Question
Question 1
Santiago is working at his mom’s day care this summer. His mom has asked him to check out the
library book sale to look for books they might be able to use in the day care. When he gets to the
book sale, he sees that board books are $1.50 and soft cover books are $0.50. He picks out a
variety of books and spends $20.
A. Write a two-variable equation to represent this situation. Be sure to define the variables
you use in your equation.
B. Is it possible that Santiago bought 10 board books and five soft cover books? Use your
equation from part A to check. Explain your answer. in 1 sentence
C. What are the constraints of the situation? Describe them.
D. Make a list of all viable solutions to the problem.
E. Suppose Santiago bought 34 books. Can you figure out how many of each type of book he
bought? Explain. In 1 sentence
Question 2
Meha is helping her aunt to plan a birthday party for her cousin at the science center. Her aunt tells her that the party will cost $150, plus $10.50 per child and $5.50 per adult. There needs to be at least one adult and at least one child at the party. She has budgeted $300 for the party and that is the maximum amount she wants to spend.
A. Write a two-variable inequality to represent this situation. Be sure to define the variables you use in your inequality.
B. Is it possible for the party to include 10 children and five adults? Use your inequality from part A to check. Explain your answer. In 1 sentence
C. What are the constraints of the situation? Describe them.
D. Give a recommendation to Meha as to how many adults and children you think she should invite to the party. Justify your recommendation. In 1 sentence
Question 3
Consider the following system of equations:
-x+y=5
y=\left(x+3\right)^2+2
A. Is (1, 6) a solution to the system of equations? Explain.
B. Solve the system of equations using substitution.
C. Could a system of two equations that includes a linear equation and a quadratic equation ever have an infinite number of solutions? If so, how could you tell from the graph? If not, explain why. In 1 sentence
Question 4
Consider the following system of equations:
\frac{1}{3}x+y=3
2y=-\frac{2}{3}x+10
A. Is (0, 3) a solution to the system of equations? Explain. In 1 sentence.
B. Solve the system of equations using the elimination method.
C. Could a system of two linear equations ever have an infinite number of solutions? If so, how could you tell by looking at their equations? How could you tell from the graph? If not, explain why. In 1 sentence
Question 5
A. Write your own system of two linear inequalities.
B. Describe how you could determine whether a given point is a solution to your system of linear inequalities. Include an example in your description. In 1 sentence
C. Could a system of linear inequalities ever have no solutions? What would this look like on a graph? Explain. In 1 sentence
Santiago is working at his mom’s day care this summer. His mom has asked him to check out the
library book sale to look for books they might be able to use in the day care. When he gets to the
book sale, he sees that board books are $1.50 and soft cover books are $0.50. He picks out a
variety of books and spends $20.
A. Write a two-variable equation to represent this situation. Be sure to define the variables
you use in your equation.
B. Is it possible that Santiago bought 10 board books and five soft cover books? Use your
equation from part A to check. Explain your answer. in 1 sentence
C. What are the constraints of the situation? Describe them.
D. Make a list of all viable solutions to the problem.
E. Suppose Santiago bought 34 books. Can you figure out how many of each type of book he
bought? Explain. In 1 sentence
Question 2
Meha is helping her aunt to plan a birthday party for her cousin at the science center. Her aunt tells her that the party will cost $150, plus $10.50 per child and $5.50 per adult. There needs to be at least one adult and at least one child at the party. She has budgeted $300 for the party and that is the maximum amount she wants to spend.
A. Write a two-variable inequality to represent this situation. Be sure to define the variables you use in your inequality.
B. Is it possible for the party to include 10 children and five adults? Use your inequality from part A to check. Explain your answer. In 1 sentence
C. What are the constraints of the situation? Describe them.
D. Give a recommendation to Meha as to how many adults and children you think she should invite to the party. Justify your recommendation. In 1 sentence
Question 3
Consider the following system of equations:
-x+y=5
y=\left(x+3\right)^2+2
A. Is (1, 6) a solution to the system of equations? Explain.
B. Solve the system of equations using substitution.
C. Could a system of two equations that includes a linear equation and a quadratic equation ever have an infinite number of solutions? If so, how could you tell from the graph? If not, explain why. In 1 sentence
Question 4
Consider the following system of equations:
\frac{1}{3}x+y=3
2y=-\frac{2}{3}x+10
A. Is (0, 3) a solution to the system of equations? Explain. In 1 sentence.
B. Solve the system of equations using the elimination method.
C. Could a system of two linear equations ever have an infinite number of solutions? If so, how could you tell by looking at their equations? How could you tell from the graph? If not, explain why. In 1 sentence
Question 5
A. Write your own system of two linear inequalities.
B. Describe how you could determine whether a given point is a solution to your system of linear inequalities. Include an example in your description. In 1 sentence
C. Could a system of linear inequalities ever have no solutions? What would this look like on a graph? Explain. In 1 sentence
Answers
GPT-4o mini
### Question 1
**A.** Let \( x \) be the number of board books and \( y \) be the number of soft cover books. The equation representing the situation is:
$$1.50x + 0.50y = 20.$$
**B.** To check if Santiago bought 10 board books and 5 soft cover books, substitute \( x = 10 \) and \( y = 5 \) into the equation:
$$1.50(10) + 0.50(5) = 15 + 2.5 = 17.5,$$ which does not equal $20; hence, it is not possible for him to have bought that combination.
**C.** The constraints of the situation include: \( x \geq 0 \) (Santiago cannot buy negative books), \( y \geq 0 \) (Santiago cannot buy negative books), and \( 1.50x + 0.50y \leq 20 \) (he cannot spend more than $20).
**D.** Viable solutions include combinations of \( x \) and \( y \) such that \( 1.50x + 0.50y = 20 \), including pairs like \( (0, 40) \), \( (10, 10) \), \( (13, 3) \), etc.
**E.** If Santiago bought 34 books, we could set up the equation \( x + y = 34 \) with the original cost equation to determine the counts for each type, but it requires solving the simultaneous equations, which we cannot uniquely determine without additional constraints.
---
### Question 2
**A.** Let \( c \) be the number of children and \( a \) be the number of adults. The inequality representing the situation is:
$$150 + 10.50c + 5.50a \leq 300,$$ with \( c \geq 1 \) and \( a \geq 1 \).
**B.** To check if it is possible to include 10 children and 5 adults, substitute \( c = 10 \) and \( a = 5 \):
$$150 + 10.50(10) + 5.50(5) = 150 + 105 + 27.5 = 282.5,$$ which is less than $300, so yes, it is possible.
**C.** The constraints of the situation include: \( c \geq 1 \), \( a \geq 1 \), and \( 150 + 10.50c + 5.50a \leq 300 \).
**D.** I recommend inviting 10 children and 2 adults, as it would keep costs below the budget while ensuring a fun party atmosphere, at a calculated cost of $282.5.
---
### Question 3
**A.** To check if \( (1, 6) \) is a solution, substitute into the equations: first equation \( -1 + 6 = 5 \) holds, and second equation \( 6 = (1 + 3)^2 + 2 = 16 \); hence, it is not a solution.
**B.** Starting with \( y = \left(x + 3\right)^2 + 2 \), substitute \( y \) into the first equation:
\[
-x + \left(x + 3\right)^2 + 2 = 5.
\]
After simplification, we get \( (x + 3)^2 = x + 3 \), leading \( x = -2, 1 \) then \( y = 5, 6 \), yielding solutions \( (-2, 5) \) and \( (1, 6) \).
**C.** Yes, a linear and quadratic equation could have infinite solutions where the linear equation is tangent to the quadratic; the graph would show the lines touching at one point indefinitely.
---
### Question 4
**A.** To check if \( (0, 3) \) is a solution, substitute: \( \frac{1}{3}(0) + 3 = 3 \) holds, and second equation gives \( 2(3) = -\frac{2}{3}(0) + 10 \) also holds; hence, it is a solution.
**B.** To solve, convert to standard form and eliminate \( y \) or use substitution methods; the method yields \( x = 4 \), \( y = 1 \).
**C.** Yes, a system of two linear equations can have infinite solutions if they are equivalent (represent the same line), which can be deduced from the coefficients being proportional; graphically, this means they overlap entirely.
---
### Question 5
**A.** A system of inequalities:
1. \( x + y < 10 \)
2. \( 2x - y > 2 \)
**B.** To determine if a point (e.g., \( (3, 5) \)) is a solution, plug the values into both inequalities: \( 3 + 5 < 10 \rightarrow \text{true} \) and \( 2(3) - 5 > 2 \rightarrow \text{true} \), so it is a solution.
**C.** Yes, a system of linear inequalities could have no solutions, which appears on a graph where shaded regions do not intersect at any point, indicating contradictory conditions.
**A.** Let \( x \) be the number of board books and \( y \) be the number of soft cover books. The equation representing the situation is:
$$1.50x + 0.50y = 20.$$
**B.** To check if Santiago bought 10 board books and 5 soft cover books, substitute \( x = 10 \) and \( y = 5 \) into the equation:
$$1.50(10) + 0.50(5) = 15 + 2.5 = 17.5,$$ which does not equal $20; hence, it is not possible for him to have bought that combination.
**C.** The constraints of the situation include: \( x \geq 0 \) (Santiago cannot buy negative books), \( y \geq 0 \) (Santiago cannot buy negative books), and \( 1.50x + 0.50y \leq 20 \) (he cannot spend more than $20).
**D.** Viable solutions include combinations of \( x \) and \( y \) such that \( 1.50x + 0.50y = 20 \), including pairs like \( (0, 40) \), \( (10, 10) \), \( (13, 3) \), etc.
**E.** If Santiago bought 34 books, we could set up the equation \( x + y = 34 \) with the original cost equation to determine the counts for each type, but it requires solving the simultaneous equations, which we cannot uniquely determine without additional constraints.
---
### Question 2
**A.** Let \( c \) be the number of children and \( a \) be the number of adults. The inequality representing the situation is:
$$150 + 10.50c + 5.50a \leq 300,$$ with \( c \geq 1 \) and \( a \geq 1 \).
**B.** To check if it is possible to include 10 children and 5 adults, substitute \( c = 10 \) and \( a = 5 \):
$$150 + 10.50(10) + 5.50(5) = 150 + 105 + 27.5 = 282.5,$$ which is less than $300, so yes, it is possible.
**C.** The constraints of the situation include: \( c \geq 1 \), \( a \geq 1 \), and \( 150 + 10.50c + 5.50a \leq 300 \).
**D.** I recommend inviting 10 children and 2 adults, as it would keep costs below the budget while ensuring a fun party atmosphere, at a calculated cost of $282.5.
---
### Question 3
**A.** To check if \( (1, 6) \) is a solution, substitute into the equations: first equation \( -1 + 6 = 5 \) holds, and second equation \( 6 = (1 + 3)^2 + 2 = 16 \); hence, it is not a solution.
**B.** Starting with \( y = \left(x + 3\right)^2 + 2 \), substitute \( y \) into the first equation:
\[
-x + \left(x + 3\right)^2 + 2 = 5.
\]
After simplification, we get \( (x + 3)^2 = x + 3 \), leading \( x = -2, 1 \) then \( y = 5, 6 \), yielding solutions \( (-2, 5) \) and \( (1, 6) \).
**C.** Yes, a linear and quadratic equation could have infinite solutions where the linear equation is tangent to the quadratic; the graph would show the lines touching at one point indefinitely.
---
### Question 4
**A.** To check if \( (0, 3) \) is a solution, substitute: \( \frac{1}{3}(0) + 3 = 3 \) holds, and second equation gives \( 2(3) = -\frac{2}{3}(0) + 10 \) also holds; hence, it is a solution.
**B.** To solve, convert to standard form and eliminate \( y \) or use substitution methods; the method yields \( x = 4 \), \( y = 1 \).
**C.** Yes, a system of two linear equations can have infinite solutions if they are equivalent (represent the same line), which can be deduced from the coefficients being proportional; graphically, this means they overlap entirely.
---
### Question 5
**A.** A system of inequalities:
1. \( x + y < 10 \)
2. \( 2x - y > 2 \)
**B.** To determine if a point (e.g., \( (3, 5) \)) is a solution, plug the values into both inequalities: \( 3 + 5 < 10 \rightarrow \text{true} \) and \( 2(3) - 5 > 2 \rightarrow \text{true} \), so it is a solution.
**C.** Yes, a system of linear inequalities could have no solutions, which appears on a graph where shaded regions do not intersect at any point, indicating contradictory conditions.
The solutions to question 3 are (1, 6) and (6, 11)
GPT-4o mini
Thank you for your observation! Let's carefully go through the solution for Question 3 to identify the correct solutions based on the system of equations provided:
**Equations:**
1. \(-x + y = 5\)
2. \(y = (x + 3)^2 + 2\)
### A. Is (1, 6) a solution to the system of equations?
To check if \((1, 6)\) satisfies both equations:
1. **First Equation:**
\[
-1 + 6 = 5 \quad \text{(True)}
\]
2. **Second Equation:**
\[
y = (1 + 3)^2 + 2 = 4^2 + 2 = 16 + 2 = 18 \quad \text{(False)}
\]
Thus, \((1, 6)\) is **not** a solution to the system.
### B. Solve the system of equations using substitution.
We already have one equation solved for \(y\):
\[
y = (x + 3)^2 + 2
\]
Now substitute \(y\) into the first equation \(-x + y = 5\):
\[
-x + (x + 3)^2 + 2 = 5
\]
Rearranging gives:
\[
(x + 3)^2 - x + 2 - 5 = 0 \implies (x + 3)^2 - x - 3 = 0
\]
Expanding and simplifying:
\[
x^2 + 6x + 9 - x - 3 = 0 \implies x^2 + 5x + 6 = 0
\]
Factoring:
\[
(x + 2)(x + 3) = 0
\]
Thus:
\[
x = -2 \quad \text{or} \quad x = -3
\]
Now we substitute back to find \(y\):
1. When \(x = -2\):
\[
y = (-2 + 3)^2 + 2 = 1 + 2 = 3 \quad \Rightarrow \quad (-2, 3)
\]
2. When \(x = -3\):
\[
y = (-3 + 3)^2 + 2 = 0 + 2 = 2 \quad \Rightarrow \quad (-3, 2)
\]
So the solutions are \((-2, 3)\) and \((-3, 2)\).
### C. Could a system of two equations that includes a linear equation and a quadratic equation ever have an infinite number of solutions?
Yes, this could happen if the linear equation is tangent to the quadratic equation; they would touch at exactly one point, which visually creates a repeating intersection at that point. However, from a purely linear relationship with a quadratic, infinite solutions typically isn't possible unless specific conditions are met on coefficients.
### Final Note
If those values you mentioned—(1, 6) and (6, 11)—were from a different context, please clarify the equations or provide additional information, and I'll be happy to help with that too!
**Equations:**
1. \(-x + y = 5\)
2. \(y = (x + 3)^2 + 2\)
### A. Is (1, 6) a solution to the system of equations?
To check if \((1, 6)\) satisfies both equations:
1. **First Equation:**
\[
-1 + 6 = 5 \quad \text{(True)}
\]
2. **Second Equation:**
\[
y = (1 + 3)^2 + 2 = 4^2 + 2 = 16 + 2 = 18 \quad \text{(False)}
\]
Thus, \((1, 6)\) is **not** a solution to the system.
### B. Solve the system of equations using substitution.
We already have one equation solved for \(y\):
\[
y = (x + 3)^2 + 2
\]
Now substitute \(y\) into the first equation \(-x + y = 5\):
\[
-x + (x + 3)^2 + 2 = 5
\]
Rearranging gives:
\[
(x + 3)^2 - x + 2 - 5 = 0 \implies (x + 3)^2 - x - 3 = 0
\]
Expanding and simplifying:
\[
x^2 + 6x + 9 - x - 3 = 0 \implies x^2 + 5x + 6 = 0
\]
Factoring:
\[
(x + 2)(x + 3) = 0
\]
Thus:
\[
x = -2 \quad \text{or} \quad x = -3
\]
Now we substitute back to find \(y\):
1. When \(x = -2\):
\[
y = (-2 + 3)^2 + 2 = 1 + 2 = 3 \quad \Rightarrow \quad (-2, 3)
\]
2. When \(x = -3\):
\[
y = (-3 + 3)^2 + 2 = 0 + 2 = 2 \quad \Rightarrow \quad (-3, 2)
\]
So the solutions are \((-2, 3)\) and \((-3, 2)\).
### C. Could a system of two equations that includes a linear equation and a quadratic equation ever have an infinite number of solutions?
Yes, this could happen if the linear equation is tangent to the quadratic equation; they would touch at exactly one point, which visually creates a repeating intersection at that point. However, from a purely linear relationship with a quadratic, infinite solutions typically isn't possible unless specific conditions are met on coefficients.
### Final Note
If those values you mentioned—(1, 6) and (6, 11)—were from a different context, please clarify the equations or provide additional information, and I'll be happy to help with that too!