Let x be the number of half dollar and y, the number of quarter dollar
Thus,
[tex]x+y=17\text{ -----eq i)}[/tex]Half dollar is 50cent, quarter dollar is 25cents,
Thus,
[tex]0.5x+0.25y=5.50\text{ -----eq i}i)[/tex]Solving the two(2) equations simultaneously; we have:
From eq i)
[tex]\begin{gathered} x+y=17 \\ x=17-y\text{ -----eq }iii) \end{gathered}[/tex]Put eq iii) into ii), we have:
[tex]\begin{gathered} 0.5x+0.25y=5.50 \\ 0.5(17-y)+0.25y=5.50 \\ 8.5-0.5y+0.25y=5.50 \\ 8.5-0.25y=5.50 \\ 8.5-5.50=0.25y \\ 0.25y=3 \\ y=\frac{3}{0.25} \\ y=12 \end{gathered}[/tex]From eq iii)
[tex]\begin{gathered} x=17-y \\ x=17-12 \\ x=5 \end{gathered}[/tex]Hence, there are 5 half dollars and 12 quarter dollars inside the cup