Respuesta :
Answer: No Solutions
Step-by-step explanation:
Define Variables:
May choose any letters.
\text{Let }d=
Let d=
\,\,\text{the number of dimes}
the number of dimes
\text{Let }q=
Let q=
\,\,\text{the number of quarters}
the number of quarters
\text{\textquotedblleft at most 25 coins"}\rightarrow \text{25 or fewer coins}
“at most 25 coins"→25 or fewer coins
Use a \le≤ symbol
Therefore the total number of coins, d+qd+q, must be less than or equal to 25:25:
d+q\le 25
d+q≤25
\text{\textquotedblleft a minimum of \$4.45"}\rightarrow \text{\$4.45 or more}
“a minimum of $4.45"→$4.45 or more
Use a \ge≥ symbol
One dime is worth $0.10, so dd dimes are worth 0.10d.0.10d. One quarter is worth $0.25, so qq quarters are worth 0.25q.0.25q. The total 0.10d+0.25q0.10d+0.25q must be greater than or equal to \$4.45:$4.45:
0.10d+0.25q\ge 4.45
0.10d+0.25q≥4.45
\text{Plug in }\color{green}{17}\text{ for }d\text{ and solve each inequality:}
Plug in 17 for d and solve each inequality:
Isabella has 17 dimes
\begin{aligned}d+q\le 25\hspace{10px}\text{and}\hspace{10px}&0.10d+0.25q\ge 4.45 \\ \color{green}{17}+q\le 25\hspace{10px}\text{and}\hspace{10px}&0.10\left(\color{green}{17}\right)+0.25q\ge 4.45 \\ q\le 8\hspace{10px}\text{and}\hspace{10px}&1.70+0.25q\ge 4.45 \\ \hspace{10px}&0.25q\ge 2.75 \\ \hspace{10px}&q\ge 11 \\ \end{aligned}
d+q≤25and
17+q≤25and
q≤8and
0.10d+0.25q≥4.45
0.10(17)+0.25q≥4.45
1.70+0.25q≥4.45
0.25q≥2.75
q≥11
\text{It is not possible to have }q\le 8\text{ AND to have }q\ge 11\text{.}
It is not possible to have q≤8 AND to have q≥11.
\text{Therefore there is NO SOLUTION}
Therefore there is NO SOLUTION
