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

3 Answers
May 9, 2018

Answer:

#x = 6#

Explanation:

The distributive property is shown here:
cdn.virtualnerd.com

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!

May 9, 2018

Answer:

#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#

May 9, 2018

Answer:

#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)#