College

Which equation, when solved, results in a different value of [tex]$x$[/tex] than the other three?

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 :

Let's solve each equation step by step.

1. For the first equation
[tex]$$8.3 = -0.6x + 11.3,$$[/tex]
subtract [tex]$11.3$[/tex] from both sides:
[tex]$$8.3 - 11.3 = -0.6x,$$[/tex]
which simplifies to:
[tex]$$-3 = -0.6x.$$[/tex]
Dividing both sides by [tex]$-0.6$[/tex], we get
[tex]$$x = \frac{-3}{-0.6} = 5.$$[/tex]

2. For the second equation
[tex]$$11.3 = 8.3 + 0.6x,$$[/tex]
subtract [tex]$8.3$[/tex] from both sides:
[tex]$$11.3 - 8.3 = 0.6x,$$[/tex]
yielding
[tex]$$3 = 0.6x.$$[/tex]
Dividing by [tex]$0.6$[/tex]:
[tex]$$x = \frac{3}{0.6} = 5.$$[/tex]

3. For the third equation
[tex]$$11.3 - 0.6x = 8.3,$$[/tex]
subtract [tex]$11.3$[/tex] from both sides:
[tex]$$-0.6x = 8.3 - 11.3,$$[/tex]
which gives:
[tex]$$-0.6x = -3.$$[/tex]
Dividing by [tex]$-0.6$[/tex], we have:
[tex]$$x = \frac{-3}{-0.6} = 5.$$[/tex]

4. For the fourth equation
[tex]$$8.3 - 0.6x = 11.3,$$[/tex]
subtract [tex]$8.3$[/tex] from both sides:
[tex]$$-0.6x = 11.3 - 8.3,$$[/tex]
resulting in:
[tex]$$-0.6x = 3.$$[/tex]
Dividing by [tex]$-0.6$[/tex]:
[tex]$$x = \frac{3}{-0.6} = -5.$$[/tex]

Thus, the first three equations all yield [tex]$x = 5$[/tex], while the fourth equation gives [tex]$x = -5$[/tex]. Therefore, the fourth equation results in a different value of [tex]$x$[/tex].