Respuesta :
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:
[tex]\left \{ {{25q+10d=1,310} \atop {q=3d+15}} \right.[/tex]
Applying the Substitution method, we can substitute the second equation into the first one and solve for "d":
[tex]25(3d+15)+10d=1,310\\\\75d+375+10d=1,310\\\\85d=1,310-375\\\\d=\frac{935}{85}\\\\d=11[/tex]
Finally, we must substitute the value of "d" into the second equation to find the value of "q". Then:
[tex]q=3(11)+15\\\\q=33+15\\\\q=48[/tex]
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 : https://brainly.com/question/12826246
- Elimination Method : https://brainly.com/question/11233927
- Sum of The Ages : https://brainly.com/question/11240586
Answer details
Grade: High School
Subject: Mathematics
Chapter: Simultaneous Linear Equations
Keywords: Simultaneous , Elimination , Substitution , Method , Linear , Equations