Solve [tex]\( 5ax + 3ax = 4ax + 12 \)[/tex] for [tex]\( x \)[/tex]. Assume [tex]\( a \neq 0 \)[/tex].

A. [tex]\( x = \frac{1}{a} \)[/tex]
B. [tex]\( x = \frac{3}{a} \)[/tex]
C. [tex]\( x = a \)[/tex]
D. [tex]\( x = 3a \)[/tex]



Answer :

To solve the equation [tex]\(5ax + 3ax = 4ax + 12\)[/tex] for [tex]\(x\)[/tex]:

1. Combine like terms on both sides:
[tex]\[ 5ax + 3ax = 8ax \][/tex]
So, the equation becomes:
[tex]\[ 8ax = 4ax + 12 \][/tex]

2. Move all terms involving [tex]\(x\)[/tex] to one side:
Subtract [tex]\(4ax\)[/tex] from both sides:
[tex]\[ 8ax - 4ax = 12 \][/tex]
Simplifying the left-hand side, we get:
[tex]\[ 4ax = 12 \][/tex]

3. Solve for [tex]\(x\)[/tex]:
Divide both sides of the equation by [tex]\(4a\)[/tex]:
[tex]\[ x = \frac{12}{4a} \][/tex]
Simplifying the fraction, we find:
[tex]\[ x = \frac{3}{a} \][/tex]

Therefore, the solution is [tex]\( x = \frac{3}{a} \)[/tex], which corresponds to option B.

So, the correct answer is:
[tex]\[ \boxed{\frac{3}{a}} \][/tex]