How do you do these linear equations?

Question

2(3x+6)+8=6x

and

3(x+2)+1=2x+7+x

i dont know how to do these because the answers come out weird with the x's canceled out help?

Answer 1

EQ 1 is a contradiction, it has no solution.

EQ 2 is an identity and has an infinite number of solutions.

                  

Answer 2

2(3x+6) + 8 = 6x

6x +12 +8 = 6x

6x + 20 = 6x

20 = 6x - 6x

20 = 0   Uhm, Houston, we have a problem.

3(x+2)+1 = 2x +7 +x

3x + 6 +1 = 3x + 7

3x + 7 = 3x +7  Well that is certainly true

3x -3x = 7 -7

0 = 0 Also True, no real solution since any value works for X.

Answer 3

1)

2(3x + 6) + 8 = 6x

2*3x + 2*6 + 8 = 6x

6x + 12 + 8 = 6x

6x - 6x = -12 - 8

0 = -20

(no solutions)

2)

3(x + 2) + 1 = 2x + 7 + x

3*x + 3*2 + 1 = 2x + x + 7

3x + 6 + 1 = 3x + 7

3x - 3x = -7 + 7

0 = 0

(infinite solutions)

Answer 4

Your first equation has no x value that will make the equation true.  [When  you simplify you end up with 20 = 0; which is an untrue statement.]

Your second equation is true for any value of x.  When it's simplified you get 3x+7 = 3x+7.

Answer 5

Basically if the x's cancel out, it means that for ANY value of x, the equation will always be satisfied or that for NO values will the equation will always be satisfied.

This might be a sign that there's a problem with the equation, so you can check for any previous errors.