#1

I've been trying to get this one for awhile but am completely vexed. Using the elimination method.

9x-5y=1

-18x+10y=1

I mean whenever I multiply 5 and 2, to solve for x, everything ends up canceling out.

Help would be much appreciated.

9x-5y=1

-18x+10y=1

I mean whenever I multiply 5 and 2, to solve for x, everything ends up canceling out.

Help would be much appreciated.

#2

y = (1-9x)/5

-18x + 10[(1-9x)/5] = 1

-18x + 2(1-9x) = 1

-18x + 2 - 18x = 1

-36x = -1

x = 1/36

amidoinitright?

EDIT: shit, no i'm not... yeah, should be

y = -(1-9x)/5

yeah in that case x would cancel out... i guess there just is no x in that, but... i dont like math sooo...

-18x + 10[(1-9x)/5] = 1

-18x + 2(1-9x) = 1

-18x + 2 - 18x = 1

-36x = -1

x = 1/36

amidoinitright?

EDIT: shit, no i'm not... yeah, should be

y = -(1-9x)/5

yeah in that case x would cancel out... i guess there just is no x in that, but... i dont like math sooo...

*Last edited by SPBY at Oct 27, 2009,*

#3

Its linearally dependent. You have a free variable.

9x-5y=1

9x=5y+1

x=5/9y+1/9

9x-5y=1

9x=5y+1

x=5/9y+1/9

#4

these two systems of eqns have no solution. Basically if you picture these equations as lines in a 2d coordinate system, they never intersect. If you plot the two lines, you should see no intersection.

#5

these two systems of eqns have no solution. Basically if you picture these equations as lines in a 2d coordinate system, they never intersect. If you plot the two lines, you should see no intersection.

+1

If you solve each equation for y, then set the two equal to each other, you get the statement 1 = -2. Obviously, this is not true, therefore there is no intersection.

#6

awesome thanks a bunch that's what i was thinking

EDIT: I'm going to A&M too! We should be friends

EDIT: I'm going to A&M too! We should be friends

*Last edited by fav13andac1)c at Oct 27, 2009,*