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Travis has a total of $60 in dimes and quarters. If he could switch the numbers of dimes with the number of quarters, he would have $87. How many of each coin dose he have?

Respuesta :

perixwinkle
For this, L and T are going to be the two amounts that the dimes and quarters had. We're are going to use systems (two equations) to solve this and we're are going to use elimination.
The equations for this would be:
.1T+.25L=60
.25T+.1L=87
To eliminate, let's multiply the second equation by -4 and the first by 10 so we can eliminate T. 
1T+2.5L=600
-1T-.4L= -348
2.1L=252
252÷2.1= 120 = L
One of the amounts is 120, let's plug it in to find the other amount.
.25T+.1(120)= 87
.25T+12=87
.25T=75
75÷.25=300=T
Let's see if we're right by plugging it in.
300(.25)+ 120(.1)
75+12
87
300(.1)+ 120(.25)
30+30
60
He has 300 dimes and 120 quarters.

The number of dimes is 300 and the number of quarters is 120.

1 dimes = $0.1

1 quarters = $0.25

He has a total of $60 in dimes and quarters.

let

the number of dimes = x

the number of quarters = y

Therefore, the initial total will be

  • 0.1x + 0.25y = 60

Now, assuming the numbers of dimes and quarters are switched. Therefore, the total will be as follows

  • 0.1y + 0.25x = 87

Combine the 2 equations,

0.1x + 0.25y = 60

0.25x  + 0.1y = 87

0.1x = 60 - 0.25y

x = 600 - 2.5y

0.25( 600 - 2.5y) + 0.1y = 87

150 - 0.625y + 0.1y = 87

-0.525 y = -63

y = 120

0.1x + 0.25(120) = 60

0.1x + 30 = 60

0.1x = 30

x = 30 / 0.1

x = 300

number of dimes = 300

number of quarters = 120

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