Create a linear system to model this situation. Then use substitution to solve the linear system to solve the problem.
- Bobbie has been saving dimes & quarters to buy a new toy. She has a total of 28 dimes and quarters, with a value of $4.30. How many of each type of coin does Bobbie have?
dimes --- x
quarters ---- 28-x
10x + 25(28-x) = 430
etc
To create a linear system to model this situation, let's use the following variables:
Let "d" represent the number of dimes Bobbie has.
Let "q" represent the number of quarters Bobbie has.
From the problem statement, we can create two equations:
1. The total number of dimes and quarters Bobbie has is 28:
d + q = 28
2. The total value of the dimes and quarters Bobbie has is $4.30.
The value of one dime is $0.10, so the value of all the dimes is 0.10d.
The value of one quarter is $0.25, so the value of all the quarters is 0.25q.
Hence, the equation for the total value is:
0.10d + 0.25q = 4.30
Now, we have a system of two linear equations:
d + q = 28
0.10d + 0.25q = 4.30
To solve this system using substitution, we can solve one of the equations for one variable, and then substitute that expression into the other equation.
Let's solve the first equation for d:
d = 28 - q
Now, substitute this value of d into the second equation:
0.10(28 - q) + 0.25q = 4.30
Simplify and solve for q:
2.80 - 0.10q + 0.25q = 4.30
0.15q = 1.50
q = 10
Now substitute the value of q back into the first equation to find d:
d + 10 = 28
d = 18
Therefore, Bobbie has 18 dimes and 10 quarters to buy a new toy.