Algebra 2 Worksheets | Systems of Equations and Inequalities ... - Free Printable

Problem 1: Exam Points and Questions

Question: An exam worth 245 points contains 45 questions. Some questions are worth 7 points, and the others are worth 2 points. How many 7-point and 2-point questions are on the test?

#### Solution:
Let:
- \( x \) = number of 7-point questions
- \( y \) = number of 2-point questions

We have two pieces of information:
1. The total number of questions is 45.
\[
x + y = 45
\]
2. The total points for the exam is 245.
\[
7x + 2y = 245
\]

We now solve this system of equations.

##### Step 1: Solve the first equation for \( y \).
\[
y = 45 - x
\]

##### Step 2: Substitute \( y = 45 - x \) into the second equation.
\[
7x + 2(45 - x) = 245
\]

##### Step 3: Simplify and solve for \( x \).
\[
7x + 90 - 2x = 245
\]
\[
5x + 90 = 245
\]
\[
5x = 155
\]
\[
x = 31
\]

##### Step 4: Substitute \( x = 31 \) back into \( y = 45 - x \).
\[
y = 45 - 31 = 14
\]

#### Final Answer:
The exam has 31 questions worth 7 points each and 14 questions worth 2 points each.
\[
\boxed{31 \text{ and } 14}
\]

---

Problem 2: Plane Speed and Wind Speed

Question: Traveling to Vietnam, a plane encounters a tailwind and averages 289 mph. However, on the return trip, the plane, now traveling against the same wind, averages 257 mph. Find the speed of the wind and the speed of the plane in no wind.

#### Solution:
Let:
- \( p \) = speed of the plane in no wind (in mph)
- \( w \) = speed of the wind (in mph)

When the plane travels with the tailwind, its effective speed is:
\[
p + w = 289
\]

When the plane travels against the wind, its effective speed is:
\[
p - w = 257
\]

We now solve this system of equations.

##### Step 1: Add the two equations to eliminate \( w \).
\[
(p + w) + (p - w) = 289 + 257
\]
\[
2p = 546
\]
\[
p = 273
\]

##### Step 2: Substitute \( p = 273 \) into one of the original equations to solve for \( w \).
Using \( p + w = 289 \):
\[
273 + w = 289
\]
\[
w = 16
\]

#### Final Answer:
The speed of the plane in no wind is 273 mph, and the speed of the wind is 16 mph.
\[
\boxed{273 \text{ and } 16}
\]

---

Problem 3: Students in Vans and Buses

Question: For a trip, one high school rented and filled 8 vans and 9 buses with 228 students. Another high school instead fit its 124 students into 4 vans and 5 buses. With each bus and van seating the same number of students, how many students can a bus carry? How many students can a van carry?

#### Solution:
Let:
- \( v \) = number of students a van can carry
- \( b \) = number of students a bus can carry

We have two pieces of information:
1. One high school used 8 vans and 9 buses to carry 228 students.
\[
8v + 9b = 228
\]
2. Another high school used 4 vans and 5 buses to carry 124 students.
\[
4v + 5b = 124
\]

We now solve this system of equations.

##### Step 1: Eliminate one variable by making the coefficients of \( v \) the same.
Multiply the second equation by 2:
\[
2(4v + 5b) = 2(124)
\]
\[
8v + 10b = 248
\]

##### Step 2: Subtract the first equation from this new equation to eliminate \( v \).
\[
(8v + 10b) - (8v + 9b) = 248 - 228
\]
\[
b = 20
\]

##### Step 3: Substitute \( b = 20 \) back into one of the original equations to solve for \( v \).
Using \( 4v + 5b = 124 \):
\[
4v + 5(20) = 124
\]
\[
4v + 100 = 124
\]
\[
4v = 24
\]
\[
v = 6
\]

#### Final Answer:
A van can carry 6 students, and a bus can carry 20 students.
\[
\boxed{6 \text{ and } 20}
\]

---

Problem 4: Movie Ticket Prices

Question: Tickets at a particular movie theater have different rates for adults and children. On Friday, the theater sold 4 adult tickets and 7 child tickets for $83. The next day, the theater sold 5 adult tickets and 6 child tickets for $90. What is the price for the adult ticket and the price for the child ticket?

#### Solution:
Let:
- \( a \) = price of an adult ticket (in dollars)
- \( c \) = price of a child ticket (in dollars)

We have two pieces of information:
1. On Friday, 4 adult tickets and 7 child tickets cost $83.
\[
4a + 7c = 83
\]
2. On Saturday, 5 adult tickets and 6 child tickets cost $90.
\[
5a + 6c = 90
\]

We now solve this system of equations.

##### Step 1: Eliminate one variable by making the coefficients of \( a \) the same.
Multiply the first equation by 5 and the second equation by 4:
\[
5(4a + 7c) = 5(83) \quad \Rightarrow \quad 20a + 35c = 415
\]
\[
4(5a + 6c) = 4(90) \quad \Rightarrow \quad 20a + 24c = 360
\]

##### Step 2: Subtract the second equation from the first to eliminate \( a \).
\[
(20a + 35c) - (20a + 24c) = 415 - 360
\]
\[
11c = 55
\]
\[
c = 5
\]

##### Step 3: Substitute \( c = 5 \) back into one of the original equations to solve for \( a \).
Using \( 4a + 7c = 83 \):
\[
4a + 7(5) = 83
\]
\[
4a + 35 = 83
\]
\[
4a = 48
\]
\[
a = 12
\]

#### Final Answer:
The price of an adult ticket is $12, and the price of a child ticket is $5.
\[
\boxed{12 \text{ and } 5}
\]

---

Final Answers:

1. \(\boxed{31 \text{ and } 14}\)
2. \(\boxed{273 \text{ and } 16}\)
3. \(\boxed{6 \text{ and } 20}\)
4. \(\boxed{12 \text{ and } 5}\)