Question 7998: Can you please help me? I am trying to help my daughter with 8th grade algebra. Her question is this/
The sum of three numbers is 120. the second of the numbers is 8 less than the first, and the third is 4 more than the first. What are the first number(s)?
This sounds like a riddle to me
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! First, let the three numbers be x, y, and z.
X is the first number.
y is the second number.
z is the third number.
From the description of the "riddle", we have:
 The sum of three numbers is 120.
 The second number, y, is (=) 8 less than (subtract 8 from) the first number (x).
 The third number, z, is (=) 4 more than (add 4 to) the first number (x).
So, intead of writing we can replace the y and the z with their equivalent expressions from above:
 Now we can solve for x, the first number, and once we have that, we can easily find the 2nd and 3rd numbers.
Of course, your problem asks only for the first number(s)?
Let's solve the equation:
Collect like-terms. Add up the x's and the numbers.
 Add 4 to both sides of the =.
 Now divide both sides by 3.
x = 41 1/3 That's 41 and one third.
Check:
x+(x-8)+(x+4) = 120
41 1/3 + 33 1/3 + 45 1/3 = 120
|
|
|
