ok this is really confusing me
1. Leo held a garade sale. He priced all the items at a dim or a quarter. His sales totaled $23.95 for 139 coins. How many dimes and quarters did he have?
your supposed to solve my graphine. subsitution or elimination
d dimes and q quarters
d+q = 139
10 d + 25 q = 2395
I would solve the first equation for d = (139 - q)
then substitute that value for d in the second equation
however you could just as well multiply the first equation, both sides, all terms, by ten. the if you subtract the first equation from the second, you eliminate d.
To solve this problem, we can use a system of equations.
Let's assume that the number of dimes is represented by 'D' and the number of quarters is represented by 'Q'.
Since Leo priced all the items at a dime or a quarter, we can form the following equation based on the total number of coins:
D + Q = 139 (Equation 1)
The total sales from the dimes and quarters is $23.95. Since a dime is $0.10 and a quarter is $0.25, we can write the equation for the total sales as:
0.10D + 0.25Q = 23.95 (Equation 2)
Now we have a system of two equations (Equation 1 and Equation 2).
To solve this system, we can use substitution or elimination method.
Let's solve it using the substitution method:
1. Solve Equation 1 for D in terms of Q:
D = 139 - Q
2. Substitute the value of D in Equation 2:
0.10(139 - Q) + 0.25Q = 23.95
Simplifying the equation:
13.9 - 0.10Q + 0.25Q = 23.95
-0.10Q + 0.25Q = 23.95 - 13.9
0.15Q = 10.05
Divide both sides by 0.15 to isolate Q:
Q = 10.05 / 0.15
Q = 67
Now we know that Leo had 67 quarters.
To find the number of dimes, substitute the value of Q in Equation 1:
D = 139 - 67
D = 72
Therefore, Leo had 72 dimes and 67 quarters.