Question

You have 47 total coins for a total of $9.95. You only have quarters and dimes. Choose the two equations that make up the system of equations that represents this situation.

Answer 1

One equation that represents the total value of the coins in cents is:

0.25x + 0.10y = 995

where x is the number of quarters and y is the number of dimes.

Another equation that represents the total number of coins is:

x + y = 47

These two equations make up a system of equations that can be used to find the number of quarters and dimes that add up to $9.95 and have a total of 47 coins.

Answer 2

`\begin{cases}x+y=47\\0.25x+0.1y=9.95\end{cases}`

Explanation:

To solve this problem, we can create and solve a system of equations.

Define the variables:

• Let x be the number of quarters.
• Let y be the number of dimes.

Values of the coins:

• The value of a quarter is $0.25.
• The value of a dime is $0.10.

Given there are a total of 47 coins:

`x+y=47`

Given there is a total of $9.95 in quarters and dimes:

`0.25x+0.1y=9.95`

Therefore, the system of equations that represents the problem is:

`\begin{cases}x+y=47\\0.25x+0.1y=9.95\end{cases}`

To solve the system of equations, rewrite the first equation to isolate y:

`y=47-x`

Substitute this into the second equation to eliminate y:

`0.25x+0.1(47-x)=9.95`

Solve the equation for x to find the number of quarters:

`\begin{aligned}0.25x+0.1(47-x)&=9.95\\0.25x+4.7-0.1x&=9.95\\0.15x+4.7&=9.95\\0.15x&=5.25\\x&=35\end{aligned}`

Therefore, there are 35 quarters.

Substitute the found value of x into the first equation and solve for y:

`\begin{aligned}35+y&=47\\y&=12\end{aligned}`

Therefore, there are 12 dimes.