# Thread: simultanious equations again

1. ## simultanious equations again

Consider the equations $12x^2-4xy+11y^2=64$ and $16^2x-9xy+11y^2=78$

By letting $y=mx$, show that $7m^2+12m-4=0$ and thus solve the equation

okay, can you help me with this equation and also can someone tell me why it is neccesary to put in the form of y=mx?

2. It's necessary because that's what you have to do in this task
If you replace $y$ with $mx$ in both equations and solve for $m$, you will get the equation $7m^2 + 12 m - 4$ which is quadratic. To solve a quadratic equation, you need to calculate its discriminant $D = b^2 - 4ac$ where $a=7, b=12, c=-4$ are the coefficients of the equation. The solutions are $m_{1}=\frac{-b + \sqrt{D}}{2a}, m_{2}=\frac{-b -\sqrt{D}}{2a}$

3. Hello, requal!

That substitution is a useful method for solving. .Did you try it?

We can, of course, solve it head-on if you prefer . . .

Solve: . $\begin{array}{cccc}12x^2-4xy+11y^2&=&64 & {\color{blue}[1]} \\ 16x^2-9xy+11y^2&=&78 & {\color{blue}[2]} \end{array}$

Subtract [2] - [1]: . $4x^3 - 5xy \:=\:14$

Solve for $y\!:\;\;y \;=\;\frac{4x^2-14}{5}\;\;{\color{blue}[3]}$

Substitute into [1]: . $12x^2 - 4x\left(\frac{4x^2-14}{5x}\right) + 11\left(\frac{4x^2-14}{5x}\right)^2 \:=\:64$

. . which simplifies to: . $9x^4 - 58x^2 + 49 \:=\:0$

. . which factors: . $(x^2-1)(9x^2 - 49) \:=\:0$

. . and has roots: . $x \;=\;\pm1,\:\pm\tfrac{7}{3}$

Substitute into [3]: . $y \;=\;\mp2,\;\pm\tfrac{2}{3}$

Answers: . $(\pm1,\:\mp2),\;\left(\pm\frac{7}{3},\:\pm\frac{2} {3}\right)$