# rearranging equations into y=mx+c

• Sep 17th 2008, 11:50 AM
vanpopta
rearranging equations into y=mx+c
Hello, i am new to this site, an i was not sure where to post (as i am not in the US schooling system), i am doing AS level maths in the UK if this helps?

i need to rearrange equations to y=mx+c

5x+y=4

x-2y+6=0

x+2y=3

y-6=2(x+4) <-- No idea whatsoever.

I hope you can help me :S
• Sep 17th 2008, 12:01 PM
courteous
$\displaystyle 5x+y=4 \Rightarrow y=-5x+4$
$\displaystyle x-2y+6=0 \Rightarrow y=\frac{1}{2}x+3$
$\displaystyle x+2y=3 \Rightarrow y=-\frac{1}{2}x+\frac{3}{2}$
$\displaystyle y-6=2(x+4) \Rightarrow y=2x+14$
• Sep 17th 2008, 12:02 PM
11rdc11
Here is an example

$\displaystyle x-2y+6=0$

$\displaystyle -2y=-x-6$

$\displaystyle y= \frac{1}{2}x +3$
• Sep 17th 2008, 12:03 PM
11rdc11
Quote:

Originally Posted by courteous
$\displaystyle 5x+y=4 \Rightarrow y=-5x+4$
$\displaystyle x-2y+6=0 \Rightarrow y=\frac{1}{2}x+6$
$\displaystyle x+2y=3 \Rightarrow y=-\frac{1}{2}x+\frac{3}{2}$
$\displaystyle y-6=2(x+4) \Rightarrow y=2x+14$

• Sep 17th 2008, 12:28 PM
vanpopta
Thank-you so much :)

Now i shall struggle along with the other questions!!
• Sep 18th 2008, 01:18 AM
vanpopta
Extra question
I got to a later question, which was State whether the following were parallel, perpendicular, or neither, I managed to do the parallel ones, so i need now to know whether these were perpendicular or not

Is $\displaystyle y=3x+4$

perpencicular to

$\displaystyle y=2-\frac{1}{3}x$

and

$\displaystyle x+3y-2=0$
• Sep 18th 2008, 03:23 AM
courteous
Quote:

Originally Posted by vanpopta
I got to a later question, which was State whether the following were parallel, perpendicular, or neither, I managed to do the parallel ones, so i need now to know whether these were perpendicular or not

Is $\displaystyle y=3x+4$

perpencicular to

$\displaystyle y=2-\frac{1}{3}x$

and

$\displaystyle x+3y-2=0$

General equation is $\displaystyle y=kx+n$, where $\displaystyle k$ is the slope of the line and $\displaystyle n$ is the point of line intersecting the y-axis.
So, the slopes are $\displaystyle y=3x+4 \Rightarrow 3$,
$\displaystyle y=2-\frac{1}{3}x \Rightarrow -\frac{1}{3}$,
$\displaystyle x+3y-2=0$ (now you know how to rearrange this(Wink)) $\displaystyle \Rightarrow -\frac{1}{3}$.

When $\displaystyle k$ equals the lines are parrarel (or the same), and when $\displaystyle k_1*k_2=-1$ (where the $\displaystyle tan$ is undefined, that is, denominator is 0; more at here) the lines are perpendicular. Now, you only need to compare those ks.
• Sep 18th 2008, 03:25 AM
sqleung
For $\displaystyle y = 3x + 4$, you will notice it has a slope of 3.

Now, a line is perpendicular to the above line if its slope is the inverse reciprocal - ie. slope of $\displaystyle \frac{-1}{3}$

$\displaystyle y = 2 - \frac{-1}{3}x$ does indeed have a slope of $\displaystyle \frac{-1}{3}$ and so, it is perpendicular to the line.

Also:

$\displaystyle x + 3y - 2 = 0$

$\displaystyle x + 3y = 2$

$\displaystyle 3y = 2 - x$

$\displaystyle y = \frac{2}{3} - \frac{1}{3}x$

And its slope is also $\displaystyle \frac{-1}{3}$ and hence, it's perpendicular to the line.
• Sep 18th 2008, 01:08 PM
vanpopta
Thanks a lot guys :)

That helped me, for the next 5 questions!

However i got stuck again.

Line q has the equation $\displaystyle y=6x+1$

(a) Give the equation of the line through$\displaystyle (1,0)$ parallel to q
(b) Give he equation of the line through$\displaystyle (3,1)$ perpendicular to q

I think (a) may well be $\displaystyle y=6x-6$

However, i am not confident of that one bit...
• Sep 18th 2008, 01:32 PM
11rdc11
Quote:

Originally Posted by vanpopta
Thanks a lot guys :)

That helped me, for the next 5 questions!

However i got stuck again.

Line q has the equation $\displaystyle y=6x+1$

(a) Give the equation of the line through$\displaystyle (1,0)$ parallel to q
(b) Give he equation of the line through$\displaystyle (3,1)$ perpendicular to q

I think (a) may well be $\displaystyle y=6x-6$

However, i am not confident of that one bit...

You are correct on a)

Remember

$\displaystyle y_2 - y_1 = m(x_2 - x_1)$

so for a)

$\displaystyle y-0 = 6(x-1)$

$\displaystyle y = 6x + 6$

to do part b) remember that the perpendicular line is the negative reciprocal

$\displaystyle y-1= -\frac{1}{6}(x-3)$