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Solve this simultaneous equation

Someone pls help me with this I’m confused ;
y=3x-1
y=x ² -5
Which bit are you stuck on? Do you know how to solve quadratic equations?
Reply 2
Original post by Chopinnocturne31
Which bit are you stuck on? Do you know how to solve quadratic equations?

I’m stuck on the whole thing lol. I don’t know they’re linear simultaneous equations I don’t know what to do
Original post by thevibex
Someone pls help me with this I’m confused ;
y=3x-1
y=x ² -5

Because they're both equal to y, 3x-1=x^2-5.
Then all you have to do is rearrange into a quadratic =0 and solve.
Since y=3x-1=x^2-5
x^2-5=3x-1
x^2-3x-4=0
(x-4)(x+1)=0
So X=4 or =-1
See if you understand :smile:
They already gave you the value of y in the second equation so you replace the y in the first equation.
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Original post by davidtse
Since y=3x-1=x^2-5
........
See if you understand :smile:

Please observe the forum guidelines (sticky): hints not full solutions.
Original post by Iwanttobesmart
They already gave you the value of y in the second equation so you replace the y in the first equation.

Please observe the forum guidelines (sticky): hints not full solutions.
Reply 8
Original post by davidtse
Since y=3x-1=x^2-5
x^2-5=3x-1
x^2-3x-4=0
(x-4)(x+1)=0
So X=4 or =-1
See if you understand :smile:

Thank you so much ! I feel dumb now that I didn’t know this even though it’s simple
Original post by old_engineer
Please observe the forum guidelines (sticky): hints not full solutions.


Really really sorry... This is my first time posting in this forum and I’m not really aware of that rule. Thanks for telling me this and I’ll observe it next time :smile:
Original post by thevibex
Thank you so much ! I feel dumb now that I didn’t know this even though it’s simple


You’re welcome and NO you’re NOT dumb! I think the question is designed to be a bit tricky with the x^2 in the second equation.

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