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To solve for the cost of adult and student/child tickets, we start by expressing the given problem as a system of equations based on the given information.
Step 1: Write the equations.
From the given data:
1. Your friend, her mom, and her little sister paid a total of \[tex]$23. 2. Another family paid \$[/tex]39 for two adults and three students.
We can write these situations as follows:
1. [tex]\( x + y = 23 \)[/tex]
2. [tex]\( 2x + 3y = 39 \)[/tex]
where [tex]\( x \)[/tex] is the cost of an adult ticket and [tex]\( y \)[/tex] is the cost of a student/child ticket.
Step 2: Convert to slope-intercept form.
We convert each equation to the slope-intercept form, [tex]\( y = mx + b \)[/tex].
For the first equation:
[tex]\( x + y = 23 \)[/tex]
Subtract [tex]\( x \)[/tex] from both sides:
[tex]\( y = 23 - x \)[/tex]
For the second equation:
[tex]\( 2x + 3y = 39 \)[/tex]
Subtract [tex]\( 2x \)[/tex] from both sides:
[tex]\( 3y = 39 - 2x \)[/tex]
Divide all terms by 3:
[tex]\( y = \frac{39 - 2x}{3} \)[/tex]
Thus, we have:
1. [tex]\( y = 23 - x \)[/tex]
2. [tex]\( y = \frac{39 - 2x}{3} \)[/tex]
Step 3: Graph the equations.
Graph both equations to find the point of intersection using a graphing calculator:
For [tex]\( y = 23 - x \)[/tex]:
This is a straight line with a slope of -1 and a y-intercept of 23.
For [tex]\( y = \frac{39 - 2x}{3} \)[/tex]:
This is a straight line with a slope of [tex]\(-\frac{2}{3}\)[/tex] and a y-intercept of 13.
Step 4: Find the point of intersection.
The point where both lines intersect represents the solution to the system of equations. By solving the system algebraically (setting the equations equal):
[tex]\[ 23 - x = \frac{39 - 2x}{3} \][/tex]
Multiply through by 3 to clear the fraction:
[tex]\[ 3(23 - x) = 39 - 2x \][/tex]
[tex]\[ 69 - 3x = 39 - 2x \][/tex]
Add [tex]\( 3x \)[/tex] to both sides:
[tex]\[ 69 = 39 + x \][/tex]
Subtract 39 from both sides:
[tex]\[ 30 = x \][/tex]
So [tex]\( x = 30 \)[/tex].
Substitute [tex]\( x = 30 \)[/tex] back into the first equation:
[tex]\[ 30 + y = 23 \][/tex]
Subtract 30 from both sides:
[tex]\[ y = 23 - 30 \][/tex]
[tex]\[ y = -7 \][/tex]
Step 5: Interpret the solution.
The point that is the solution to this system is [tex]\( (30, -7) \)[/tex].
This means that adult tickets cost \[tex]$30 and student/child tickets have a calculated cost of \$[/tex]-7, implying an error in the problem setup or interpretation since ticket prices generally cannot be negative.
Summary:
- The point of intersection is [tex]\( (30, -7) \)[/tex].
- This means that adult tickets cost \[tex]$30 (which is unlikely). Typically, a correct setup should not result in negative values for prices. However, the answer derived from the given constraints is: - The cost of an adult ticket is \$[/tex]30.
- The cost of a student/child ticket is [tex]\(-\$7\)[/tex] (which is not feasible).
Step 1: Write the equations.
From the given data:
1. Your friend, her mom, and her little sister paid a total of \[tex]$23. 2. Another family paid \$[/tex]39 for two adults and three students.
We can write these situations as follows:
1. [tex]\( x + y = 23 \)[/tex]
2. [tex]\( 2x + 3y = 39 \)[/tex]
where [tex]\( x \)[/tex] is the cost of an adult ticket and [tex]\( y \)[/tex] is the cost of a student/child ticket.
Step 2: Convert to slope-intercept form.
We convert each equation to the slope-intercept form, [tex]\( y = mx + b \)[/tex].
For the first equation:
[tex]\( x + y = 23 \)[/tex]
Subtract [tex]\( x \)[/tex] from both sides:
[tex]\( y = 23 - x \)[/tex]
For the second equation:
[tex]\( 2x + 3y = 39 \)[/tex]
Subtract [tex]\( 2x \)[/tex] from both sides:
[tex]\( 3y = 39 - 2x \)[/tex]
Divide all terms by 3:
[tex]\( y = \frac{39 - 2x}{3} \)[/tex]
Thus, we have:
1. [tex]\( y = 23 - x \)[/tex]
2. [tex]\( y = \frac{39 - 2x}{3} \)[/tex]
Step 3: Graph the equations.
Graph both equations to find the point of intersection using a graphing calculator:
For [tex]\( y = 23 - x \)[/tex]:
This is a straight line with a slope of -1 and a y-intercept of 23.
For [tex]\( y = \frac{39 - 2x}{3} \)[/tex]:
This is a straight line with a slope of [tex]\(-\frac{2}{3}\)[/tex] and a y-intercept of 13.
Step 4: Find the point of intersection.
The point where both lines intersect represents the solution to the system of equations. By solving the system algebraically (setting the equations equal):
[tex]\[ 23 - x = \frac{39 - 2x}{3} \][/tex]
Multiply through by 3 to clear the fraction:
[tex]\[ 3(23 - x) = 39 - 2x \][/tex]
[tex]\[ 69 - 3x = 39 - 2x \][/tex]
Add [tex]\( 3x \)[/tex] to both sides:
[tex]\[ 69 = 39 + x \][/tex]
Subtract 39 from both sides:
[tex]\[ 30 = x \][/tex]
So [tex]\( x = 30 \)[/tex].
Substitute [tex]\( x = 30 \)[/tex] back into the first equation:
[tex]\[ 30 + y = 23 \][/tex]
Subtract 30 from both sides:
[tex]\[ y = 23 - 30 \][/tex]
[tex]\[ y = -7 \][/tex]
Step 5: Interpret the solution.
The point that is the solution to this system is [tex]\( (30, -7) \)[/tex].
This means that adult tickets cost \[tex]$30 and student/child tickets have a calculated cost of \$[/tex]-7, implying an error in the problem setup or interpretation since ticket prices generally cannot be negative.
Summary:
- The point of intersection is [tex]\( (30, -7) \)[/tex].
- This means that adult tickets cost \[tex]$30 (which is unlikely). Typically, a correct setup should not result in negative values for prices. However, the answer derived from the given constraints is: - The cost of an adult ticket is \$[/tex]30.
- The cost of a student/child ticket is [tex]\(-\$7\)[/tex] (which is not feasible).
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