College

15. Which equation, when solved, results in a different value of [tex]$x$[/tex] than the other three? Show your work:

A. [tex]$8.3 = -0.6x + 11.3$[/tex]

B. [tex]$11.3 = 8.3 + 0.6x$[/tex]

C. [tex]$11.3 - 0.6x = 8.3$[/tex]

D. [tex]$8.3 - 0.6x = 11.3$[/tex]

Answer :

We start by solving each equation for [tex]$x$[/tex].

[tex]$$\textbf{Equation A: } 8.3 = -0.6x + 11.3$$[/tex]

Subtract [tex]$11.3$[/tex] from both sides:
[tex]$$8.3 - 11.3 = -0.6x$$[/tex]
[tex]$$-3 = -0.6x$$[/tex]

Divide both sides by [tex]$-0.6$[/tex]:
[tex]$$x = \frac{-3}{-0.6} = 5$$[/tex]

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[tex]$$\textbf{Equation B: } 11.3 = 8.3 + 0.6x$$[/tex]

Subtract [tex]$8.3$[/tex] from both sides:
[tex]$$11.3 - 8.3 = 0.6x$$[/tex]
[tex]$$3 = 0.6x$$[/tex]

Divide both sides by [tex]$0.6$[/tex]:
[tex]$$x = \frac{3}{0.6} = 5$$[/tex]

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[tex]$$\textbf{Equation C: } 11.3 - 0.6x = 8.3$$[/tex]

Subtract [tex]$11.3$[/tex] from both sides:
[tex]$$-0.6x = 8.3 - 11.3$$[/tex]
[tex]$$-0.6x = -3$$[/tex]

Divide both sides by [tex]$-0.6$[/tex]:
[tex]$$x = \frac{-3}{-0.6} = 5$$[/tex]

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[tex]$$\textbf{Equation D: } 8.3 - 0.6x = 11.3$$[/tex]

Subtract [tex]$8.3$[/tex] from both sides:
[tex]$$-0.6x = 11.3 - 8.3$$[/tex]
[tex]$$-0.6x = 3$$[/tex]

Divide both sides by [tex]$-0.6$[/tex]:
[tex]$$x = \frac{3}{-0.6} = -5$$[/tex]

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In summary, Equations A, B, and C all have [tex]$x = 5$[/tex], while Equation D gives [tex]$x = -5$[/tex].

Thus, the equation that results in a different value of [tex]$x$[/tex] is:

[tex]$$\boxed{8.3 - 0.6x = 11.3}$$[/tex]