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# This maths question...

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1. The graph of y=(px-1)/(2q+1) crosses the x-axis at (4,0) and the y-axis at (0,-5). How do find the values of p and q? So far, I've only substituted the x and y values in the equation...
2. you get 2 equations with 2 unknowns ( p & q )

so then solve simultaneously.*
3. (Original post by honour)
The graph of y=(px-1)/(2q+1) crosses the x-axis at (4,0) and the y-axis at (0,-5). How do find the values of p and q? So far, I've only substituted the x and y values in the equation...
Two unknowns and two equations, simultaneous equations
4. If you substitute (4,0), you should notice something that in essence makes the q 'disappear' allowing you to solve for p and then q... (I think; I am very tired atm from AEA papers )
5. (Original post by the bear)
you get 2 equations with 2 unknowns ( p & q )

so then solve simultaneously.*
Thank you Bjorn!
6. (Original post by honour)
The graph of y=(px-1)/(2q+1) crosses the x-axis at (4,0) and the y-axis at (0,-5). How do find the values of p and q? So far, I've only substituted the x and y values in the equation...
Hope this helps lol

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7. (Original post by huzaifa-tariq)
Hope this helps lol

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This is so helpful, thanks ever so much Huzaifa!!

8. (Original post by honour)
This is so helpful, thanks ever so much Huzaifa!!

Haha, you're welcome!

9. (Original post by huzaifa-tariq)
Hope this helps lol

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why's the gradient negative ? honour is this C1?
10. (Original post by EA7_)
why's the gradient negative ? honour is this C1?
The gradient wasn't meant to be negative, but I found the correct answer in the end. And yes, this is C1.
11. (Original post by honour)
The gradient wasn't meant to be negative, but I found the correct answer in the end. And yes, this is C1.
thanks! could you show me how he wrote equation y=(px-1)/(2q+1) into mx + c
12. (Original post by EA7_)
thanks! could you show me how he wrote equation y=(px-1)/(2q+1) into mx + c
Lol sorry I messed up with the gradient. It should've been 0 - (-5) resulting in a positive 5/4 gradient and subsequently, a positive value for p.
As for the mx + c part, you can turn any equation of a line into that form. Just make the equation into some kind of form where 'x' has a coefficient and 'y' is the subject of the equation. WHATEVER is being multiplied with 'x' will be the gradient 'm' and the rest will be 'c.'

13. (Original post by huzaifa-tariq)
Lol sorry I messed up with the gradient. It should've been 0 - (-5) resulting in a positive 5/4 gradient and subsequently, a positive value for p.
As for the mx + c part, you can turn any equation of a line into that form. Just make the equation into some kind of form where 'x' has a coefficient and 'y' is the subject of the equation. WHATEVER is being multiplied with 'x' will be the gradient 'm' and the rest will be 'c.'

don't worry haha.. everyone makes mistakes....and thanks for the explanation! I understand it better now!

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