The area bounded by the line y = 2x – 2, y = -x and x-axis is given by

The correct option is None of these.

To find the area bounded by the lines $y=2x-2$, $y=-x$, and the x-axis ($y=0$), we first find the intersection points of these boundary lines.

Step 1: Find the intersection of the two lines
Set the equations equal to each other to find the x-coordinate:
$$2x-2=-x$$
Add $x$ to both sides and add $2$ to both sides:
$$3x=2\Rightarrow x=\frac{2}{3}$$
Now substitute $x=\frac{2}{3}$ back into $y=-x$:
$$y=-\frac{2}{3}$$
Thus, the point of intersection is:
$$C\left(\frac{2}{3},,,-,\frac{2}{3}\right)$$

Step 2: Find the intersections with the x-axis (where $y=0$)
For the line $y=-x$:
$$-x=0\Rightarrow x=0$$
This gives the intersection point $A\left(0,,,0\right)$.
For the line $y=2x-2$:
$$2x-2=0\Rightarrow x=1$$
This gives the intersection point $B\left(1,,,0\right)$.

Step 3: Calculate the area of the bounded triangular region
The region forms a triangle with vertices at $A\left(0,,,0\right)$, $B\left(1,,,0\right)$, and $C\left(\frac{2}{3},,,-,\frac{2}{3}\right)$.
Since $A$ and $B$ both lie on the x-axis, the base of the triangle lies along the x-axis:
$$\text{Base}=1-0=1\text{unit}$$
The height of the triangle is the perpendicular distance from vertex $C$ to the x-axis, which is the absolute value of its y-coordinate:
$$\text{Height}=\left|-\frac{2}{3}\right|=\frac{2}{3}\text{units}$$
Using the formula for the area of a triangle:
$$\text{Area}=\frac{1}{2}\times \text{Base}\times \text{Height}$$
$$\text{Area}=\frac{1}{2}\times 1\times \frac{2}{3}=\frac{1}{3}\text{sq. units}$$

Since the area is $\frac{1}{3}$ sq. units, which does not match any of the given choices ("9/2 sq. units", "43/6 sq. units", or "35/6 sq. units"), the correct answer is indeed None of these.