Answer:
There are no possible solutions
Step-by-step explanation:
Given:
Minimum number of coins Alexa has =28 coins
Number of quarters Alexa has = 17 quarters
Total amount in coins = $5.05 = 505 cents
Let number of dimes be = [tex]x[/tex] coins
So we can have two inequalities.
1) Total number of coins
[tex]x+17\geq28\\[/tex] [Since minimum number of coins=28]
2) Total value of coins
[tex]10x+(25)(17)\leq505[/tex] [As 1 dime=10 cents and 1 quarter=25 cents]
[tex]10x+425\leq505[/tex]
Solving inequality (1)
Subtracting both sides by 17.
[tex]x+17-17\geq28-17[/tex]
∴ [tex]x\geq 11[/tex]
Solving inequality (2)
Subtracting both sides by 425.
[tex]10x+425-425\leq505-425[/tex]
[tex]10x\leq 80[/tex]
Dividing both sides by 10.
[tex]\frac{10x}{10}\leq \frac{80}{10}[/tex]
∴ [tex]x\leq 8[/tex]
On combining both solutions
[tex]x\geq 11[/tex] and [tex]x\leq 8[/tex] ,
we see that there are no possible solutions as number of dimes cannot be ≥11 and ≤8 at the same time.