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