tHE SUM OF THREE NUMBERS IS 75. THE SECOND NUMBER IS 5MORE THAN THE FIRST AND THE THIRD IS THREE TIMES THE SECOND. FIND THE NUMBERS.

Don't know what all this is.

let the first be x
then the second is x+5
and the third = 3(x+5)

x + x+5 + 3(x+5) = 75

x+x+5+3x+15 = 75
5x = 55
x = 11

the numbers are 11, 16, and 48

in yo mouth

child touchery96969696969969

Why did the numbers go to therapy? Because they couldn't find their sum! Anyway, let's solve this math puzzle together.

Let's call the first number x. According to the problem, the second number is 5 more than the first, so it's x + 5. The third number is three times the second, so it's 3(x + 5).

Now, we know that the sum of the three numbers is 75. So we can set up an equation:

x + (x + 5) + 3(x + 5) = 75

Simplifying the equation,

x + x + 5 + 3x + 15 = 75
5x + 20 = 75

Subtracting 20 from both sides,

5x = 55

Dividing both sides by 5,

x = 11

So the first number is 11. Now we can find the other numbers.

The second number is 5 more than the first, so it's 11 + 5 = 16.
The third number is three times the second, so it's 3(16) = 48.

Therefore, the three numbers are 11, 16, and 48. So they finally found their sum!

To find the numbers, let's assign variables to each number. Let's say the first number is "x", the second number is "y", and the third number is "z".

According to the given information:
1) The sum of three numbers is 75: x + y + z = 75
2) The second number is 5 more than the first: y = x + 5
3) The third number is three times the second: z = 3y

Now, let's use these equations to solve for the numbers.

First, substitute equation (2) in equation (3):
z = 3(x + 5)

Next, substitute equations (2) and (3) in equation (1):
x + (x + 5) + 3(x + 5) = 75

Simplify the equation:
x + x + 5 + 3x + 15 = 75
5x + 20 = 75
5x = 75 - 20
5x = 55

Divide both sides of the equation by 5:
x = 55 / 5
x = 11

Now, substitute the value of x back into equation (2):
y = x + 5
y = 11 + 5
y = 16

Finally, substitute the values of x and y back into equation (3):
z = 3y
z = 3(16)
z = 48

Therefore, the three numbers are 11, 16, and 48.

Let A,B and C be the numbers.

1.A%2BB%2BC=75
B=5%2B4A
C=2A
Substitute into eq. 1,
A%2B5%2B4A%2B2A=75
7A%2B5=75
7A=70
did i help confusing aye?


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B=5%2B4%2810%29
highlight%28+B=45%29
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C=2%2810%29
highlight%28+C=20%29