Jose has $6.70 in an assortment of coins. if he has 3 more quarters than pennies,11 more dimes than pennies, and 2.5 times as many nickels as pennies, how many of each type of coin does he have?in advance.

He has 10 pennies, 13 quarters , 21 dimes and 25 nickels.

Explanation

Suppose, the number of pennies is

As he has 3 more quarters than pennies, 11 more dimes than pennies and 2.5 times as many nickels as pennies....

So, the number of quarters ,

the number of dimes and

the number of nickels

We know, the value of each penny is $0.01 , value of each quarter is $0.25 , value of each dime is $0.10 and value of each nickel is $0.05

Given that, the total value of all coins is $6.70

So the equation will be.....

So, the number of pennies is 10 , number of quarters is (10+3)= 13 , number of dimes is (10+11)= 21 and number of nickels is (2.5×10)= 25

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1