Consider the following system of equations: y=2x−2 6x+3y=2 The graph of these equations consists of two lines that: 1. intersect at more than one point. 2. intersect in an infinite number of points. 3. intersect at exactly one point. 4. do not intersect.

3.  Intersect at exactly one point.  ( (2/3), (-2/3) )

Step-by-step explanation:

To make the comparison of these lines easier, let's rewrite the 2nd equation into slope-intercept form, as the 1st equation is in slope-intercept form.

[1] y = 2x - 2

[2] 6x + 3y = 2 ==> 3y = 2 - 6x ==> y = -2x + (2/3)

[2] y = -2x + (2/3)

So now that we have both equations in slope-intercept form, we can see that the two equations are both linear, have different slopes, and have different y-intercepts.

Since these equations have both different slopes and different y-intercepts, we know that the lines will cross at least one point.  We can confirm that the lines only cross at a single point using the fact that both equations are linear, meaning there will only be one point of crossing.  To find that point, we can simply set the equations equal to each other.

y = 2x - 2

y = -2x + (2/3)

2x - 2 = -2x + (2/3)

4x = (8/3)

x = (8/12) = (2/3)

And plug this x value back into one of the equations:

y = 2x - 2

y = 2(2/3) - 2

y = (4/3) - (6/3)

y = (-2/3)

Thus these lines only cross at the point ( (2/3), (-2/3) ).

Cheers.

what is the original number? you forgot to add it.