Jane has 15 coins which include $2-coins and $5-coins only. If the total value of $2-coins is less than that of $5-coins by $33, how many $5-coins does Jane have?
Hello
let x be the amount of $2-coins
y the amount of $5-coins
Jane has 15 coins, therefor
x+y = 15
Suppose you have x=6 $2-coins, then its worth is 6*2 =$12, and you have y=9 coins (actually this is the solution) then you have 9*5 = $45. You know the difference between the worth of all your $2 coins and 5$-coins is 33$.
Hence
2x = 5y-33
(2 and 5 because of the worth of x and y)
You need to solve the equations
x+y = 15
2x = 5y-33
Regards,
Rapha
You're welcome
this is the first equation: x+y=15, solving for x => x = 15-y
now we know x, therefor:
2(15-y)+33 = 5y
and calculate this
30 - 2y +33 = 5y // +2y
63 = 7y // :7
63/7 = y
9 = y
Now our x depends on y, because we got x = 15-y. We know y exactly and thus
x = 15-9 = 6
Done.
Rapha