Let be "q" the number of quarters Bethany has and "d" the number of dimes she has.
You know that 1 quarter is $0.25 and 1 dime is $0.10. Then, since Bethany has $2.80 in quarters and dimes, you can set up the following equation:
[tex]0.25q+0.10d=2.80[/tex]Knowing that the number of dimes is 7 less than the number of quarters, you can set up the second equation:
[tex]d=q-7[/tex]Then you have the following System of equations:
[tex]\mleft\{\begin{aligned}0.25q+0.10d=2.80 \\ d=q-7\end{aligned}\mright.[/tex]In order to solve it, you can use the Substitution method:
1. Substitute the second equation into the first equation.
2. Solve for "q".
Then:
[tex]\begin{gathered} 0.25q+0.10d=2.80 \\ 0.25q+0.10(q-7)=2.80 \\ 0.25q+0.10q-0.7=2.80 \\ 0.35q=2.80+0.7 \\ q=\frac{3.5}{0.35} \\ q=10 \end{gathered}[/tex]3. Substitute the value of "q" into the second equation and evaluate, in order to find the value of "d":
[tex]\begin{gathered} d=q-7 \\ d=10-7 \\ d=3 \end{gathered}[/tex]The answer is:
- Number of dimes: 3
- Number of quarters: 10