Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been on this for about an hour, researched everywhere, but I cannot find a viable solution.

Question: The straight line $mx = 5y + 4$ has the same gradient as the line $7x + 6y + 5 = 0$, Find the value of $m$.

I've tried: $mx - 5y + 4 = 7x + 6y + 5$ but can't get it right :(.

Any viable solution for this?

I tried to change it to the $y = mx + b$ format, I got this:

$5y = mx - 4$ and $6y = 7x - 5$, this is the part I'm stuck on.

share|cite|improve this question
$5y=mx-4$ isn't exactly in the $y=mx+b$ format. You have to get rid of the $5$ on the left hand side. The same for the second equation. – celtschk Aug 16 '12 at 13:58
up vote 2 down vote accepted

Your equations should (your equation $6y=7x-5$ is slightly off) be $$ \tag{1}5y=mx-4 $$ and $$\tag{2} 6y=-7x-5.$$ You want $y$ isolated on the left hand side in each of these equations. So, divide both sides of equation $(1)$ by $5$: $$\tag{3} y={m\over5}x-{4\over5}; $$ and divide both sides of equation $(2)$ by $6$: $$\tag{4} y={-7\over6}x-{5\over6}. $$ The slope (gradient) of the line given by $(3)$ is $m/5$ and the slope of the line given by $(4)$ is $-7/6$. From this information, you should be able to solve for $m$.

share|cite|improve this answer
Thank you so much :) I should be able to solve the whole chapter now. – user38052 Aug 16 '12 at 14:17

You have two lines :

1) $7x+6y+5=0$ with $m_{1}=-\frac{7}{6}$


2) $mx-5y+4=0$ with $m_{2}=\frac{m}{5}$

so you have to resolve following equation: $-\frac{7}{6}=\frac{m}{5}$ $\Rightarrow$ $m=-\frac{35}{6}$. I noted with $m_{1}$ the slope for first line and $m_{2}$ the slope for second line.

share|cite|improve this answer
You should try the following link: – Iuli Aug 16 '12 at 14:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.