Question

Joshua has $1.20 worth of nickels and dimes. He has 6 more nickels than dimes. Graphically solve a system of equations in order to determine the number of nickels, x, commax, and the number of dimes, y, commay, that Joshua has.

Answer 1

To solve the problem, we will set up a system of equations based on the information provided.

Let:
- \( x \) = number of nickels
- \( y \) = number of dimes

From the information given, we can create the following equations:

1. **Value Equation**: The total value of the nickels and dimes is $1.20.
- The value of nickels is \( 0.05x \) (since each nickel is worth $0.05).
- The value of dimes is \( 0.10y \) (since each dime is worth $0.10).
- Thus, the equation for the total value can be written as:
\[
0.05x + 0.10y = 1.20
\]
Multiplying the entire equation by 100 to eliminate decimals gives:
\[
5x + 10y = 120
\]
or simplifying:
\[
x + 2y = 24 \quad \text{(1)}
\]

2. **Nickels & Dimes Relationship**: Joshua has 6 more nickels than dimes. This can be expressed as:
\[
x = y + 6 \quad \text{(2)}
\]

Now we have a system of two equations:
1. \( x + 2y = 24 \)
2. \( x - y = 6 \)

Now, we can solve this system of equations graphically or algebraically. For graphical solving, we need to rearrange both equations into slope-intercept form (i.e., \( y = mx + c \)):

From equation (2):
\[
x - y = 6 \implies y = x - 6 \quad \text{(3)}
\]

From equation (1):
\[
x + 2y = 24 \implies 2y = 24 - x \implies y = -\frac{1}{2}x + 12 \quad \text{(4)}
\]

Now we can arrange equations (3) and (4) for graphing:

1. **Graph of Equation (3)**: \( y = x - 6 \) (a straight line with slope 1 and y-intercept -6).
2. **Graph of Equation (4)**: \( y = -\frac{1}{2}x + 12 \) (a straight line with slope -1/2 and y-intercept 12).

### Finding the intersection:
To find the intersection point algebraically:

1. Substitute equation (3) into equation (1):
\[
x + 2(x - 6) = 24
\]
Simplifying gives:
\[
x + 2x - 12 = 24
\]
\[
3x - 12 = 24
\]
\[
3x = 36
\]
\[
x = 12
\]

2. Substitute \( x = 12 \) back into equation (2) to find \( y \):
\[
12 = y + 6 \implies y = 6
\]

### Conclusion:
Joshua has 12 nickels and 6 dimes. In summary:
- Number of nickels, \( x = 12 \)
- Number of dimes, \( y = 6 \)