Question 1206119
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As another tutor says specifically, a problem like this is usually easier to set up using three variables; but often then solving the problem is more difficult.<br>
I generally prefer setting up the problem using a single variable, which usually makes solving the problem easier.  Let's see what happens....<br>
x = amount he started with<br>
amount he gave to Bob: {{{(2/5)x+5}}}<br>
amount he had left: {{{x-((2/5)x+5)=(3/5)(x)-5}}}<br>
amount he gave to Karen: {{{(3/4)((3/5)(x)-5)+6}}}<br>
amount he had left: {{{((3/5)(x)-5)-(3/4)((3/5)(x)-5)+6=(1/4)((3/5)(x)-5)-6}}}<br>
The amount he had left was $44 less than the amount he gave Karen:<br>
{{{(1/4)((3/5)(x)-5)-6=(3/4)((3/5)(x)-5)+6-44=(3/4)((3/5)(x)-5)-38}}}
{{{32=(1/2)((3/5)(x)-5)}}}
{{{64=(3/5)(x)-5}}}
{{{(3/5)x=69}}}
{{{x=69(5/3)=115}}}<br>
The amount he started with was $115.<br>
The amount he gave to Bob was {{{2/5)x+5=46+5=51}}}<br>
ANSWER: $51<br>
CHECK:
start: $115
give to Bob: (2/5)$115+5 = $51
left: $115-$51 = $64
give to Karen: (3/4)$64+$6 = $48+$6 = $54
left: $64-$54 = $10<br>
The amount he had left is $44 less than what he gave to Karen -- correct<br>
It seems the algebra gets complicated enough that in this problem using a single variable is not an easier way to go.<br>
When solving similar problems, try solving using a single variable and using three variables, and find what method works best FOR YOU.<br>