# Simultaneous equations GCSEWatch

#1
Y= 11X + 3
Y= 4x^2

Please walk me through step by step, I will positive rate your profile
1
7 years ago
#2
you have to link them together so 11x+3=4x^2...rearrange giving 4x^2-11x-3...times -3 and 4 together giving -12, then solve the quadratic....4x^2-11x-12, two numbers that add to give -11 and times to give -12....so that will be 4x^2+1x-12x-3....factorise them -giving 4x(x-3)+1(x-3)..so your final answer should be (4x+1) and (x-3)...x=3 and x=-1/4, sub into any y equations giving y=11(3)+3, y=36 and y=11(-1/4)+3 giving your other answer ...thats how i think you should do it..anyone want to clarify my answer it might be wrong
1
7 years ago
#3
Y= 11X + 3
Y= 4x^2

Please walk me through step by step, I will positive rate your profile
Have you heard of solving by substitution?
0
7 years ago
#4
Rearrange the first one to give : 3 = y - 11x
Substitute in the second one : 4x^2 - 11x = 3
Rearrange to get a quadratic equation : 4x^2 - 11x - 3 = 0
Factorize the quadratic : 4x^2 -12x + 1x - 3
4x(x-3) / -1(x-3)
(4x -1)(x-3)
Then solve normally: x = 1/4 or x = 3
0
7 years ago
#5
Oops I forgot the y bit but it'll be the same as ,,,, above
0
7 years ago
#6
Right, you put theme equal to each other:
11x+3=4x^2
Rearrange:
4x^2-11x-3=0
Factorise into 2 brackets:
(4x+1)(x-3)=0
Then split it up:
4x+1=0
x-3=0

and work out x

and sub it back into the y equation
0
7 years ago
#7
(Original post by ,,,,)
you have to link them together so 11x+3=4x^2...rearrange giving 4x^2-11x-3...times -3 and 4 together giving -12, then solve the quadratic....4x^2-11x-12, two numbers that add to give -11 and times to give -12....so that will be 4x^2+1x-12x-3....factorise them -giving 4x(x-3)+1(x-3)..so your final answer should be (4x+1) and (x-3)...x=3 and x=-1/4, sub into any y equations giving y=11(3)+3, y=33 and y=11(-1/4)+3 giving your other answer ...thats how i think you should do it..anyone want to clarify my answer it might be wrong
Why one earth would you do that?
0
7 years ago
#8
do what?
0
7 years ago
#9
(Original post by ,,,,)
do what?
times the -3 and 4 together bit
0
7 years ago
#10
you negged me didnt you LOL....urm well that's how i was taught?
0
7 years ago
#11
Just sub one of the values into the other, you get

This is a quadratic equation, you can use the quadratic formula or factorise it

Sub those in to the linear equation to find the corresponding y values.
0
#12
(Original post by Stormwhite)
Just sub one of the values into the other, you get

This is a quadratic equation, you can use the quadratic formula or factorise it

Sub those in to the linear equation to find the corresponding y values.
The Y bit is what I'm stuck on... lol
0
7 years ago
#13
(Original post by ,,,,)
you negged me didnt you LOL....urm well that's how i was taught?
lmao no I didn't! :') I have repped too many people today already

Oh, ok, it just seemed a strange way to do it!
0
7 years ago
#14
The Y bit is what I'm stuck on... lol
Well just use your x values, and sub them into the original equation, to work out y...?
0
7 years ago
#15
LOOL ahh never mind then...i guess theres more than one way of doing things
1
7 years ago
#16
(Original post by proud nd luvin it)
Well just use your x values, and sub them into the original equation, to work out y...?
LOL...come on ipad user its not that hard! use 3 and -1/4 in one of the above equations
2
7 years ago
#17
Y= 11X + 3
Y= 4x^2

Please walk me through step by step, I will positive rate your profile
Ok, well. There is one linear equation in there (doesn't have a power to the X). So you substitute the Y = 11X + 3 into the other equation Y= 4x^2.

This would become 4x^2 = 11X + 3.

You would rearrange it to get it all to one side of the quation and it would become 4x^2 - 11X - 3 = 0.

You would then factorise (you know how to do this, right?)

(4X + 1) (X - 3) = 0

And to make each bracket equal zero (you have to get X as the subject of the equation):

4X + 1 = 0
4X = -1
Therefore X = -1/4

and:

X - 3 = 0
X = 3

Substitute both values into the equation (the linear is easiest to do), so:

Y = 11X + 3

For X = -1/4, Y = 11(-1/4) + 3 = 1/4

For X = 3, Y = 11(3) + 3 = 36

So the answer is X = -1/4 and Y = 1/4, and X = 3 and Y = 36

Haven't done them in ages, so please correct me if I'm wrong!

Please quote me if you want anything explaining that didn't make sense. I have some useful websites you can look at if you still don't get it.
0
7 years ago
#18
(Original post by Dizzy in my Head)
Ok, well. There is one linear equation in there (doesn't have a power to the X). So you substitute the Y = 11X + 3 into the other equation Y= 4x^2.

This would become 4x^2 = 11X + 3.

You would rearrange it to get it all to one side of the quation and it would become 4x^2 - 11X - 3 = 0.

You would then factorise (you know how to do this, right?)

(4X + 1) (X - 3) = 0

And to make each bracket equal zero (you have to get X as the subject of the equation):

4X + 1 = 0
4X = -1
Therefore X = -1/4

and:

X - 3 = 0
X = 3

Substitute both values into the equation (the linear is easiest to do), so:

Y = 11X + 3

For X = -1/4, Y = 11(-1/4) + 3 = 1/4

For X = 3, Y = 11(3) + 3 = 36

So the answer is X = -1/4 and Y = 1/4, and X = 3 and Y = 36

Haven't done them in ages, so please correct me if I'm wrong!
0
7 years ago
#19
(Original post by ,,,,)
*you're
1
7 years ago
#20
(Original post by ,,,,)
Haha, great! I took A level Maths, if I didn't get that right I think I would fail life as a mathematician.

Thank you!
0
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