Question 1208177
<pre>
                 PRICE          QTY         COST

CARS              p              3d          3pd

DOLLS             p+10           d          d(p+10)

Total                                        1972
</pre>

"She paid $476 more for the cars than the dolls."

{{{3pd-d(p+10)=476}}}



Very interesting to notice the sum of two expressions and then the difference of 
those same two expressions.

{{{system(3pd+d(p+10)=1972,3pd-d(p+10)=476)}}}

-

{{{3pd+pd+10d=1972}}}
{{{4pd+10d=1972}}}
{{{2pd+5d=986}}}
-
{{{3pd-pd-10d=476}}}
{{{2pd-10d=476}}}
{{{pd-5d=238}}}


system: {{{system(2pd+5d=986,pd-5d=238)}}}


Easier if kept in the form {{{system(2pd+5d=986,2pd-10d=476)}}} because now easy to eliminate for pd
to solve for d.



{{{15d=510}}}

{{{highlight(d=34)}}}----------This is how many dolls bought.


{{{pd-5d=238}}}
{{{pd=238+5d}}}
{{{p=(238+5d)/d}}}

{{{p=(238+5*34)/34}}}

{{{p=12}}}---------car price

Doll is $10 more than a car, so {{{p+10=highlight(22)}}}.  Price of a doll





-----------------------


d=34 so p=12.

check:
3pd+d(p+10)=1972 ?
3*12*34+34(12+10)
1224+748
1972
checks correctly.