the sum of anumber x and twice another is 20.if the product of these numbers is not more than 48 , whate are all possible values of x?
one number ---- x
the other number --- y
x + 2y = 20
x = 20-2y or y = (20-x)/2
xy ≤ 48
(20-2y)(y) ≤ 48
-2y^2 + 20y - 48 ≤ 0
y^2 - 10y + 24 ≥ 0
(y-6)(y-4) ≥ 0
y ≤ 4 OR y ≥ 6
(20-x)/2 ≤ 4 OR (20-x)/2 ≥ 6
20-x ≤ 8 OR 20-x ≥ 12
-x ≤ -12 OR -x ≥ -8
x ≥ 12 OR x ≤ 8
{X:xless than or equal to 8}U{x: x is geater than or equal to 12}
very good
Well, well, looks like we have a math problem! Let's see if we can solve it with a touch of humor.
So, we have two numbers, let's call them x and y. The problem says that the sum of x and twice y is 20. Alright, let's write that down: x + 2y = 20.
Now, it also says that the product of these numbers is not more than 48. Okay, let's write that too: xy ≤ 48.
To find the possible values of x, let's start with the second equation. We want xy to be less than or equal to 48. That means either x or y (or both) must be less than or equal to 48. See what I did there? But hold on, there's more!
Now, let's go back to the first equation. We have x + 2y = 20. If we assume y is equal to 0, we can solve for x: x + 2(0) = 20, which gives us x = 20.
So, one possible value of x is 20. That's a pretty straightforward one.
But wait, there's more humor! You see, we could also assume that x is equal to 0. In that case, we'd have 0 + 2y = 20, which gives us y = 10.
So, another possible value of x is 0. That's like getting a free clown nose!
To summarize, the possible values of x are 0 and 20. I hope that brings a smile to your face, just like a clown at a circus!
To find the possible values of x, we can break down the problem into two equations and then solve them simultaneously.
Let's say the first number is y.
According to the problem, the sum of x and twice y is 20, which can be represented as:
x + 2y = 20 (Equation 1)
The product of x and y should not exceed 48, which can be represented as:
xy ≤ 48 (Equation 2)
Now, we will solve these equations simultaneously to find the possible values of x.
Step 1: Solve Equation 1 for x in terms of y.
x = 20 - 2y
Step 2: Substitute x in Equation 2.
(20 - 2y)y ≤ 48
Step 3: Simplify the equation.
20y - 2y^2 ≤ 48
Step 4: Rearrange the equation.
2y^2 - 20y + 48 ≥ 0
Step 5: Factor the quadratic equation.
(y - 4)(y - 6) ≥ 0
Now, we have two cases to consider:
Case 1: (y - 4)(y - 6) > 0
Solving this inequality, we find the possible values for y:
y > 6 or y < 4
Step 6: Substitute the value of y in Equation 1 to find the values of x.
For y > 6:
x = 20 - 2y
x = 20 - 2(7)
x = 20 - 14
x = 6
For y < 4:
x = 20 - 2y
x = 20 - 2(3)
x = 20 - 6
x = 14
Case 2: (y - 4)(y - 6) = 0
Solving this equation, we find the value of y:
y = 4 or y = 6
Step 7: Substitute the values of y in Equation 1 to find the values of x.
For y = 4:
x = 20 - 2y
x = 20 - 2(4)
x = 20 - 8
x = 12
For y = 6:
x = 20 - 2y
x = 20 - 2(6)
x = 20 - 12
x = 8
Therefore, the possible values of x are 6, 12, and 8.
(x)(y) ≤ 48 And x+2y=20
(20-2y)(y) ≤ 48 x=20-2y or y=(20-x)/2 choose one of the two
20y-2y^2 ≤ 48 (20-x)/2 ≥ 6 or (20-x)/2 ≤ 4
-2y^2+20y-48 ≥ 0 20-x ≥ 12 20-x ≤ 8
y^2-10y+48 ≥ 0 -x ≥ 12-20 -x ≤ 8-20
y^2-6y-4y+48 ≥ 0 x ≤ 8 x ≥ 12
y(y-6)-4(y-6) ≥ 0
(y-6)(y-4) ≥ 0 S,S = {x : x ≤ 8}U{x : x ≥ 12}
y ≥ 6 and y ≤ 4 THANK YOU!!!!