# Math Help - solving 2 linear equations

1. ## solving 2 linear equations

this should be easy but keep getting different answers every time I do it.

$2x + \frac{5}{y} = -11$

$4x-\frac{3}{y}=\frac{21}{3}$

multiplied eq1 by 3 and eq2 by 5 to get

$6x+\frac{15}{y} = -33$
$20x-\frac{15}{y}=\frac{105}{3}$

$26x = -33 + \frac{105}{3}$

multiply thru by 3

$78x = -99 + 105 = 6$

$x = \frac{6}{78}$

before deriving y using x it is hard to believe that this value is correct??

2. Originally Posted by bigwave
this should be easy but keep getting different answers every time I do it.

$2x + \frac{5}{y} = -11$

$4x-\frac{3}{y}=\frac{21}{3}$

multiplied eq1 by 3 and eq2 by 5 to get

$6x+\frac{15}{y} = -33$
$20x-\frac{15}{y}=\frac{105}{3}$

$26x = -33 + \frac{105}{3}$

multiply thru by 3

$78x = -99 + 105 = 6$

$x = \frac{6}{78}$

before deriving y using x it is hard to believe that this value is correct??
$x = \frac {1}{13} \;\; y = - \frac {13}{29}$

now just put that x in any of the first 2 equations and you will get y

Edit: perhaps it would be easier to multiply first one with -2 and add them ... but it's the same result

3. ok, just to finish it out

$x=\frac{6}{78}=\frac{1}{13}$

$2(\frac{1}{13}) + \frac{5}{y} = -11$

$2y + 65 = -143y$

$145y = -65$

$y = \frac{-65}{145} = -\frac{13}{29}$

4. yes you have done it correctly
always look for the simpler way to solve , so there will be less room for the error

5. actually, it was sugested that when fractions show in simultanious equations that it is best to multiply or cancel them out as an easy way to do it... however I see your point in that -2 would of left the fractions but the x would of canceled... quess the y would not of been that hard to get... yes i always seem to introduce error with a lot of steps

6. One "difficulty", by the way, is that these are not "linear equations"!

7. Originally Posted by HallsofIvy
One "difficulty", by the way, is that these are not "linear equations"!
Hah! I initially thought the same thing. But then I decided that one could reasonably argue that they are linear in the variables x and 1/y .....

8. ## when I ploted them they were not lines

yep, you are right, when I ploted them they were not lines. I presume this is because the y is in the denominator making it rational resulting a graph with an asymptotes.
I did try to make this into the $y=mx+b$ but of course cannot.
however the text did say they were linear equations but doesn't look like it.
interesting catch not sure where the red flag was.