Answer:
The jar has 32 dimes and 18 quarters
Step-by-step explanation:
To solve this problem we can create a system of linear equations in terms of two-variables (say [tex]x[/tex] and [tex]y[/tex]) and solve it. To begin let us analyze the problem further. We know that the values of each coin type are:
Dimes ( [tex]x[/tex] ) = $0.10
Quarters( [tex]y[/tex] ) = $0.25
Total Value = $7.70
Total Coins = 50
Now let us set up our system of equations as:
[tex]0.10x+0.25y=7.70[/tex] Eqn.(1)
[tex]x+y=50[/tex] Eqn.(2)
Lets take Eqn.(2) and rerrange it to solve for [tex]x[/tex] as:
[tex]x=50-y[/tex] Eqn.(3)
Now lets plug this, in Eqn.(1) so we get the value of [tex]y[/tex] as:
[tex]0.10(50-y)+0.25y=7.70\\\\5-0.10y+0.25y=7.70\\\\-0.10y+0.25y=7.70-5\\\\0.15y=2.7\\\\[/tex]
[tex]y=\frac{2.70}{0.15} \\\\y=18[/tex]
Plugging in [tex]y=18[/tex] back in Eqn.(3) we finally have:
[tex]x=50-18\\x=32[/tex]
Thus we conclude that in the jar the coins are:
Dimes ( [tex]x[/tex] ) [tex]=32[/tex]
Quarters( [tex]y[/tex] ) [tex]=18[/tex]