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    (Original post by willg66)
    Oh okay so so if y = 5x + 3 rearrange to 5x = y +3 if y is -6 then 5x = -3 so -3/5 to get x = -0.6 not 33 I'm failing to understand how everyone's got 33. I solved by making them equal to each other so when you solved the quadratic after making them equal you'd solve for the y values not the x values then substitute in to find X well that's what my teacher said.


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    I think it was y= 5x -3 not plus
    And you put x Into this equation you get 5(-6) -3 which is y= -33 and I did this with the top equation and it gave me the same value. It should have worked if you put it into any equation I think.

    I might not be right but I'm guessing that because that's what the majority put it could be the right answer idk
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    (Original post by Mr_Elmo)
    Yeah i got 175 and then something on decimal as well, but 175km is perfectly fine
    i got 173km will i get marked down for that?!
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    What did everyone get for the sin and trig triangle question??
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    (Original post by sabaaaa)
    What did everyone get for the sin and trig triangle question??
    Something like 61.7 idk


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    (Original post by Kristina.24)
    I think it was y= 5x -3 not plus
    And you put x Into this equation you get 5(-6) -3 which is y= -33 and I did this with the top equation and it gave me the same value. It should have worked if you put it into any equation I think.

    I might not be right but I'm guessing that because that's what the majority put it could be the right answer idk
    What I did was set both equations equal to each other that gave a quadratic in terms of x when arranged to something like 2x^2+11x+2 or something along them lines however when solving this equation you'd be solving for Y not X even though the equation is written in X you'd be solving for Y as its Y = so on etc. Then you'd use the quadratic equation to find Y and input these values to get X. This how I've been taught either I'm wrong or half of the county has done the mistake a lot of us do -- my teacher will have a copy of the test next week or maybe even tomorrow so I'll ask her to get the answer and I'll put them on here


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    (Original post by willg66)
    What I did was set both equations equal to each other that gave a quadratic in terms of x when arranged to something like 2x^2+11x+2 or something along them lines however when solving this equation you'd be solving for Y not X even though the equation is written in X you'd be solving for Y as its Y = so on etc. Then you'd use the quadratic equation to find Y and input these values to get X. This how I've been taught either I'm wrong or half of the county has done the mistake a lot of us do -- my teacher will have a copy of the test next week or maybe even tomorrow so I'll ask her to get the answer and I'll put them on here


    Posted from TSR Mobile
    When you use the quadratic equation it's to find the x values. You use it when the equation can't be factorised or can't be solved through completing the square. Then when you find those values for x you input them into the equation which is y= whatever

    I guess you're right though like it may have given you the right values
 
 
 
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