Find three consecutive integers such that nine less than three times the third is seven more than the sum of the first and second.
ahh...those algebra days...
so, anyways, here's what u do...
1) identify the variables:
Let the 1st integer be "x"
Let the 2nd integer be "(x+1)"
Let the 3rd integer be "(x+2)"
2) now, set up the equation...
3(x+2) - 9 = x + (x+1)
3) SOLVE away!!! :D
3x + 6 - 9 = 2x + 1
3x - 3 = 2x + 1
x = 4
4) NOW the three consecutive integers will be...
{ 4, 5, 6, }
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YAY, the answer :D
now, DON'T FORGET TO HERP THE DERP!!!
solve as above, but don't forget the "seven more..."
3(x+2) - 9 = x + (x+1) + 7
To solve this problem, let's break it down into steps:
Step 1: Define the three consecutive integers
Let's assume that the first integer is "x", the second integer is "x + 1", and the third integer is "x + 2".
Step 2: Write the equation based on the problem statement
According to the problem statement, "nine less than three times the third is seven more than the sum of the first and second." This can be expressed as:
3(x + 2) - 9 = (x + x + 1) + 7
Step 3: Simplify and solve the equation
Start by simplifying both sides of the equation:
3x + 6 - 9 = 2x + 1 + 7
3x - 3 = 2x + 8
Simplify further:
3x - 2x = 8 + 3
x = 11
Step 4: Find the three consecutive integers
Now that we have found the value of "x" (which is 11), we can substitute it back into our initial assumption.
The three consecutive integers are:
First integer: x = 11
Second integer: x + 1 = 11 + 1 = 12
Third integer: x + 2 = 11 + 2 = 13
So, the three consecutive integers that satisfy the given condition are 11, 12, and 13.