( (-2, 3) \)

##### Problem 4:
\[
\begin{aligned}
& -2x + 3y = 11 \\
& x = 4y - 3
\end{aligned}
\]
- Substitute \( x = 4y - 3 \) into the first equation:
\[
-2(4y - 3) + 3y = 11
\]
\[
-8y + 6 + 3y = 11
\]
\[
-5y + 6 = 11
\]
\[
-5y = 5
\]
\[
y = -1
\]
- Substitute \( y = -1 \) back into \( x = 4y - 3 \):
\[
x = 4(-1) - 3 = -4 - 3 = -7
\]
- Solution: \( (-7, -1) \)

##### Problem 5:
\[
\begin{aligned}
& y = 6x - 5 \\
& y = -x + 9
\end{aligned}
\]
- Set the two expressions for \( y \) equal to each other:
\[
6x - 5 = -x + 9
\]
\[
6x + x = 9 + 5
\]
\[
7x = 14
\]
\[
x = 2
\]
- Substitute \( x = 2 \) into \( y = 6x - 5 \):
\[
y = 6(2) - 5 = 12 - 5 = 7
\]
- Solution: \( (2, 7) \)

##### Problem 6:
\[
\begin{aligned}
& -3x + y = 7 \\
& 5x + 2y = 3
\end{aligned}
\]
- Solve the first equation for \( y \):
\[
y = 3x + 7
\]
- Substitute \( y = 3x + 7 \) into the second equation:
\[
5x + 2(3x + 7) = 3
\]
\[
5x + 6x + 14 = 3
\]
\[
11x + 14 = 3
\]
\[
11x = -11
\]
\[
x = -1
\]
- Substitute \( x = -1 \) back into \( y = 3x + 7 \):
\[
y = 3(-1) + 7 = -3 + 7 = 4
\]
- Solution: \( (-1, 4) \)

##### Problem 7:
\[
\begin{aligned}
& x - y = 11 \\
& 3x + 10y = -6
\end{aligned}
\]
- Solve the first equation for \( x \):
\[
x = y + 11
\]
- Substitute \( x = y + 11 \) into the second equation:
\[
3(y + 11) + 10y = -6
\]
\[
3y + 33 + 10y = -6
\]
\[
13y + 33 = -6
\]
\[
13y = -39
\]
\[
y = -3
\]
- Substitute \( y = -3 \) back into \( x = y + 11 \):
\[
x = -3 + 11 = 8
\]
- Solution: \( (8, -3) \)

##### Problem 8:
\[
\begin{aligned}
& -4x + y = 4 \\
& 2x + 2y = 13
\end{aligned}
\]
- Solve the first equation for \( y \):
\[
y = 4x + 4
\]
- Substitute \( y = 4x + 4 \) into the second equation:
\[
2x + 2(4x + 4) = 13
\]
\[
2x + 8x + 8 = 13
\]
\[
10x + 8 = 13
\]
\[
10x = 5
\]
\[
x = \frac{1}{2}
\]
- Substitute \( x = \frac{1}{2} \) back into \( y = 4x + 4 \):
\[
y = 4\left(\frac{1}{2}\right) + 4 = 2 + 4 = 6
\]
- Solution: \( \left(\frac{1}{2}, 6\right) \)

##### Problem 9:
\[
\begin{aligned}
& x + y = 1 \\
& 5x - 4y = -7
\end{aligned}
\]
- Solve the first equation for \( x \):
\[
x = 1 - y
\]
- Substitute \( x = 1 - y \) into the second equation:
\[
5(1 - y) - 4y = -7
\]
\[
5 - 5y - 4y = -7
\]
\[
5 - 9y = -7
\]
\[
-9y = -12
\]
\[
y = \frac{4}{3}
\]
- Substitute \( y = \frac{4}{3} \) back into \( x = 1 - y \):
\[
x = 1 - \frac{4}{3} = \frac{3}{3} - \frac{4}{3} = -\frac{1}{3}
\]
- Solution: \( \left(-\frac{1}{3}, \frac{4}{3}\right) \)

##### Problem 10:
\[
\begin{aligned}
& -5x + 3y = 11 \\
& x - 2y = 2
\end{aligned}
\]
- Solve the second equation for \( x \):
\[
x = 2y + 2
\]
- Substitute \( x = 2y + 2 \) into the first equation:
\[
-5(2y + 2) + 3y = 11
\]
\[
-10y - 10 + 3y = 11
\]
\[
-7y - 10 = 11
\]
\[
-7y = 21
\]
\[
y = -3
\]
- Substitute \( y = -3 \) back into \( x = 2y + 2 \):
\[
x = 2(-3) + 2 = -6 + 2 = -4
\]
- Solution: \( (-4, -3) \)

##### Problem 11:
\[
\begin{aligned}
& x + 9y = -1 \\
& 2x + 4y = 5
\end{aligned}
\]
- Solve the first equation for \( x \):
\[
x = -1 - 9y
\]
- Substitute \( x = -1 - 9y \) into the second equation:
\[
2(-1 - 9y) + 4y = 5
\]
\[
-2 - 18y + 4y = 5
\]
\[
-2 - 14y = 5
\]
\[
-14y = 7
\]
\[
y = -\frac{1}{2}
\]
- Substitute \( y = -\frac{1}{2} \) back into \( x = -1 - 9y \):
\[
x = -1 - 9\left(-\frac{1}{2}\right) = -1 + \frac{9}{2} = \frac{-2}{2} + \frac{9}{2} = \frac{7}{2}
\]
- Solution: \( \left(\frac{7}{2}, -\frac{1}{2}\right) \)

##### Problem 12:
\[
\begin{aligned}
& -5x + y = 35 \\
& 3x + 2y = -21
\end{aligned}
\]
- Solve the first equation for \( y \):
\[
y = 5x + 35
\]
- Substitute \( y = 5x + 35 \) into the second equation:
\[
3x + 2(5x + 35) = -21
\]
\[
3x + 10x + 70 = -21
\]
\[
13x + 70 = -21
\]
\[
13x = -91
\]
\[
x = -7
\]
- Substitute \( x = -7 \) back into \( y = 5x + 35 \):
\[
y = 5(-7) + 35 = -35 + 35 = 0
\]
- Solution: \( (-7, 0) \)

##### Problem 13:
This is a word problem, not a system of equations. We solve it separately below.

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Word Problem (Problem 13):

A math test is worth 100 points and has 30 problems. Each problem is worth either 3 points or 4 points. How many 4-point problems are there?

Let:
- \( x \): Number of 3-point problems
- \( y \): Number of 4-point problems

We have two equations:
\[
\begin{aligned}
& x + y = 30 \quad \text{(total number of problems)} \\
& 3x + 4y = 100 \quad \text{(total points)}
\end{aligned}
\]

- Solve the first equation for \( x \):
\[
x = 30 - y
\]
- Substitute \( x = 30 - y \) into the second equation:
\[
3(30 - y) + 4y = 100
\]
\[
90 - 3y + 4y = 100
\]
\[
90 + y = 100
\]
\[
y = 10
\]
- Substitute \( y = 10 \) back into \( x = 30 - y \):
\[
x = 30 - 10 = 20
\]

Thus, there are 10 four-point problems.

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Final Answer:

\[
\boxed{10}
\]