How would you solve this sumultaneous equation?

Question

y = x ²

2x-4y= -12

Could you provide workings as well, but please put your answer first so I can attempt it reach this answer.

With trial and error I found that:
x = 2
y= 4

However, I would like a method because I cannot continue doing trial and error - unless that's the method =?

Answer 1

2x - 4x² = - 12
4x² - 2x - 12 = 0
2x² - x - 6 = 0
(2x + 3)(x - 2) = 0
x = - 3/2 , x = 2
y = 9/4 , y = 4

Answer 2

Both the above answerers have got it right, but I don't think their explanations are very clear. Here's my attempt:

The first equation is y = x2
The second equation is 2x - 4y = -12

Divide the second eq. by the common factor: x - 2y = -6

Substitute x2 for y: x - 2x2 = -6

Reduce the right hand side to 0: x - 2x2 + 6 = 0

Change all the signs and put the terms in proper order:
2x2 - x - 6 = 0

Find the factors by trial and error: they are (2x + 3)(x - 2) = 0

Therefore either 2x + 3 = 0 or x - 2 = 0

If 2x + 3 = 0 then x = -1.5 and y = 2.25

If x - 2 = 0 then x = 2 and y = 4

(In both cases you get the value of y by squaring the value of x. Don't forget that the square of a negative number is positive.)

This might be too simple an explanation for an intending Oxford undergraduate (and eventually graduate!) but I hope it helps.

Answer 3

When x =2, y = 4 , and when x= -3/2, y=9/4.
If y =x^2
then 2(x) - 4(x^2) = -12
which gives 4(x^2) -2(x) -12 =0
or 2x^2 -x -6 =0
or (2x+3)(x-2) = 0

so x = -3/2 or x = 2
so y = (-3/2)^2 = 9/4 or (2^2) or 4

Therefore (x,y) = (-3/2 , 9/4) or (2 , 4)

Answer 4

y=x^2-------------------------(1)
2x-4y=-12--------------------(2)
Subsituting for y in (2) we get
2x-4(x^2)=-12
or 2x-4x^2+12=0
or 4x^2-2x-12=0
or 2x^2-x-6=0
or 2x^2-4x+3x-6=0
or 2x(x-2)+3(x-2)=0
or (x-2)(2x+3)=0
so either x=2 or 2x=-3 or x=-3/2
subsituting in (1),we get
for x=2,y=2^2=4
& for x=-3/2,y=9/4 ans

Answer 5

(-3/2,9/4) and (2,4)

sub 1 into 2
2x-4x²=-12
2x²-x-6=0
(2x+3)(x-2)=0
x=-3/2, 2
(-3/2,9/4) and (2,4)