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Answer the questions about Equations A and B:

Equation A: [tex]\(2x - 1 + 3x = 0\)[/tex]
Equation B: [tex]\(5x - 1 = 0\)[/tex]

1) How can we get Equation B from Equation A?

Choose one answer:
A. Add/subtract the same quantity to/from both sides.
B. Add/subtract a quantity to/from only one side.
C. Rewrite one side (or both) by combining like terms.
D. Rewrite one side (or both) using the distributive property.

2) Based on the previous answer, are the equations equivalent? In other words, do they have the same solution?

Choose one answer:
A. Yes
B. No

Sagot :

GinnyAnsweridn
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.
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