Question 1184378
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Often you want to try to set up solving a problem using a single variable, since it will almost always make the solution easier.  But the way the information is given in this problem makes that difficult if not impossible.<br>
So use four variables and convert each piece of given information into an equation in those variables.<br>
(1) The number of Audrey’s stickers was 70% fewer than the total number of stickers of Beth, Corrine and Dione<br>
Read "70% fewer" as "30% as much":
{{{A=.3(B+C+D)}}}<br>
(2) The ratio of the number of Beth’s stickers to the total number of stickers of Audrey, Corrine and Dione was 4:9<br>
Convert the ratio x:y=4:9 to 9x=4y:
{{{9B=4(A+C+D)}}}<br>
(3) The ratio of the number of Corrine’s stickers to the total number of stickers of Audrey, Beth and Dione was 3:23<br>
Do the same as with (2):
{{{23C=3(A+B+D)}}}<br>
(4) After Audrey lost 57 stickers and Dione lost 81 stickers,  the ratio of the number of Audrey’s stickers to the number of Dione’s stickers became 1:3<br>
Again the same thing:
{{{3(A-57)=D-81}}}<br>
There are our four equations in A, B, C, and D.  Because of the kinds of equations we have, a purely algebraic solution is going to be rather messy.  Probably the easiest way to solve the system is to convert each equation to the form pA+qB+rC+sD=t and solve using matrices.<br>
(1) {{{A=.3(B+C+D)}}}  ==>  {{{A-.3B-.3C-.3D=0}}}
(2) {{{9B=4(A+C+D)}}}  ==>  {{{4A-9B+4C+4D=0}}}
(3) {{{23C=3(A+B+D)}}}  ==>  {{{3A+3B-23C+3D=0}}}
(4) {{{3(A-57)=D-81}}}  ==>  {{{3A+0B+0C-D=90}}}<br>
The matrix is then<br>
{{{matrix(4,5,1,-.3,-.3,-.3,0,4,-9,4,4,0,3,3,-23,3,0,3,0,0,-1,90)}}}<br>
Solving the matrix gives....<br>
ANSWERS:
Audrey 60
Beth 80
Corrine 30
Dione 90<br>
CHECK:
(1) A=60; .3(B+C+D) = .3(80+30+90) = .3(200) = 60
(2) 9B=720; 4(A+C+D)=4(60+30+90)=4(180)=720
(3) 23C=690; 3(A+B+D)=3(60+80+90)=3(230)=690
(4) 3(A-57)=3(60-57)=3(3) = 9; D-81=90-9=81<br>