Which of the following equations has an infinate 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 it D
How can 6y-2 = 2y-2 have more than one solution?
Check C. It is true for any value of x.
Oh yeah it is C, thanks Steve.
To determine which equation has an infinite number of solutions, we need to solve each equation and see if there is a variable that cancels out, resulting in the equation being always true.
Let's solve each equation step by step:
A. 3x - 3 = -4x
First, let's combine like terms by adding 4x to both sides:
3x + 4x - 3 = -4x + 4x
7x - 3 = 0
Then, let's isolate x by adding 3 to both sides:
7x - 3 + 3 = 0 + 3
7x = 3
Finally, divide both sides by 7 to solve for x:
7x/7 = 3/7
x = 3/7
B. 2y + 4 - y = 16
First, simplify the equation by combining like terms:
y + 4 = 16
Next, isolate y by subtracting 4 from both sides:
y + 4 - 4 = 16 - 4
y = 12
C. 7x + 5 = 4x + 5 + 3x
Combine like terms by adding 4x and 3x on the right side:
7x + 5 = 7x + 5
Subtract 7x from both sides:
7x - 7x + 5 = 7x - 7x + 5
5 = 5
The equation simplifies to 5 = 5, which is always true, but it does not lead to an infinite number of solutions. This equation has a unique solution, not an infinite number of solutions.
D. 6y - 2 = 2(y - 1)
Let's distribute the 2 on the right side:
6y - 2 = 2y - 2
Next, let's combine like terms by subtracting 2y from both sides:
6y - 2y - 2 = 2y - 2y - 2
4y - 2 = -2
Then, isolate y by adding 2 to both sides:
4y - 2 + 2 = -2 + 2
4y = 0
Finally, divide both sides by 4 to solve for y:
4y/4 = 0/4
y = 0
The equation 6y - 2 = 2(y - 1), which corresponds to option D, has a unique solution of y = 0, not an infinite number of solutions.
Therefore, the equation with an infinite number of solutions is not D, but rather option C, 7x + 5 = 4x + 5 + 3x.