i have 30 coins made up of 5c and 20c pieces that total $3.45. how many of each do i have?????
number of 5c pieces --- x
number of 20c pieces --- 30-x
5x + 20(30-x) = 345
Expand
5x+600−20x=345
Simplify 5x+600−20x to −15x+600
−15x+600=345
Subtract 600 from both sides
−15x=345−600
Simplify 345−600 to −255
−15x=−255
Multiply both sides by −1
15x=255
Divide both sides by 15
x=25515
Simplify 255/15 to 17
x=17
To solve this problem, we can use a system of equations to represent the given information. Let's assign variables to represent the number of 5c coins and 20c coins.
Let's say x represents the number of 5c coins and y represents the number of 20c coins.
We know that there are a total of 30 coins, so the first equation is:
x + y = 30
We also know that the total value of the coins is $3.45, which can be expressed as:
(0.05)x + (0.20)y = 3.45
To solve this system of equations, we can use the substitution or elimination method.
Let's use the elimination method to eliminate the variables. Multiply the first equation by 20 to eliminate x:
20x + 20y = 600
Now subtract this equation from the second equation:
(0.05)x + (0.20)y - (20x + 20y) = 3.45 - 600
-19.95x = -596.55
Divide both sides of the equation by -19.95:
x = -596.55 / -19.95
x ≈ 29.9
Since we cannot have a fraction of a coin, let's round x to the nearest whole number:
x ≈ 30
Now substitute the value of x back into the first equation to solve for y:
30 + y = 30
y = 30 - 30
y = 0
Therefore, you have approximately 30 coins of 5c and 0 coins of 20c.
number of 5c pieces --- x
number of 20c pieces --- 30-x
5x + 20(30-x) = 345
solve for x