the sum of a number x and twice another is 20. If the produt of these numbers is not more than 48, what are all possible value of x ?
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To solve this problem, we can set up a system of equations based on the given information.
Let's call the first number x, and the second number y.
From the problem statement, we know that the sum of x and twice y is 20. This can be written as the equation:
x + 2y = 20 ----(Equation 1)
We are also given that the product of these numbers (xy) is not more than 48. Mathematically, we can write this as:
xy ≤ 48 ----(Equation 2)
Now, let's solve this system of equations:
1. First, we'll simplify Equation 1 by subtracting x from both sides:
2y = 20 - x
2y = 20 - x ----(Equation 3)
2. Next, we'll solve Equation 3 for y by dividing both sides by 2:
y = (20 - x) / 2 ----(Equation 4)
3. Now, let's substitute Equation 4 into Equation 2:
x * [(20 - x) / 2] ≤ 48
Now, we can simplify this inequality:
x(20 - x) ≤ 96
20x - x^2 ≤ 96
x^2 - 20x + 96 ≥ 0 ----(Equation 5)
4. We can solve Equation 5 by factoring or using the quadratic formula. The factored form is:
(x - 12)(x - 8) ≥ 0
Now, we have two critical points, x = 12 and x = 8. These points divide the number line into three intervals:
Interval 1: x < 8
Interval 2: 8 ≤ x ≤ 12
Interval 3: x > 12
5. To determine the possible values of x, we need to find out in which intervals the inequality holds true. Since we want x to be a whole number, we can ignore the first interval (x < 8).
For the second interval (8 ≤ x ≤ 12), let's test a few values within that range. If we plug x = 8 into the inequality, we get:
(8 - 12)(8 - 8) ≥ 0
-4*0 ≥ 0
0 ≥ 0 (True)
So, x = 8 is a possible value.
For the third interval (x > 12), let's test a few values. If we plug x = 13 into the inequality, we get:
(13 - 12)(13 - 8) ≥ 0
1*5 ≥ 0
5 ≥ 0 (True)
So, x = 13 is also a possible value.
Therefore, the possible values of x that satisfy both conditions are x = 8 and x = 13.