The sum of three positive numbers is 33. The second number is 3 greater than the first and the third is 4 times the first. Find the three numbers.
x+y+z = 33
y = x+3
z = 4x
Now substitute y and z to get
x+(x+3)+(4x) = 33
6x = 30
x = 5
so the numbers are 5,8,20
Let's solve this problem step by step.
Step 1: Set up the equation.
Let's call the first number x.
The second number is 3 greater than the first, so it is x + 3.
The third number is 4 times the first, so it is 4x.
The sum of all three numbers is 33, so we can write the equation as:
x + (x + 3) + 4x = 33.
Step 2: Simplify the equation.
Combine like terms:
6x + 3 = 33.
Step 3: Solve for x.
To isolate x, subtract 3 from both sides of the equation:
6x = 30.
Divide both sides of the equation by 6:
x = 5.
Step 4: Find the other two numbers.
Now that we know the value of x is 5, we can substitute it back into the expressions we created earlier:
First number = x = 5.
Second number = x + 3 = 5 + 3 = 8.
Third number = 4x = 4(5) = 20.
Therefore, the three numbers are 5, 8, and 20.