Respuesta :
Given data:
Total: $41.70
She has one more than three times as many dimes (d) as she has nickels (n):
[tex]1+3d=n[/tex]she has five times as many quarters(q) as nickels(n):
[tex]5q=n[/tex]1 d is equivalient to $0.10
1 n is equivalent to $0.05
1 q is equivalent to $0.25
[tex]0.10d+0.05n+0.25q=41.70[/tex]Use the next equations to solve the problem
[tex]\begin{gathered} 0.10d+0.05n+0.25q=41.70 \\ 1+3d=n \\ 5q=n \end{gathered}[/tex]1. Solve d in the second equation:
[tex]\begin{gathered} 1+3d=n \\ 3d=n-1 \\ d=\frac{n}{3}-\frac{1}{3} \end{gathered}[/tex]2. Solve q in the third equation:
[tex]\begin{gathered} 5q=n \\ q=\frac{n}{5} \end{gathered}[/tex]3. Substitute the d and q in the first equation by the values you get in steps 1 and 2:
[tex]0.10(\frac{n}{3}-\frac{1}{3})+0.05n+0.25(\frac{n}{5})=41.70[/tex]4. Solve n:
[tex]\begin{gathered} \frac{0.10n}{3}-\frac{0.10}{3}+0.05n+\frac{0.25n}{5}=41.70 \\ \\ \text{add (0.10/3) in both sides of the equation:} \\ \frac{0.10n}{3}-\frac{0.10}{3}+\frac{0.10}{3}+0.05n+\frac{0.25n}{5}=41.70+\frac{0.10}{3} \\ \\ \frac{0.10n}{3}+0.05n+\frac{0.25n}{5}=\frac{125.1+0.10}{3} \\ \\ \frac{0.10n}{3}+0.05n+\frac{0.25n}{5}=\frac{125.2}{3} \\ \\ \frac{0.10n+0.15n}{3}+\frac{0.25n}{5}=\frac{125.2}{3} \\ \\ \frac{0.25n}{3}+\frac{0.25n}{5}=\frac{125.2}{3} \\ \\ \frac{1.25n+0.75n}{15}=\frac{125.2}{3} \\ \\ \frac{2n}{15}=\frac{125.2}{3} \\ \\ \text{Multiply both sides of the equation by 15} \\ 15(\frac{2n}{15})=15(\frac{125.2}{3}) \\ \\ 2n=626 \\ \\ \text{Divide both sides of the equation into 2:} \\ \frac{2}{2}n=\frac{626}{2} \\ \\ n=313 \end{gathered}[/tex]5. Use the value of n=313 to solve d and q:
[tex]\begin{gathered} d=\frac{n}{3}-\frac{1}{3} \\ \\ d=\frac{313}{3}-\frac{1}{3} \\ \\ d=\frac{312}{3} \\ \\ d=104 \end{gathered}[/tex][tex]undefined[/tex]