Answer :
Sure! Let's complete the two-column proof for this equation [tex]\(\frac{x}{6}+2=15\)[/tex] to show that [tex]\(x\)[/tex] is indeed 78.
| Statements | Reasons |
|-----------------------------|-----------------------------------------|
| 1. [tex]\(\frac{x}{6}+2=15\)[/tex] | Given |
| 2. [tex]\(\frac{x}{6} = 13\)[/tex] | Subtract 2 from both sides |
| 3. [tex]\(x = 78\)[/tex] | Multiply both sides by 6 |
Let's break it down further with justifications for each step:
1. Given: The initial equation is [tex]\(\frac{x}{6} + 2 = 15\)[/tex].
2. Subtract 2 from both sides: To isolate the term with [tex]\(x\)[/tex], subtract 2 from each side of the equation:
[tex]\[ \frac{x}{6} + 2 - 2 = 15 - 2 \][/tex]
Simplifying both sides, we get:
[tex]\[ \frac{x}{6} = 13 \][/tex]
3. Multiply both sides by 6: To completely solve for [tex]\(x\)[/tex], multiply each side of the equation by 6:
[tex]\[ 6 \left(\frac{x}{6}\right) = 13 \cdot 6 \][/tex]
This simplifies to:
[tex]\[ x = 78 \][/tex]
Thus, by the end of these steps, we have proven that [tex]\(x = 78\)[/tex].
| Statements | Reasons |
|-----------------------------|-----------------------------------------|
| 1. [tex]\(\frac{x}{6}+2=15\)[/tex] | Given |
| 2. [tex]\(\frac{x}{6} = 13\)[/tex] | Subtract 2 from both sides |
| 3. [tex]\(x = 78\)[/tex] | Multiply both sides by 6 |
Let's break it down further with justifications for each step:
1. Given: The initial equation is [tex]\(\frac{x}{6} + 2 = 15\)[/tex].
2. Subtract 2 from both sides: To isolate the term with [tex]\(x\)[/tex], subtract 2 from each side of the equation:
[tex]\[ \frac{x}{6} + 2 - 2 = 15 - 2 \][/tex]
Simplifying both sides, we get:
[tex]\[ \frac{x}{6} = 13 \][/tex]
3. Multiply both sides by 6: To completely solve for [tex]\(x\)[/tex], multiply each side of the equation by 6:
[tex]\[ 6 \left(\frac{x}{6}\right) = 13 \cdot 6 \][/tex]
This simplifies to:
[tex]\[ x = 78 \][/tex]
Thus, by the end of these steps, we have proven that [tex]\(x = 78\)[/tex].
