Question

How do you solve `2(x-5+2)=6` using the distributive property?

Answer 1

`x = 6`

Explanation:

The distributive property is shown here:

Although the image has two terms inside the parenthesis rather than three terms in the question, we can apply it to this question.

The `color(blue)2` multiplies to everything inside the parenthesis:
`(2 * x) + (2 * -5) + (2 * 2) = 6`

Now simplify by multiplying:
`2x - 10 + 4 = 6`

Combine `color(blue)(-10 + 4)`:
`2x - 6 = 6`

Add `color(blue)6` to both sides:
`2x - 6 quadcolor(blue)(+quad6) = 6 quadcolor(blue)(+quad6)`

`2x = 12`

Divide both sides by `color(blue)2`
`(2x)/color(blue)2 = 12/color(blue)2`

`x = 6`

Hope this helps!

Answer 2

`x=6`

Explanation:

Since the `2` is outside the parentheses, that is the number that is going to be distributed. You'll multiply `2` by `x`, `2` by `-5`, and then `2` by `2`. Therefore, you should get `2x`, `-10`, and `4`.

Put those together and you have your new equation:

`2x-10+4=6`

Now, you can add `-10` and `4` to get `-6`, so your equation will be

`2x-6=6`

Add `6` to both sides. Then you'll get the equation

`2x=12`

Now you have a division by `2` on both sides to get your answer that

`x=6`

Answer 3

`x = 6`

Explanation:

Distributive Property is when you get rid of the brackets so what happens is...

You got the equation...
`color(red)(2(x - 5 + 2)) = color(blue)(6)`
You take the one with brackets...
`color(red)(2(x - 5 + 2))`
Then you times the number outside the brackets with everything inside like this...
`color(red)(2)color(blue)((x - 5 + 2)`

`color(red)(2) xx color(blue)x = 2x`
`color(red)2 xx color(blue)(-5) = -10`
`color(red)2 xx color(blue)(2)= 4`

`2x - 10 + 4`

so `color(red)(2(x - 5 + 2))` becomes `color(red)(2x - 10 + 4)`

Now you got: `2x - 10 + 4 = 6`
With this, you can now find the '`color(green)(x)`'

`color(orange)(2x - 10 + 4 = 6`
`2x color(red)(- 10 + 4) = 6`
adding the together...
`2x color(red)(- 6) = 6`

`color(orange)(2x - 6 = 6`
`2x color(red)(- 6) = color(red)(6`
moving `-6` to the other side...
`2x = color(red)(6) color(red)(+ 6`
`2x = color(red)(12`

`color(orange)(2x = 12`
`color(red)(2)x = color(red)(12`
moving `2` to the other side...
`x = color(red)(12)/color(red)(2` (12 divided by 2)
`x = color(red)(6)`