# Thread: rearranging equations into y=mx+c

1. ## 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

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

3. Here is an example

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

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

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

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

5. Thank-you so much

Now i shall struggle along with the other questions!!

6. ## 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

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

7. 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) $\displaystyle \Rightarrow -\frac{1}{3}$.

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.

8. 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.

9. 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...

10. 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

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

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