Question

How do you solve the simultaneous equations `y = 3x^2- 10` and `13x - y = 14`?

Answer 1

Substitute the expression for `y` from the first equation into the second to yield a quadratic in `x`. Solve for `x` and hence find `y`.

`(x, y) = (4, 38)` or `(1/3, -29/3)`

Explanation:

Substitute the expression for `y` from the first equation into the second to get:

`14 = 13x-(3x^2-10) = 13x-3x^2+10`

Subtract the right hand side from the left to get:

`3x^2-13x+4 = 0`

Use a version of the AC Method to factorize this:

Let `A=3`, `B=13`, `C=4`

Find a pair of factors of `AC = 3*4 = 12` whose sum is `B=13`.

The pair `B1=12`, `B2=1` works.

Then for each of the pairs `(A, B1)` and `(A, B2)` divide by the HCF (highest common factor) to find a pair of coefficients of a factor (choosing suitable signs):

`(A, B1) = (3, 12) -> (1, 4) -> (x-4)`
`(A, B2) = (3, 1) -> (3, 1) -> (3x-1)`

So:

`0 = 3x^2-13x+4 = (x-4)(3x-1)`

which has roots `x = 4` and `x=1/3`

If `x = 4` then `y = 3x^2-10 = 48-10 = 38`

If `x = 1/3` then `y = 3x^2-10 = 3*1/9-10 = 1/3-10 = 1/3-30/3 = -29/3`