Two-variable Equations & Inequalities Portfolio Worksheet
Question 1
A. Write a two-variable equation to represent this situation. Define the variables.
Let:
- \( x \) = number of hard cover books
- \( y \) = number of soft cover books
The equation representing the total cost is: \[ 2.00x + 1.00y = 30 \]
B. Is it possible that Santiago bought 10 hard cover books and 5 soft cover books? Use your equation from part A to check. Explain your answer.
Substituting \( x = 10 \) and \( y = 5 \) into the equation: \[ 2.00(10) + 1.00(5) = 20 + 5 = 25 \]
Since \( 25 \neq 30 \), it is not possible that Santiago bought 10 hard cover books and 5 soft cover books.
C. Graph the solution set for the equation you wrote in part A on a piece of graph paper. Note: At this point, do not yet consider any constraints given by the situation.
[Insert Graph Here - E.g., The line 2x + y = 30 intercepts the x-axis at (15, 0) and the y-axis at (0, 30).]
D. What are the constraints of the situation? Describe them.
The constraints are:
- \( x \geq 0 \) (Santiago cannot buy negative numbers of hard cover books)
- \( y \geq 0 \) (Santiago cannot buy negative numbers of soft cover books)
- \( 2x + y \leq 30 \) (The total cost cannot exceed $30)
E. Use your graph to help you list three viable solutions to the problem when \( 2.00x + 1.00y = 30 \).
Three possible solutions could be:
- \( (0, 30) \) - 0 hard cover books and 30 soft cover books.
- \( (10, 10) \) - 10 hard cover books and 10 soft cover books.
- \( (5, 20) \) - 5 hard cover books and 20 soft cover books.
F. Suppose Santiago bought 26 books. Can you figure out how many of each type of book he bought? Explain. Write a system of equations and solve. Use Demos Calculator to check your work by adding the equation \( x + y = 26 \) to the graph above to help answer the question.
We now have a system of two equations:
- \( 2x + y = 30 \)
- \( x + y = 26 \)
Now we can use substitution or elimination to solve.
From equation 2: \[ y = 26 - x \]
Substituting into equation 1: \[ 2x + (26 - x) = 30 \] \[ 2x + 26 - x = 30 \] \[ x + 26 = 30 \] \[ x = 4 \]
Substituting \( x = 4 \) back into equation 2: \[ 4 + y = 26 \] \[ y = 22 \]
Santiago bought 4 hard cover books and 22 soft cover books.
Question 2
A. Is (1, 6) a solution to the system of equations? Explain.
To determine if (1, 6) is a solution, substitute \( x = 1 \) and \( y = 6 \) into both equations of the system.
If both equations are satisfied, then (1, 6) is a solution.
B. Solve the system of equations using substitution.
Let’s say our system is: \[ \begin{aligned}
- & \quad x + 2y = 13 \
- & \quad 3x - y = 4 \end{aligned} \]
Using substitution, we can solve these equations. Let's isolate \( y \) from equation 1: \[ 2y = 13 - x \] \[ y = \frac{13 - x}{2} \]
Substituting into equation 2: \[ 3x - \left( \frac{13 - x}{2} \right) = 4 \]
Multiply everything by 2 to eliminate the fraction: \[ 6x - (13 - x) = 8 \] \[ 6x - 13 + x = 8 \] \[ 7x - 13 = 8 \] \[ 7x = 21 \] \[ x = 3 \]
Substituting \( x = 3 \) back to find \( y \): \[ y = \frac{13 - 3}{2} = 5 \]
The solution is \( (3, 5) \).
C. Graph the system of equations on a piece of graph paper. On the graph, label the solution(s) to the system of equations.
[Insert Graph Here - E.g., The intersection point (3, 5) should be marked clearly on the graph.]
D. 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.
Yes, a system with a linear and a quadratic equation can have an infinite number of solutions, particularly if the linear equation is tangent to the quadratic curve. This can be observed on the graph when the line touches the curve at exactly one point, which indicates that they share infinitely many points at that specific location. However, this is not the case for generic linear and quadratic systems, which typically yields a finite number of solutions or none at all.
[End of Worksheet]
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