To determine which of the given equations has no solutions, let's analyze each one using algebraic manipulation. Given that [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are three different numbers:
Option A: [tex]\(a x = b x + c\)[/tex]
Start by isolating the terms involving [tex]\(x\)[/tex] on one side:
[tex]\[a x - b x = c\][/tex]
Factor out [tex]\(x\)[/tex] from the left side of the equation:
[tex]\[(a - b) x = c\][/tex]
Since [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are different numbers, [tex]\(a - b\)[/tex] is non-zero. To solve for [tex]\(x\)[/tex], divide both sides by [tex]\(a - b\)[/tex]:
[tex]\[x = \frac{c}{a - b}\][/tex]
This equation has a solution [tex]\(x = \frac{c}{a - b}\)[/tex].
Option B: [tex]\(a x + b = a x + c\)[/tex]
Start by isolating the terms involving [tex]\(x\)[/tex] on one side:
[tex]\[a x + b - a x = c\][/tex]
Simplify by removing [tex]\(a x\)[/tex] from both sides:
[tex]\[b = c\][/tex]
Here, [tex]\(b = c\)[/tex] is stated to be a contradiction because [tex]\(b\)[/tex] and [tex]\(c\)[/tex] are different numbers. As a result, there cannot be any values of [tex]\(x\)[/tex] that satisfy this equation. Thus, option B has no solutions.
Option C: [tex]\(a x + b = a x + b\)[/tex]
This equation is obviously true for all values of [tex]\(x\)[/tex]. Both sides of the equation are identical:
[tex]\[a x + b = a x + b\][/tex]
Because both sides are the same, this equation holds for any value of [tex]\(x\)[/tex]. Therefore, option C has infinitely many solutions.
In conclusion, the equation that has no solutions is:
[tex]\[ B. \ a x + b = a x + c \][/tex]
