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Question about finding interception of two graphs (simultaneous equations)



It is to do with this question.

I completed part a), where by rearranging the equation into y = mx + c, I got y = -(2/3)x + (26/3).

This allowed me to find the gradient of l2, which being negative reciprocal, had to be (3/2)x.

Because the line was going through the (0,0), I concluded that the y-intercept in y = mx + c, would simply be + 0. Hence, logically, the equation would be y = (3/2)x (although I'm not sure about my explanation - I had to look in the mark scheme to check what the answer is first and then developed my own understanding based on the information provided).


In part b), I understood that the y-intercept would clearly (due to my previous work in part a)) be at (0, 26/3).

Then I tried to find point C and this is when the problem had to occur - I thought that by making the two equations equal, I would find the x value.
So I'd change y = -(2/3)x + (26/3) and y = (3/2)x into:
-(2/3)x + (26/3) = (3/2)x

However, by doing this, I got the wrong answer (I actually got 26/25 for some reason).

I tried a different method. Through substitution of y = (3/2)x into the initial equation, I got 2x + 3((3/2)x) = 26 and solved it to get 4, which turned out to be the right answer.

My question is: why didn't the first try of making the equations equal work? Was it some calculation error? It got me worried, because that'd be the method I'd use in the exam and clearly, I'd get the answer wrong.

Cheers

btw. I tried the first method 3 times now, twice getting the working out wrong, but at 3rd try I'm pretty sure I made no mistakes, yet still got the wrong answer.
I did (5/2)x = 26/3
5x = 2(26/3)
5x = 52/3
x = 52/15
52/15 =/= 4 : (
Reply 1
Original post by frostyy
-(2/3)x + (26/3) = (3/2)x

btw. I tried the first method 3 times now, twice getting the working out wrong, but at 3rd try I'm pretty sure I made no mistakes, yet still got the wrong answer.
I did (5/2)x = 26/3
5x = 2(26/3)
5x = 52/3
x = 52/15
52/15 =/= 4 : (

How did you get 5/2 ?
Reply 2
Original post by notnek
How did you get 5/2 ?

(3/2)x + (2/3)x
Reply 3
Original post by frostyy
(3/2)x + (2/3)x

32+2352\displaystyle \frac{3}{2} + \frac{2}{3} \neq \frac{5}{2}

Try it again.
Reply 4
Original post by notnek
32+2352\displaystyle \frac{3}{2} + \frac{2}{3} \neq \frac{5}{2}

Try it again.


omfg

thanks
Reply 5
Original post by frostyy
(3/2)x + (2/3)x


Take a calculator and check that.

32+23=96+46=136\frac{3}{2} + \frac{2}{3} = \frac{9}{6} + \frac{4}{6} = \frac{13}{6}

Edit: nevermind. :lol:

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