Solve for [tex]$x$[/tex]. State any restrictions on the variable.

[tex]\[a x-5=b x+7\][/tex]

a. [tex]$x=\frac{12}{a-b} ; a \neq b$[/tex]

b. [tex]$x=\frac{12}{a+b} ; a \neq -b$[/tex]

c. [tex][tex]$x=\frac{a+b}{2} ;$[/tex][/tex] no restrictions

d. [tex]$x=\sqrt{\frac{12}{a-b}} ; a \neq b$[/tex]



Answer :

To solve the equation [tex]\(a x - 5 = b x + 7\)[/tex] for [tex]\(x\)[/tex], follow these steps:

### Step 1: Isolate terms containing [tex]\(x\)[/tex].
Subtract [tex]\(b x\)[/tex] from both sides to bring all [tex]\(x\)[/tex]-terms to one side of the equation:
[tex]\[a x - b x - 5 = 7\][/tex]

### Step 2: Simplify the equation.
Combine the [tex]\(x\)[/tex]-terms on the left-hand side:
[tex]\[(a - b)x - 5 = 7\][/tex]

### Step 3: Isolate [tex]\(x\)[/tex].
Add 5 to both sides to move the constant term to the right-hand side:
[tex]\[(a - b)x = 7 + 5\][/tex]
[tex]\[(a - b)x = 12\][/tex]

### Step 4: Solve for [tex]\(x\)[/tex].
Divide both sides by [tex]\((a - b)\)[/tex]:
[tex]\[x = \frac{12}{a - b}\][/tex]

### Step 5: State any restrictions on the variable.
For the solution to be valid, the denominator cannot be zero. Therefore, [tex]\(a - b \neq 0\)[/tex]. This implies that [tex]\(a \neq b\)[/tex].

### Step 6: Compare with the provided choices.
The provided choices are:

a. [tex]\(x = \frac{12}{a - b}; a \neq b\)[/tex]

b. [tex]\(x = \frac{12}{a + b}; a \neq -b\)[/tex]

c. [tex]\(x = \frac{a + b}{2};\)[/tex] no restrictions

d. [tex]\(x = \sqrt{\frac{12}{a - b}}; a \neq b\)[/tex]

The correct choice is:

a. [tex]\(x = \frac{12}{a - b}; a \neq b\)[/tex]