Twice a number added to half of itself equal 20. Find the number

Let assume number = x

2x + x/2 = 20
4x + x = 40
5x = 40
X = 8

2x+x/2=20

3x/2=20
3x=20×2
3x=40
X=40/3

Twice a number added to half of equals 20 find the number

Dash

let the number we dont know = x

2x+x/2=20
now multply 2 from both sides
2(2x)+x=2(20)
4x+x=40
now add 4x to x
5x=40
now divide both sides from 5 to eliminate coifficient
5x/5=40/5
now we get 8 as an answer
x+8
also subscribe my youtube channel
armustafa423 gamer
plsssssssssss

Let assume number = x

2x + x/2 = 20
now multply 2 from both sides
2(2x+x/2)=2(20)
4x+2x/2=40
4x + x = 40
5x = 40
X = 8

To find the number, we can set up an equation based on the given information. Let's call the number "x".

According to the given information, "Twice a number added to half of itself equal 20". Mathematically, we can express this as:

2x + (1/2)x = 20

Now, we need to solve this equation to find the value of x.

To do that, we can simplify the equation by combining like terms:

(2 + 1/2)x = 20

To remove the fraction, we can multiply both sides of the equation by the denominator of the fraction, 2:

(2 + 1/2)x * 2 = 20 * 2

Simplifying further:

(5/2)x = 40

Now, we can isolate x by dividing both sides of the equation by 5/2:

x = 40 / (5/2)

To divide by a fraction, we can multiply by its reciprocal:

x = 40 * (2/5)

Simplifying the product:

x = 16

Therefore, the number is 16.

2x + x/2 = 20

13