SIMULTANEOUS EQUATION ; y+2x=3 and 2x^2-3xy=14

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Hello Livvie.
We have to solve for x here,.. you start by substituting one of the equation in the other, in such a way that it makes it eaiser to solve.

y + 2x = 3 ---------------(1)
2x^2 - 3xy = 14---------(2)

From (1)
y + 2x = 3
y = 3 - 2x

Replace y in (2)
2x^2 - 3xy = 14
2x^2 - 3x(3 - 2x) = 14
2x^2 - 9x + 6x^2 = 14
2x^2 + 6x^2 - 9x - 14 = 0
8x^2 - 9x - 14 = 0

Factorise:
8x^2 - 9x - 14 = 0
8x^2 + 7x - 16x - 14 = 0
(8x + 7)(x - 2) = 0

x = -7/8 or 2 <<<<--------

Replace values of x in (1) to find y:
y - 7/4 = 3
y = 19/4

y + 4 = 3
y = -1

y = 19/4 or -1 <<<<--------

Hope this helps
Yuv :)

Rearrange the first equation so that y is in terms of x:
y + 2x =3 becomes y = 3 - 2x

Then substitute 3 - 2x for the value of y in the second equation:
2x^2 - 3x (3 - 2x) = 14

And multiply out the brackets to simplify:
2x^2 - 9x + 6x^2 = 14

So.... 2x^2 - 9x + 6x^2 - 14 =0

This can be simplified again to
:
8x^2 - 9x - 14 =0

Then factorize the equation:
(8x + 7) (x - 2) = 0

Now, in order for the equation to equal 0 one set of brackets must equal 0 (because any number multiplied by 0 is 0):
....So, if (x - 2) = 0, x must equal 2
and if (8x +7) = 0, x must equal - 7/8

We can now find y by substituting these values for x into one of the original equations:
....when x - 2, y = 3 - 2 (2) = 3 - 4 = -1
when x = -7/8, y = 3 -2 (-7/8) = 3 + 7/4 = 19/4

y+2x=3 and 2x^2-3xy=14

y = 3 - 2x
Substitute into the other eq'n for 'y'
Hence
2x^2 - 3x(3 - 2x) = 14
2x^2 - 9x + 6x^2 = 14
8x^2 - 9x - 14 = 0
Factorise
(8x + 7)(x - 2) = 0
Hence
x = - 7/8 & x = 2
Substitute these values back into the linear eq'n y + 2x = 3