Question 1208177
<br>
He bought 3 times as many cars as dolls:
let x = # of dolls
then 3x = # of dolls<br>
Each doll cost $10 more than each car:
let y = cost of each car
then y+10 = cost of each doll<br>
total cost of the cars: (3x)(y) = 3xy
total cost of the dolls: (x)(y+10)=xy+10x<br>
The total cost was $1972; the cost of the cars was $476 more than the cost of the dolls:
{{{3xy+(xy+10x)=4xy+10x=1972}}}
{{{3xy-(xy+10x)=2xy-10x=476}}}<br>
Eliminate the xy term from the two equations:
{{{4xy+10x=1972}}}
{{{4xy-20x=952}}}
{{{30x=1020}}}
{{{x=34}}}<br>
He bought x=34 dolls and 3x=102 cars<br>
Substitute x=34 in one of the equations to solve for y:
{{{2xy-10x=68y-340=476}}}
{{{68y=816}}}
{{{y=12}}}<br>
The cost of each car was y = $12; the cost of each doll was y+10 = $22<br>
ANSWER: $22<br>
CHECK:
cost of 34 dolls and 102 cars: 34($22)+102($12) = $748+$1224 = $1972
difference between cost of cars and cost of dolls: $1224-$748 = $476<br>