Question 137404
3y = x + 5 ---------------------eq1
3x - 9y = -10 -------------------eq2

We'll solve it three ways:

1.  From eq1, we see that x=3y-5 (we subtracted 5 from each side)
Now substitute this into eq2 and we get:

3(3y-5)-9y=-10 or
9y-15-9y=10-------------------------NO solution!!!


2.  Multiply eq1 by 3 and we get 9y=3x+15, substitute this into eq2

3x-(3x+15)=-10 or
3x-3x-15=-10-------------------------NO solution!!!!!!


3. re-write eq2 as follows:

3x-3*(3y)=-10 now substitute 3y=x+5 into this:

3x-3(x+5)=-10
3x-3x-15=-10--------------------------NO solution!!!!

The equations are inconsistent---they have no solution. Why?????

Write each equation in slope/intercept form(y=mx+b where m is the slope and b is the y-intercept)

eq1 3y=x+5 or (we divided both sides by 3)
y=(1/3)x  +  5/3--------------------slope is 1/3


eq2  3x-9y=-10  subtract 3x from each side

-9y=-3x-10  multiply each side by -1
9y=3x+10  divide each term by 9

y=(1/3)x+10/9---------------------------slope is 1/3

Both equations have a slope of 1/3 which means the lines are parallel


Hope this helps---ptaylor