Which of the following equations has an infinite number of solutions?

A. 3x – 3 = –4x
B. 2y + 4 – y = 16
C. 7x + 5 = 4x + 5 + 3x
D. 6y – 2 = 2(y – 1)

Is C correct?

yes

Thanks!

To determine which equation has an infinite number of solutions, we need to analyze each equation and see if there is a possibility for any value to satisfy it.

Let's examine each option:

A. 3x – 3 = –4x.
To solve this equation, we can combine like terms by adding 4x to both sides:
3x + 4x – 3 = –4x + 4x.
This simplifies to:
7x – 3 = 0.
Now, we can isolate x by adding 3 to both sides:
7x = 3.
Finally, we can solve for x by dividing both sides by 7:
x = 3/7.

B. 2y + 4 – y = 16.
To solve this equation, we can combine like terms by subtracting y from both sides:
2y – y + 4 = 16.
Simplifying further results in:
y + 4 = 16.
Then, subtracting 4 from both sides, we have:
y = 12.

C. 7x + 5 = 4x + 5 + 3x.
To solve this equation, we can combine like terms by adding 4x and 3x to both sides:
7x + 4x + 3x + 5 = 4x + 5 + 3x + 3x.
Simplifying, we get:
14x + 5 = 7x + 5.
Subtracting 7x from both sides, we obtain:
7x + 5 = 5.
Now, subtracting 5 from both sides, we have:
7x = 0.
Finally, dividing both sides by 7, we get:
x = 0.
So, the equation C is incorrect, as it has a unique solution.

D. 6y – 2 = 2(y – 1).
To solve this equation, we can simplify the right-hand side by distributing the 2:
6y – 2 = 2y – 2.
Now, moving all variables to one side, we have:
6y – 2y = -2 + 2.
This simplifies to:
4y = 0.
Dividing both sides by 4, we obtain:
y = 0.

Thus, the equation with an infinite number of solutions is A. 3x – 3 = –4x, which simplifies to x = 3/7. Therefore, C is incorrect.