The equations 5x+2y=48 and 3x+2y=32 represent the money collected from school concert ticket sales during two class periods. If x represents the cost for each adult ticket nd y represents the cost for each student ticket, what is the cost for each adult ticket?
a. $20
b. $10
c. $8
d. $4
set the two equations up as simlutaneous equations and solve by substitution.
what do you mean by simultaneous equations?
The type of promblem you are dealing with is a set (in this case 2) simultaneous equations. You use the two equations TOGETHER to solve for x and y. Now do you get it?
but how do you set it up as simultaneous equations?
read this it explains it. If you still don't understand I will be happy to help.
algebra-help.info/simultaneous-equations.php
well, i got x=-8, so the answer should be c?
yep x=8
thanks
how did you get 8? I keep getting 16 for x.
To find the cost for each adult ticket, we need to solve the given system of equations:
5x + 2y = 48 ...(Equation 1)
3x + 2y = 32 ...(Equation 2)
We can solve this system of equations using the method of elimination. Here are the steps:
Step 1: Multiply Equation 1 by 3 and Equation 2 by 5 to make the coefficients of "y" in both equations the same. This will help us eliminate the variable "y" when we subtract the equations.
3*(5x + 2y) = 3*48
5*(3x + 2y) = 5*32
Simplifying:
15x + 6y = 144 ...(Equation 3)
15x + 10y = 160 ...(Equation 4)
Step 2: Subtract Equation 3 from Equation 4 to eliminate "x".
(15x + 10y) - (15x + 6y) = 160 - 144
15x - 15x + 10y - 6y = 16
4y = 16
Step 3: Solve for "y" by dividing both sides of the equation by 4.
4y/4 = 16/4
y = 4
Now that we have found the value of y, we can substitute it back into either Equation 1 or Equation 2 to solve for the value of x. Let's use Equation 2:
3x + 2*4 = 32
3x + 8 = 32
3x = 32 - 8
3x = 24
Divide both sides by 3 to solve for "x":
3x/3 = 24/3
x = 8
So, the cost for each adult ticket is $8. Therefore, the correct answer is option c. $8.