Answer
$(3,6),\left(\frac{1}{3} ,-\frac{98}{9}\right)$
Work Step by Step
$y=x^2+3x-12$ ........................ eq (1)
$y=4x^2-7x-9$............................ eq (2)
Comparing equation (1) and equation (2)
$4x^2-7x-9=x^2+3x-12$
$4x^2-7x-9-x^2-3x+12=0$
$4x^2-x^2-7x-3x-9+12=0$
$3x^2-10x+3=0$
$3x^2-9x-x+3=0$
$3x(x-3)-1(x-3)=0$
$(3x-1)(x-3)=0$
$\Longrightarrow$
$3x-1=0$ Or $(x-3)=0$
$3x-1=0$ $\Longrightarrow$ $3x=1$ $\Longrightarrow$ $x=\frac{1}{3}$
Putting in equation (1)
$y=(\frac{1}{3})^2+3\times \frac{1}{3}-12$
$y=-\frac{98}{9}$
Solution
$(\frac{1}{3},-\frac{98}{9})$
Now
$x-3=0$ $\Longrightarrow$ $x=3$
Putting in equation (1)
$y=3^2+3\times3-12=6$
Solution
$(3,6)$
Check
Putting $(\frac{1}{3},-\frac{98}{9})$ in equation (2)
$-\frac{98}{9}=4(\frac{1}{3})^2-7(\frac{1}{3})-9$
$-\frac{98}{9}=\frac{4}{9} -\frac{7}{3}-9$
$-\frac{98}{9}=-\frac{98}{9}$
LHS=RHS
Putting $(3,6)$ in equation (2)
$6=4(3)^2-7(3)-9$
$6=6$
LHS=RHS
