From the question, we can say that:
[tex]\begin{gathered} D+Q=20 \\ D\times10+Q\times25=275 \end{gathered}[/tex]where D stands for the number of dimes and Q for the number of quarters.
If we isolate Q in the first equation/relation we wrote above, and substitute it in the second equation, we will be able to find the number D, as follows:
[tex]\begin{gathered} D+Q=20\to Q=20-D \\ \\ D\times10+(20-D)\times25=275\to10D+500-25D=275\to \\ \to500-15D=275\to500-275=15D\to15D=225\to D=\frac{225}{15} \\ \to D=15 \end{gathered}[/tex]Now, we can substitute the value of D in the first equation, as follows:
[tex]\begin{gathered} D+Q=15+Q=20\to Q=20-15\to \\ \to Q=5 \end{gathered}[/tex]From the present solution, we conclude that the number of dimes equals 15, and the number of quarters is 5.
