SOLUTION: There were 310 phones and tabs for sale in a shop.All the phones in the shop cost $2484 more than all the tabs. After 1/3 of the phones and 10/11 of the tabs were sold, there were

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: There were 310 phones and tabs for sale in a shop.All the phones in the shop cost $2484 more than all the tabs. After 1/3 of the phones and 10/11 of the tabs were sold, there were       Log On


   

Question 802659: There were 310 phones and tabs for sale in a shop.All the phones in the shop cost $2484 more than all the tabs. After 1/3 of the phones and 10/11 of the tabs were sold, there were thrice as many phones as tabs left.If each phone cost $714 more than each tab,how much did each tab cost?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
x= number of phones
y= number of tabs
We use the information given to set a system of linear equations to find x and y.
x%2By=310
After 1/3 of the phones were sold, there are x-%281%2F3%29x=%282%2F3%29x left.
After 10/11 of the tabs were sold, there are y-%2810%2F11%29y=%281%2F11%29y left.
Since %282%2F3%29x is thrice as many as %281%2F11%29y,
%282%2F3%29x=3%2A%281%2F11%29y
%282%2F3%29x=%283%2F11%29y
33%2A%282%2F3%29x=33%2A%283%2F11%29y
22x=9y
y=%2822%2F9%29x
So x%2B%2822%2F9%29x=310-->%2831%2F9%29x=310-->x=310%2A9%2F31-->x=90
So y=%2822%2F9%2990-->y=220

t= price of a tab, in $
so t%2B714= price of a phone, in $
90%28t%2B714%29= how much all the phones originally in the shop cost
$220t%2B2484%29= $2484 more than what all the tabs originally in the shop cost
220t%2B2484=90%28t%2B714%29
220t%2B2484=90t%2B64260
220t-90t=64260-2484
130t=31776
t=31776%2F130 --> highlight%28t=475.2%29