Which equation has infinitely many solutions?
A ) (4x + 3)/3 = 4x + 5 (10x + 8)/2 = 5x + 4
B ) ) 8x - 5 3 - 5/6 = 2x - 2
D ) (8x - 5)/2 = 4x - 6
its B. i just picked C and got it wrong and its B.
To determine which equation has infinitely many solutions, we need to examine each equation and see if there is a possibility of all values of x making the equation true.
Let's analyze each equation:
A) (4x + 3)/3 = 4x + 5
To solve this equation, we can start by multiplying both sides of the equation by 3 to eliminate the fraction:
4x + 3 = 3(4x + 5)
4x + 3 = 12x + 15
-8x = 12
x = -1.5
The value of x is -1.5, which means this equation has a specific solution. It does not have infinitely many solutions.
B) (8x - 5) / (3 - 5/6) = 2x - 2
To solve this equation, we can start by simplifying the fraction on the left side:
(8x - 5) / (3 - 5/6) = 2x - 2
(8x - 5) / (13/6) = 2x - 2
To eliminate the fraction, we can multiply both sides of the equation by (13/6):
(8x - 5) = (13/6)(2x - 2)
(8x - 5) = (26/6)x - (26/6)(2)
(8x - 5) = (13/3)x - (26/3)
We can simplify the equation further by combining like terms:
8x - (13/3)x = 5 - (26/3)
(8 - 13/3)x = 15/3 - 26/3
(8 - 13/3)x = -11/3
To isolate x, we divide both sides of the equation by (8 - 13/3):
x = -11/3 ÷ (8 - 13/3)
x = -11/3 ÷ (24/3 - 13/3)
x = -11/3 ÷ 11/3
x = -1
The value of x is -1, which means this equation has a specific solution. It does not have infinitely many solutions.
D) (8x - 5)/2 = 4x - 6
To solve this equation, we can start by multiplying both sides of the equation by 2 to eliminate the fraction:
(8x - 5)/2 = 4x - 6
2((8x - 5)/2) = 2(4x - 6)
8x - 5 = 8x - 12
We can observe that the variable x cancels out on both sides of the equation:
-5 = -12
This equation simplifies to -5 = -12, which is false. This implies that there are no valid solutions for x that make the equation true. Therefore, this equation has infinitely many solutions.
In conclusion, the equation (8x - 5)/2 = 4x - 6 has infinitely many solutions among the given options.
well, not D, since
8x-5 ≠ 8x-12
now work the others, to see which produces the same thing on both sides of the equation