Certainly! Let's tackle these questions one step at a time.
### Question 1: How can we get Equation [tex]$B$[/tex] from Equation [tex]$A$[/tex]?
Equation [tex]$A$[/tex] is: [tex]\( 2x - 1 + 3x = 0 \)[/tex]
To simplify this, we combine the like terms on the left side of the equation.
- The term [tex]\(2x\)[/tex] is a like-term with [tex]\(3x\)[/tex].
- When we add [tex]\(2x\)[/tex] and [tex]\(3x\)[/tex], we get [tex]\(5x\)[/tex].
So, combining like terms, Equation [tex]$A$[/tex] becomes:
[tex]\[ 5x - 1 = 0 \][/tex]
This new form matches Equation [tex]$B$[/tex]. Therefore, the correct answer to the first question is:
(C) Rewrite one side (or both) by combining like terms
### Question 2: Are the equations equivalent? Do they have the same solution?
Since we derived Equation [tex]$B$[/tex] directly from Equation [tex]$A$[/tex] by combining like terms, both equations should have the same solution.
To confirm, let's solve both equations.
For Equation [tex]$A:
Starting from the simplified form derived, which is:
\[ 5x - 1 = 0 \]
We solve for \(x\):
\[ 5x = 1 \]
\[ x = \frac{1}{5} \]
For Equation $[/tex]B:
Equation [tex]$B$[/tex] is already in the form:
[tex]\[ 5x - 1 = 0 \][/tex]
So solving it follows the same steps:
[tex]\[ 5x = 1 \][/tex]
[tex]\[ x = \frac{1}{5} \][/tex]
Since both equations yield the same solution, they are equivalent. Hence, the correct answer to the second question is:
(A) Yes
In summary:
1. The way to get from Equation [tex]$A$[/tex] to Equation [tex]$B$[/tex] is by combining like terms.
2. Yes, the equations are equivalent and have the same solution.
