Let

q

and

d

be the number of quarters and dimes in Sam's pocket, respectively. Since he has a total of

12

coins, we can set the following equation.

q + d = 12

We know each quarter is worth

0.25

dollars, and each dime is worth

0.10

dollars. To find out how much money Sam has in his pocket, we multiply

q

by

0.25

and

d

by

0.10,

then add them up.

Total = 0.25 q + 0.10 d

Also we are told that Sam has a total of

1.95

dollars. We can substitute that into the equation above.

0.25 q + 0.10 d = 1.95

Therefore, we've obtained a system of two equations.

{q + d = 12 0.25 q + 0.10 d = 1.95 (I) (II)

We can solve this system using the Elimination Method, but first we will multiply the second equation by

10 .

{q + d = 12 0.25 q + 0.10 d = 1.95 (I) (II)

$(II):$ $LHS \cdot 10 = RHS \cdot 10$

{q + d = 12 2.5 q + d = 19.5

$(II):$ Subtract $(I)$

{q + d = 12 2.5 q + d - (q + d) = 19.5 - 12

$(II):$ Distribute $(- 1)$

{q + d = 12 2.5 q + d - q - d = 19.5 - 12

$(II):$ Subtract terms

{q + d = 12 1.5 q = 7.5

$(II):$ $LHS / 1.5 = RHS / 1.5$

{q + d = 12 q = 5

The number of quarters in Sam's pocket is

5 .

Substituting

q = 5

into the first equation, we will find the value of

d .

{q + d = 12 q = 5

$(I):$ $q = 5$

{5 + d = 12 q = 5

$(I):$ $LHS - 5 = RHS - 5$

{d = 7 q = 5

Therefore, Sam has

7

dimes in his pocket along with the

5

quarters.

Exercise 64 Page 302

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