The x-intercept of the line that passes through the intersection of the lines x + 2y = 4 and 2x + 3y = 6, and is perpendicular to the line 3x — y = 2 is

The x-intercept of the line that passes through the intersection of the

lines x + 2y = 4 and 2x + 3y = 6, and is perpendicular to the line 3x — y = 2 is

Let us find the intersection point of the lines 

x + 2y = 4 and 2x + 3y = 6

Multiplying (x + 2y = 4) by 2 and subtracting with 2x + 3y = 6

2x+4y = 8

2x+3y = 6

y = 2

So, x = 0

Therefore, the point of intersection is (0,2).

We know that a line that passes through (0,2) is perpendicular to the line 3x-y = 2

The slope of the line 3x-y = 2

y = 3x-2

Slope is 3

We know the product of slopes of perpendicular lines is $$m_1m_2=-1$$

Therefore, the line perpendicular to 3x - y = 2 will have a slope of -1/3

Thus, the line passing through (0,2) with a slope of -1/3 will be

$$y-2\ =-\dfrac{1\left(x-0\right)}{3}$$

3y-6 = -x

3y+x = 6

Thus, the x-intercept of this line will be when y = 0

so, x = 6