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1. The equation [tex]$4(x-2)=100$[/tex] is a true equation for a particular value of [tex]$x$[/tex]. Explain why [tex]$2(x-2)=50$[/tex] is also true for the same value of [tex][tex]$x$[/tex][/tex].

2. To solve the equation [tex]$7.5d=2.5d$[/tex], Lin divides each side by [tex]$2.5d$[/tex], and Elena subtracts [tex][tex]$2.5d$[/tex][/tex] from each side.

a. Will both moves lead to the solution? Explain your reasoning.

b. What is the solution?

Answer :

Sure, let's break down the solution step-by-step:

### Part 1: Explain why [tex]\( 2(x-2)=50 \)[/tex] is also true for the same value of [tex]\( x \)[/tex] in [tex]\( 4(x-2)=100 \)[/tex]

1. Solve the equation [tex]\( 4(x - 2) = 100 \)[/tex]:
[tex]\[
4(x - 2) = 100
\][/tex]
Divide both sides of the equation by 4:
[tex]\[
x - 2 = 25
\][/tex]
Add 2 to both sides to solve for [tex]\( x \)[/tex]:
[tex]\[
x = 25 + 2
\][/tex]
[tex]\[
x = 27
\][/tex]

2. Verify [tex]\( 2(x - 2) = 50 \)[/tex] with [tex]\( x = 27 \)[/tex]:
Substitute [tex]\( x = 27 \)[/tex] into the equation [tex]\( 2(x - 2) \)[/tex]:
[tex]\[
2(27 - 2)
\][/tex]
Simplifies to:
[tex]\[
2 \times 25
\][/tex]
Which equals:
[tex]\[
50
\][/tex]
Therefore, [tex]\( 50 = 50 \)[/tex] confirms that [tex]\( 2(x-2) = 50 \)[/tex] is also true for [tex]\( x = 27 \)[/tex].

### Part 2: Solving the equation [tex]\( 7.5d = 2.5d \)[/tex]

Part 2A: Will both moves lead to the solution?

1. Lin's move (dividing each side by [tex]\( 2.5d \)[/tex]):
[tex]\[
7.5d = 2.5d
\][/tex]
Divide both sides by [tex]\( 2.5d \)[/tex] (assuming [tex]\( d \neq 0 \)[/tex]):
[tex]\[
\frac{7.5d}{2.5d} = \frac{2.5d}{2.5d}
\][/tex]
This simplifies to:
[tex]\[
3 = 1
\][/tex]
However, [tex]\( 3 = 1 \)[/tex] is clearly not true, indicating all [tex]\( d \)[/tex] values except [tex]\( 0 \)[/tex] lead to a contradiction.

2. Elena's move (subtracting [tex]\( 2.5d \)[/tex] from both sides):
[tex]\[
7.5d - 2.5d = 2.5d - 2.5d
\][/tex]
Which simplifies to:
[tex]\[
5d = 0
\][/tex]
Dividing both sides by 5 gives:
[tex]\[
d = 0
\][/tex]

Therefore, Lin's method indicates a contradiction (except for the special case [tex]\( d = 0 \)[/tex]), while Elena's method directly leads to the solution [tex]\( d = 0 \)[/tex].

Part 2B: What is the solution?

Based on Elena's method:
[tex]\[
5d = 0
\][/tex]
Therefore:
[tex]\[
d = 0
\][/tex]

Conclusion:
- For Part 1, [tex]\( x = 27 \)[/tex].
- For Part 2:
- Both methods show that [tex]\( d = 0 \)[/tex] when considering a valid [tex]\( d \)[/tex] that does not lead to contradictions.

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