Question

kate begins solving the equation \\(\frac{2}{3}(6x - 3) = \frac{1}{2}(6x - 4)\\). her work is correct and is shown below.\\(\frac{2}{3}(6x - 3) = \frac{1}{2}(6x - 4)\\)\\(4x - 2 = 3x - 2\\)when she adds 2 to both sides, the equation \\(4x = 3x\\) results. which solution will best illustrate what happens to \\(x\\)?\\(\circ\\) the equation has infinite solutions.\\(\circ\\) the equation has one solution: \\(x = 0\\).\\(\circ\\) the equation has one solution: \\(x = \frac{4}{3}\\).\\(\circ\\) the equation has no solution.

Explanation:

Step1: Start with the equation after adding 2

We have the equation \(4x = 3x\).

Step2: Subtract \(3x\) from both sides

Subtracting \(3x\) from both sides gives \(4x - 3x=3x - 3x\), which simplifies to \(x = 0\). Wait, no, wait. Wait, actually, let's re - examine. Wait, the equation is \(4x=3x\). If we subtract \(3x\) from both sides, we get \(4x - 3x=3x - 3x\), so \(x = 0\)? Wait, no, that's not right. Wait, let's go back. The original equation after Kate's step is \(4x-2 = 3x - 2\). Then she adds 2 to both sides: \(4x-2 + 2=3x - 2+2\), so \(4x=3x\). Now, subtract \(3x\) from both sides: \(4x-3x=3x - 3x\), so \(x = 0\)? Wait, no, that would mean one solution. But wait, let's check the original equation. Let's plug \(x = 0\) into the original equation \(\frac{2}{3}(6x - 3)=\frac{1}{2}(6x - 4)\). Left - hand side: \(\frac{2}{3}(0 - 3)=\frac{2}{3}\times(-3)=- 2\). Right - hand side: \(\frac{1}{2}(0 - 4)=\frac{1}{2}\times(-4)=-2\). So \(x = 0\) is a solution. But wait, let's see the equation \(4x = 3x\). If we consider the general form of a linear equation \(ax + b=cx + d\). In this case, \(a = 4\), \(c = 3\), \(b = 0\), \(d = 0\). Since \(a
eq c\) (4≠3), the equation has exactly one solution. And when we solve \(4x=3x\), we subtract \(3x\) from both sides to get \(x = 0\).

Answer:

The equation has one solution: \(x = 0\) (the option corresponding to "The equation has one solution: \(x = 0\)").