Dave Bowers collects U.S. gold coins. He has a collection of 54 coins. Some are $10 coins, and the rest are $20 coins. If the face value of the coins is $780, how many of each denomination does he have?
X = $10-coins.
54-x = $20-coins.
10x + 20(54-x) = 780.
10x + 1080 - 20x = 780.
-10x = -300.
X = 30.
54-x = 54-30 = 24
10x +20y =780. X + y =54 . Solve simultaneously x=30 and y=24. . . If u wana knw hw many gold coin7 itz lyk dis 10 x 30=$300and 20 x 24 =$480
To determine how many $10 and $20 coins Dave Bowers has, we can set up a system of two equations.
Let's assume the number of $10 coins is represented by the variable x, and the number of $20 coins is represented by the variable y.
1. From the given information, we know that the total number of coins is 54:
x + y = 54
2. We also know that the face value of the coins is $780:
10x + 20y = 780
Now we have a system of two equations that we can solve simultaneously to find the values of x and y.
There are multiple ways to solve this system of equations. Here, I will use the method of substitution:
From the first equation, we can express x as a function of y:
x = 54 - y
Substitute the value of x in the second equation:
10(54 - y) + 20y = 780
Now, we can simplify and solve for y:
540 - 10y + 20y = 780
10y = 240
y = 24
Substituting the value of y back into the first equation, we can find x:
x + 24 = 54
x = 30
Therefore, Dave Bowers has 30 $10 coins and 24 $20 coins in his collection.
t for ten and w for twenty
t + w = 54
10 t + 20 w = 780
solve the system