Question 205608
Let x=hourly wages of the plumber
Then x-3=hourly wages of his apprentice
Let y=number of hours the plumber worked
Then y+10 =number of hours the apprentice worked

Now we know that the amount made by the plumber=x*y=$600----eq1
We also know that amount made by the apprentice=(x-3)(y+10)=$600----eq2

So, our equation to solve is:
xy=(x-3)(y+10) expand right side using FOIL (First, Outer, Inner, Last)
xy=xy+10x-3y-30  subtract xy from each side; also add 3y and 30 to each side
xy-xy+3y+30=xy-xy+10x-3y+3y-30+30  collect like terms
3y+30=10x substitute y=600/x from eq1 
3(600/x)+30=10x  multiply each term by x
1800+30x=10x^2  divide each term by 10
x^2-3x-180=0  quadratic in standard form and it can be factored
(x-15)(x+12)=0
x=-12-------------------no good!  wages are positive
and
x=$15---------------------plumber's hourly wages
x-3=15-3=$12------------apprencice's hourly wages
substitute x=15 into eq1
15y=600
y=40  hours worked by plumber
y+10=40+10=50--------------hours worked by apprentice

CK
50*12=40*15=600
600=600=600
Hope this helps---ptaylor