Certainly! Below are the solutions and explanations for each part of the questions you've provided.
Question 1
A. Two-variable equation
Let:
- \( x \) = number of board books
- \( y \) = number of soft cover books
The equation representing Santiago's spending is: \[ 2x + y = 24 \]
B. Check if 8 board books and 8 soft cover books are possible
Substituting \( x = 8 \) and \( y = 8 \) in the equation: \[ 2(8) + 8 = 16 + 8 = 24 \] Yes, Santiago could have bought 8 board books and 8 soft cover books.
C. Graph the solution set
To graph \( 2x + y = 24 \):
- Find the intercepts:
- Setting \( x = 0 \): \( y = 24 \) (y-intercept)
- Setting \( y = 0 \): \( 2x = 24 \rightarrow x = 12 \) (x-intercept)
Plot these points (0, 24) and (12, 0) and draw the line.
D. Viable solutions
The viable solutions will be pairs \( (x, y) \) where both are non-negative integers and satisfy the equation. Examples include:
- (0, 24)
- (1, 22)
- (2, 20)
- (3, 18)
- (4, 16)
- (5, 14)
- (6, 12)
- (7, 10)
- (8, 8)
- (9, 6)
- (10, 4)
- (11, 2)
- (12, 0)
E. If he bought 20 books
\[ x + y = 20 \] Using the equations \( 2x + y = 24 \) and \( x + y = 20 \):
- Solve for \( y \): \( y = 20 - x \).
- Substitute into the first equation: \[ 2x + (20 - x) = 24 \] \[ x + 20 = 24 \rightarrow x = 4 \] \[ y = 20 - 4 = 16 \] So, he bought 4 board books and 16 soft cover books.
Question 2
A. Two-variable inequality
Let:
- \( c \) = number of children
- \( a \) = number of adults
The inequality is: \[ 150 + 10.5c + 5.5a \leq 300 \]
B. Check if 10 children and 5 adults are feasible
Substituting \( c = 10 \) and \( a = 5 \): \[ 150 + 10.5(10) + 5.5(5) = 150 + 105 + 27.5 = 282.5 \] Since \( 282.5 \leq 300 \), it is possible.
C. Graph the solution set
To graph:
- Rearrange: \( 10.5c + 5.5a \leq 150 \).
- Find intercepts:
- Setting \( c = 0 \): \( a \leq \frac{150}{5.5} \approx 27.27 \)
- Setting \( a = 0 \): \( c \leq \frac{150}{10.5} \approx 14.29 \)
Plot these points and shade the feasible region, ensuring at least one adult and child.
D. Viable solutions
A list of integer pairs \( (c, a) \) that satisfy:
- Minimum one child and one adult.
- Examples: (1, 1), (2, 1), (10, 1), etc., within budget.
Question 3
A. Check if (4,9) is a solution
Substituting \( x = 4 \) and \( y = 9 \):
- For \(-x + y = 5\): \[-4 + 9 = 5 \quad \text{True}\]
- For \( y = 4x + 2 \): \[9 = 4(4) + 2 \rightarrow 9 = 16 + 2 \rightarrow 9 \neq 18\] Thus, (4, 9) is not a solution.
B. Solve using substitution
From \( -x + y = 5 \): \[ y = x + 5 \] Substitute into \( y = 4x + 2 \): \[ x + 5 = 4x + 2 \] Rearranging gives: \[ 3 = 3x \rightarrow x = 1 \] Substituting back gives: \[ y = 1 + 5 = 6 \] Solution: \( (1, 6) \).
Question 4
A. Check if (0,3) is a solution
Substituting \( x = 0 \) and \( y = 3 \):
- For \( x + 3y = 9 \): \( 0 + 9 = 9 \)
- For \( 2x + 6y = 30 \): \( 0 + 18 = 30 \) (not true) Thus, (0, 3) is not a solution.
B. Solve using elimination
- Multiply the first equation by 2: \[ 2x + 6y = 18 \]
- Now subtract from the second equation: \[ (2x + 6y = 30) - (2x + 6y = 18) \] \[ 0 = 12 \] This indicates the system is dependent, and infinitely many solutions along the line \( x + 3y = 9 \).
Feel free to ask for clarifications or further details if needed!