# janet has \$8.55 in nickels,dimes,and quarters. she has 7 more dimes than nickels and quarters combined. how many of each coin does she have?

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## To solve this problem, we need to set up equations based on the given information.

Let's assume that Janet has x nickels, y dimes, and z quarters.

We are given that Janet has \$8.55 in total. Since each nickel is worth \$0.05, each dime is worth \$0.10, and each quarter is worth \$0.25, we can set up the following equation:

0.05x + 0.10y + 0.25z = 8.55

We are also given that Janet has 7 more dimes than the combined number of nickels and quarters. Mathematically, this can be expressed as:

y = x + z + 7

Now we have two equations to solve for the values of x, y, and z.

One way to solve these equations is by substitution. We can rewrite the second equation as:

x = y - z - 7

Now, substitute x in the first equation:

0.05(y - z - 7) + 0.10y + 0.25z = 8.55

Simplify and solve for y and z:

0.05y - 0.05z - 0.35 + 0.10y + 0.25z = 8.55

0.15y + 0.20z = 8.90

Multiply the entire equation by 100 to eliminate the decimal:

15y + 20z = 890

Next, substitute the value of x in terms of y and z from the second equation into the second equation:

y = (y - z - 7) + z + 7

Simplify:

y = y

This indicates that there is no unique solution for the value of y. However, we know that the number of coins cannot be a fraction or negative. Therefore, we can try different values for y and find the corresponding values of x and z.

For example, if we assume that y = 10, we can substitute this value into the second equation to find x:

10 = x + z + 7

3 = x + z

Now, substitute these values back into the first equation:

0.05(3) + 0.10(10) + 0.25z = 8.55

0.15 + 1 + 0.25z = 8.55

0.25z = 7.40

z = 29.6

Since z cannot be a decimal value, we can conclude that the assumption y = 10 is not valid. We can continue trying different values for y until we find a valid solution.

Alternatively, you can use trial and error or a solver tool to find the appropriate values for x, y, and z that satisfy both equations.