1. ## simulteanous equations

[HTML]as i am solving this question ... problem occurs plz help me there...

3s^2 + 2t^2 = 11 (eq 1)

3s + 2t = 1 (eq 2)

from eq 2 we get

2t=1-3s

:- t=1-3s/2 (eq 3)

put equation no 3 into eq no 1

3s^2 +2(1-3s/2)^2 =11

plz help solve further

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2. Hello, faisalmath!

$\displaystyle \begin{array}{cccc}3x^2 + 2y^2 &=& 11 & [1] \\ 3x + 2y &=& 1 & [2] \end{array}$

From [2], we get: .$\displaystyle y\:=\:\frac{1-3x}{2} \;\;[3]$

Put [3] into [1]: .$\displaystyle 3x^2 +2\left(\frac{1-3x}{2}\right)^2 \:=\:11$

plz help solve further.

We have: .$\displaystyle 3x^2 + 2\left(\frac{1-6x+9x^2}{4}\right) \:=\:11 \quad\Rightarrow\quad 3x^2 + \frac{1-6x+9x^2}{2} \:=\:11$

Multiply by 2: .$\displaystyle 6x^2 + 1 - 6x + 9x^2 \:=\:22 \quad\Rightarrow\quad 15x^2 - 6x - 21 \:=\:0$

Factor: .$\displaystyle 3(x+1)(5x-7) \:=\:0 \quad\Rightarrow\quad x \:=\:-1,\:\tfrac{5}{7}$

Substitute into [3]: .$\displaystyle y \;=\;2,\:-\tfrac{8}{5}$

Answers: .$\displaystyle (-1,2),\;\left(\frac{7}{5},\:-\frac{8}{5}\right)$