Can anyone help with this simultaneous equation (for AS level)?

Question

y = 2x² - 3x + 4, =4x +1
Please show steps of working
It is out of the PearsonLongman AS Core for OCR textbook.
Thanks

Answer 1

take all everything to one side, so it will b equal to 0. then factorise
2x squared -7x +3=0
factorise
(2x -1)(x-3)=0
so x equals 0.5 and 3

Answer 2

y = 2x² - 3x + 4 - 1
y=4x +1 - 2

well we know y =4x +1...therefore we substitute it into the 1st equation

therefore we have

4x+1 = 2x² - 3x + 4 - 1

then we rearrange it all and get it all on the same side so we have:

2x² - 3x + 4 - 1 - 4x -1 = 0

simplified: 2x² - 7x +3 = 0

then we use the quadratic equation

a=2
b= -7
c=3

therefore substituting it into the equation we have

-(-)7 +/- (root) -7² - (4 x 2 x 3) / 2 x 2

then we get 7 +/- (root) 25 / 4

then we have 7 +/- 5 / 4

then the answers we get are 12/4 = 3

and 2/4 = 1/2

therfore your answer is 3 or 1/2

Answer 3

if we have y = 2x^2 - 3x + 4
and also y = 4x+1

then we can equate the two to get

2x^2 -3x + 4 = 4x + 1

then take all the terms onto one side and equate to zero

2x^2 - 3x + 4 - 4x - 1 = 2x^2 -7x + 3 = 0

then find the roots of this quadratic

x = 3 , 1/2

that will lead us to the points of

(3, 13) and (1/2, 3) as being the solutions to both equations

Answer 4

Substitute y=4x+1 in the other given equation

4x+1=2x^2-3x+4
2x^2-7x+3=0
Factorising,
(2x-1)(x-3)=0
Therefore
2x-1=0 and x-3=0
x=1/2 x=3
y=3 y=13

Answer 5

if y = 2x² - 3x + 4 and y =4x +1 then you know that
2x² - 3x + 4 = 4x +1
2x² - 7x + 3 = 0
(2x - 1) (x - 3) = 0
so x equals either 3 or 1/2.

Answer 6

y=2x^2-3x+4=4x=1
y=4x-1
4x=1=2x^2-3x+4
0=2x^2-7x+3 re-arrange and get everything on one side o the equation then factorise
(2x - 1)(x-3)
x = 1/2,3
sub back in to find y
y = 4x-1
y=4(1/2)-1 = 1
y=4x-1
y=4(3)-1 = 11
so you end up with the co-ordinates
(1/2,1) (3,11)
these are where the parabola (curve graph in this case x^2) crosses the x-axis
+3 is where it crosses the y-axis
you can work out the co-ordinates of the vertex by completing the square
2x^2-7x+3
2(x-3.5)^2 + 3
2(x-3.5)^2 + 3 -12.25
2(x-3.5)^2 - 9.25
2(x-3.5)^2 - 18.5
so the co-ordinates now are
(3.5,-18.5)
the 3.5 is now positive because graph functions work like that, you have to make the inside of the bracket equal to 0

Answer 7

im guessing youre saying that both equations are equal

y = 2x^2 - 3x +4

y = 4x + 1

so substitute the second equation in for y in the first equation

4x + 1 = 2x^2 - 3x +4

now get all the x's on one side

4x + 3x - 2x^2 = 4 -1

simplify

7x - 2x^2 = 3

at this point you realize you have to use the quadratic formula, so set this equation equal to zero

7x - 2x^2 - 3 = 0

For the quadratic equation use:
A = -2
B = 7
C = -3

and for answers you get
3 and .5

Answer 8

y = 2x² - 3x + 4
y = 4x + 1

2x² - 3x + 4 = 4x + 1
2x² - 7x + 3 = 0
(2x - 1)(x - 3) = 0
x = 1/2 , x = 3
y = 3 , y = 13

(1/2 , 3) , (3 , 13)

Answer 9

y = 2x^2 - 3x + 4

y = 4x+1

substitute for y:
4x+1 = 2x^2 - 3x +4
take one side
0 = 2x^2 -7x +3
0 = (2x-1)(x-3)
x = 1/2 , 3

Answer 10

substitute the simpler equn in the more complex.

y = 2(4x+1)^2 - 3(4x+1) +4

This can be multiplied into a quadratic equn. Then gather together like tems (squareds, xs and numbers) and solve by factorising (putting into brackets) or by that formula you learnt when you did GCSE