Question

Peter has been saving his loose change for several days. When he counted his quarters and dimes, he found they had a total value of $13.10. The number of quarters was 15 more than 3 times the number of dimes. How many quarters and how many dimes did Peter have?

Answer 1

Peter had 48 quarters and 11 dimes

Further explanation

Simultaneous Linear Equations could be solved by using several methods such as :

• Elimination Method
• Substitution Method
• Graph Method

If we have two linear equations with 2 variables x and y , then we need to find the value of x and y that satisfying the two equations simultaneously.

Let us tackle the problem!

Let :

Number of quarters ( 25 cent coins ) = x

Number of dimes ( 10 cent coins ) = y

When he counted his quarters and dimes, he found they had a total value of $13.10.

0.25x + 0.10y = 13.10

The number of quarters was 15 more than 3 times the number of dimes.

x = 15 + 3y

If we would like to use the Substitution Method , then second equations above could be substituted into first equations.

0.25x + 0.10y = 13.10

0.25 (15 + 3y) + 0.10y = 13.10

3.75 + 0.75y + 0.10y = 13.10

0.85y = 13.10 - 3.75

0.85y = 9.35

y = 9.35 / 0.85

y = 11

At last , we could find the value of x by substituting this y value into one of the two equations above :

x = 15 + 3y

x = 15 + 3(11)

x = 15 + 33

x = 48

Learn more

• Perimeter of Rectangle : brainly.com/question/12826246
• Elimination Method : brainly.com/question/11233927
• Sum of The Ages : brainly.com/question/11240586

Answer details

Grade: High School

Subject: Mathematics

Chapter: Simultaneous Linear Equations

Keywords: Simultaneous , Elimination , Substitution , Method , Linear , Equations

Answer 2

Answer: Peter had 48 quarters and 11 dimes.

Step-by-step explanation:

Let be "q" the number of quarters and "d" the number of dimes.

We know that $13.10 in cents is 1,310 cents. Then, we can set up the following system of equations:

$\left \{ {{25q+10d=1,310} \atop {q=3d+15}} \right.$

Applying the Substitution method, we can substitute the second equation into the first one and solve for "d":

$25(3d+15)+10d=1,310\\\\75d+375+10d=1,310\\\\85d=1,310-375\\\\d=\frac{935}{85}\\\\d=11$

Finally, we must substitute the value of "d" into the second equation to find the value of "q". Then:

$q=3(11)+15\\\\q=33+15\\\\q=48$