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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.

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    $5y=mx-4$ 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.2012-08-16

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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 $m$.

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    Thank you so much :) I should be able to solve the whole chapter now.2012-08-16
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You have two lines :

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

and

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.

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    You should try the following link: http://en.wikipedia.org/wiki/Slope2012-08-16