by barnamah » Sat 09 April, 2011 14:37
quarters is 0.25 and dime is 0.1 so
let quarter be Q and dime be D so we have:
`0.25Q+0.1D=3.40` (EQ1)
Also we have another information which is you have 22 coins of both types
Q+D=22 (EQ2)
Now we have 2 equations. We have 2 unknown Q and D and we got 2 equation so we can solve it.
First we have 0.25Q and we can multiply the EQ2 by -0.25 (both side of the equation) which looks
`-0.25Q-0.25D=-0.25(22) `
`-0.25Q-0.25D=-5.5 ` (EQ3)
Now we can add both equation EQ1 and EQ3
`0.25Q+0.1D=3.40`
`-0.25Q-0.25D=-5.5 `
================
`0-0.15D=-2.1`
Now we have `-0.15D=-2.1` dividing both side by -0.15 we have:
`(-0.15D)/-0.15=-2.1/-0.15` the -0.15 on the left cancels out and we have:
`D=-2.1/-0.15=14`
D=14 So we have 14 dimes
Now using d=14 sub it in EQ2 which is Q+D=22 we have:
`Q+14=22` subtracting both side by -14 we have :`Q+14-14=22-14` which the 14 on the left cancels out and we left with:
`Q=22-14=8`
Q=8
So you got 8 Quarters.
Let's test it 8*0.25=$2
And you got 14 dimes
14*0.1=1.4
Adding 1.4+2=3.4
Indeed the answer is right and you got 14 Dims and 8 Quarters.
Explore and know. That is asked.