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

Jul 26, 2015

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

$\left(x , y\right) = \left(4 , 38\right)$ or $\left(\frac{1}{3} , - \frac{29}{3}\right)$

Explanation:

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

$14 = 13 x - \left(3 {x}^{2} - 10\right) = 13 x - 3 {x}^{2} + 10$

Subtract the right hand side from the left to get:

$3 {x}^{2} - 13 x + 4 = 0$

Use a version of the AC Method to factorize this:

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

Find a pair of factors of $A C = 3 \cdot 4 = 12$ whose sum is $B = 13$.

The pair $B 1 = 12$, $B 2 = 1$ works.

Then for each of the pairs $\left(A , B 1\right)$ and $\left(A , B 2\right)$ divide by the HCF (highest common factor) to find a pair of coefficients of a factor (choosing suitable signs):

$\left(A , B 1\right) = \left(3 , 12\right) \to \left(1 , 4\right) \to \left(x - 4\right)$
$\left(A , B 2\right) = \left(3 , 1\right) \to \left(3 , 1\right) \to \left(3 x - 1\right)$

So:

$0 = 3 {x}^{2} - 13 x + 4 = \left(x - 4\right) \left(3 x - 1\right)$

which has roots $x = 4$ and $x = \frac{1}{3}$

If $x = 4$ then $y = 3 {x}^{2} - 10 = 48 - 10 = 38$

If $x = \frac{1}{3}$ then $y = 3 {x}^{2} - 10 = 3 \cdot \frac{1}{9} - 10 = \frac{1}{3} - 10 = \frac{1}{3} - \frac{30}{3} = - \frac{29}{3}$