Taking into account the definition of a system of linear equations, there are 6 quarters in the change purse.
<h3>System of linear equations</h3>
A system of linear equations is a set of two or more equations of the first degree, in which two or more unknowns are related.
Solving a system of equations consists of finding the value of each unknown so that all the equations of the system are satisfied. That is to say, the values of the unknowns must be sought, with which when replacing, they must give the solution proposed in both equations.
<h3>Amount of quarters in the change purse</h3>
In this case, a system of linear equations must be proposed taking into account that:
- N in the amount of nickels in the change purse.
- D in the amount of dimes in the change purse.
- Q in the amount of quarters in the change purse.
On the other hand, you know:
- A change purse contains 16 coins.
- The value of the change is $2.10.
- The number of dimes is 4 fewer than the number of quarters.
The system of equations to be solved is
Equation 1: N + D + Q= 16
Equation 2: 0.05N + 0.10D + 0.25Q= 2.10
Equation 3: D= Q - 4
There are several methods to solve a system of equations, it is decided to solve it using the substitution method, which consists of clearing one of the two variables in one of the equations of the system and substituting its value in the other equation.
Substituting equation 3 into equation 1 you get:
N + Q - 4 + Q= 16 → N +2Q= 16 +4 → N + 2Q= 20
Isolating the variable N you get:
N= 20 - 2Q
Replacing this expression and equation 3 in equation 2 you get:
0.05×(20 -2Q) + 0.10×(Q -4) + 0.25Q= 2.10
Solving:
0.05×20 - 0.05×2Q + 0.10Q - 0.10×4 + 0.25Q= 2.10
1 - 0.10Q + 0.10Q - 0.40+ 0.25Q= 2.10
- 0.10Q + 0.10Q + 0.25Q= 2.10 -1 +0.40
0.25Q= 2.10 -1 +0.40
0.25Q= 1.50
Q= 1.50÷ 0.25
<u><em>Q= 6</em></u>
Finally, there are 6 quarters in the change purse.
Learn more about system of equations:
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