Question

Use the three steps to solve the problem. Betty has 10 more dimes than quarters. If she has $3.45, how many coins does she have?

Answer 1

Step 1-------> set variables.
Let
x-----------> number of dimes
y-----------> number of quarters

Step 2: set the equation and solve for the variable
we know that
0.10 x+0.25 y=3.45---------------> 10x+25y=345 equation (1)
x=y+10 equation (2)
substituting 2 in 1
10[y+10]+25y=345----> 10y+100+25y=345
35y=345-100-----------> y=245/35=7
x=y+10--------> x=7+10=17

Step 3: plug in the value of x and y from last step into the variables.

y=7--------> number of quarters coins
x=17-------> number of dimes coins

She has (7+17)=24 coins in total

Answer 2

If we assume the quarters are x, then the dimes are 10+x
Thus, 0.25x + 0.1(10+x) = 3.45
= 0.25x +1 +0.1 x = 3.45
= 0.35 x = 3.45 -1
= 0.35x = 2.45
x = 7
Therefore, the quarters are 7, while the dimes were 17,
Thus a total of 24 coins