High School

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

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

Answer :

To solve this problem, we are given four equations, and our goal is to identify which one results in a different value of [tex]\( x \)[/tex] compared to the others. Let's review each equation.

1. Equation 1:
[tex]\[ 8.3 = -0.6x + 11.3 \][/tex]

2. Equation 2:
[tex]\[ 11.3 = 8.3 + 0.6x \][/tex]

3. Equation 3:
[tex]\[ 11.3 - 0.6x = 8.3 \][/tex]

4. Equation 4:
[tex]\[ 8.3 - 0.6x = 11.3 \][/tex]

Now, we'll solve these equations one by one for [tex]\( x \)[/tex]:

### Solving Equation 1:
[tex]\[ 8.3 = -0.6x + 11.3 \][/tex]
Subtract 11.3 from both sides:
[tex]\[ 8.3 - 11.3 = -0.6x \][/tex]
[tex]\[ -3 = -0.6x \][/tex]
Divide both sides by -0.6:
[tex]\[ x = \frac{-3}{-0.6} \][/tex]
[tex]\[ x = 5 \][/tex]

### Solving Equation 2:
[tex]\[ 11.3 = 8.3 + 0.6x \][/tex]
Subtract 8.3 from both sides:
[tex]\[ 11.3 - 8.3 = 0.6x \][/tex]
[tex]\[ 3 = 0.6x \][/tex]
Divide both sides by 0.6:
[tex]\[ x = \frac{3}{0.6} \][/tex]
[tex]\[ x = 5 \][/tex]

### Solving Equation 3:
[tex]\[ 11.3 - 0.6x = 8.3 \][/tex]
Subtract 11.3 from both sides and rearrange:
[tex]\[ -0.6x = 8.3 - 11.3 \][/tex]
[tex]\[ -0.6x = -3 \][/tex]
Divide both sides by -0.6:
[tex]\[ x = \frac{-3}{-0.6} \][/tex]
[tex]\[ x = 5 \][/tex]

### Solving Equation 4:
[tex]\[ 8.3 - 0.6x = 11.3 \][/tex]
Subtract 8.3 from both sides:
[tex]\[ -0.6x = 11.3 - 8.3 \][/tex]
[tex]\[ -0.6x = 3 \][/tex]
Divide both sides by -0.6:
[tex]\[ x = \frac{3}{-0.6} \][/tex]
[tex]\[ x = -5 \][/tex]

### Conclusion:

After solving all four equations, we find that:

- Equations 1, 2, and 3 result in [tex]\( x = 5 \)[/tex].
- Equation 4 results in [tex]\( x = -5 \)[/tex].

Therefore, the equation that results in a different value of [tex]\( x \)[/tex] is Equation 4:
[tex]\[ 8.3 - 0.6x = 11.3 \][/tex]

This equation gives a different solution compared to the others.