Question #dd25e

There are 8 adult tickets and 5 children tickets that total 177.50. There are also 12 adult tickets and 6 children tickets that total 255.00. Fnd the price of the child's ticket and the price of an adult ticket using elimination to solve two simultaneous equations.

Firstly, we'll define some variables.

Let #c=#the cost of a child's ticket.
Let #a=#the cost of an adult's ticket.

From the question:

#8a+5c=177.5" " (eq^n1)#
#12a+6c=255" "(eq^n2)#

Elimination is probably the best method to solve these simultaneous equations anyway; the alternative method would involve loads of awkward fractions which no one wants to deal with.
For elimination, we need to have two variables with the same coefficients, then we can add or subtract the equations. In this example, the coefficient of #a# OR the coefficient of #c# needs to be the same in both equations.

I will make the co-efficient of #a# the same. I will do this by multiplying every term in equation 1 by 3, and every term in equation 2 by 2.

#24a+15c=532.5" "(3xxeq^n1)#
#24a+12c=510" "(2xxeq^n2)#

See how we have #24a# in both? Now, we will do

#3(eq^n1)-2(eq^n2);#

#3c=22.5#

Look how the a's cancel out? Now, divide by 3:

#c=7.5#
So a child's ticket costs 7.5

Now, substitute this value back into any of the equations we've formed along the way. Generally go for the easiest equation. I will sub into #eq^n2#

Sub into #eq^n2#

#12a+6(7.5)=255#
#12a+45=255#
#12a=210#
#a=17.5#
So an adults ticket costs 17.5

It is worth giving your equations names so you can show your work clearly. You may not wish to write #eq^n# each time; simply writing #(1)# in brackets should be fine. This allows you to show clearly what you are doing, ie #"sub into "(2)# or #3(1)-2(2)#.

Hope this helps!