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2x + 6 - 5x = 12


i keep getting no solution...but the options are

x =

- 6

-2

2

6

if it is any one of those answers can you please explain to me how you got that answer. because i am studying for my test tomorrow and i really need to know this. thank you.!!

2007-12-03 12:06:30 · 8 answers · asked by Anonymous in Science & Mathematics Mathematics

8 answers

1. 2x + 6 - 5x = 12 original equation
2. -3x + 6 = 12 combine x terms (add)
3. -3x = 6 isolate x on one side
4. x = -2

2007-12-03 12:11:11 · answer #1 · answered by whizkid66 3 · 0 0

Simplifying
2x + 6 + -5x = 12

Reorder the terms:
6 + 2x + -5x = 12

Combine like terms: 2x + -5x = -3x
6 + -3x = 12

Solving
6 + -3x = 12

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-6' to each side of the equation.
6 + -6 + -3x = 12 + -6

Combine like terms: 6 + -6 = 0
0 + -3x = 12 + -6
-3x = 12 + -6

Combine like terms: 12 + -6 = 6
-3x = 6

Divide each side by '-3'.
x = -2

Simplifying
x = -2

2007-12-03 12:18:20 · answer #2 · answered by Anonymous · 0 0

2x + 6 - 5x = 12
combine like terms (2x and -5x)
-3x + 6 = 12
subtract 6 from both sides
-3x = 6
divide by -3
x = -2

2007-12-03 12:09:35 · answer #3 · answered by Bollywood Masti 4 · 0 0

2x + 6 - 5x = 12
if your solving for x, first you have to combine like terms.
in this case, 2x and 5x are like terms.

2x - 5x = -3x

now the problem is;
-3x + 6 = 12

you have to subract 6 from both sides to get x by itself.
-3x + 6 - 6 = 12 -6
-3x=6
then you divide both sides to get x alone.
-3x / -3 = 6/-3
x = -2.

2007-12-03 12:12:16 · answer #4 · answered by Tori 2 · 0 0

2x-5x=12-6 (Subtract 6 from both sides)
-3x=6 (Simplify)
x=-2 (Divide by -3)

2007-12-03 12:09:56 · answer #5 · answered by someone2841 3 · 0 0

2x-5x is -3x

-3x + 6 = 12
-6 -6

-3x = 6
/-3 /-3

x = -2

2007-12-03 12:09:42 · answer #6 · answered by Birdie 2 · 0 0

2x+6-5x=12
6-3x=12
-3x=12-6
x=12/-3
x=-2

2007-12-03 12:09:49 · answer #7 · answered by phillytommba 2 · 0 0

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2016-10-19 01:37:05 · answer #8 · answered by Anonymous · 0 0

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