Question 622367
I need help with these 4 problems
I need to graph the following

x = 3y - 6

2x - y = -2

{{{y = (2/3)x + 4}}}

y = 2

Thank you for your help.


1) Solve each equation for y, or put in y = mx + b form

x = 3y - 6 


x + 6 = 3y


{{{y = (1/3)x + 2}}}


Because the graph of the equation will intersect the y-axis at x = 0, and the y-intercept (b) is 2, then the graph will intersect the y-axis at coordinate point, (0, 2)


Because the graph of the equation will intersect the x-axis at y = 0, we set the y-value in the equation at 0 (zero) to get:


{{{0 = (1/3)x + 2}}}


{{{(1/3)x = - 2}}} ---- Isolating the variable


{{{x = (- 2)/(1/3)}}}


x = - 2 * 3, or - 6, which gives us an x-intercept coordinate point of (- 6, 0)


Now just join the points, (0, 2) and (- 6, 0), and you have your graph


I'm sure you can do the others by applying the same concept.


N.B.

Number 3 has already been solved for y, or is already in y = mx + b form
Number 4 is a horizontal line at y = 4. That's all it takes to graph this one.


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