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I want to show that the following system of equations does not have a solution, but I do not know how to do this

$$w_1+w_2=\frac{1}{2}$$ $$w_1s_1+w_2s_2=\frac{1}{6}$$ $$w_1t_1+w_2t_2=\frac{1}{6}$$ $$w_1s_1t_1+w_2s_2t_2=\frac{1}{24}$$ $$w_1s_1^2+w_2s_2^2=\frac{1}{12}$$ $$w_1t_1^2+w_2t_2^2=\frac{1}{12}$$ that all $t1,t2,w1,w2,s1,s2$ are unknown.

please help me, Thanks.

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this is having linear-algebra and matrices. Observe that the variables you have are in pairs, so the best you can do is use $2 \times 2$ matrices. You can split your system into pairs and find various conditions not to or to have solutions. It is a system of equations so you only need one contradiction to prove it has no solution. – user31280 May 9 '13 at 6:19
up vote 1 down vote accepted

First note that we cannot have $w_1 = 0$ or $w_2 = 0$.

Subtracting the second and third give us

$$w_1(s_1 - t_1) = -w_2(s_2 - t_2)$$

Now if $s_1 = t_1$ then $s_2 = t_2$, and we can verify there is no solution, by comparing the fourth and fifth.

Subtracting the fifth and sixth gives us

$$w_1(s_1^2 - t_1^2) = -w_2(s_2^2 - t_2^2)$$

and thus from the above, we must have that

$$s_1 + t_1 = s_2 + t_2 = x \;(\text{say})$$

Adding second and third gives us

$$(w_1 + w_2) x = \frac{1}{3}$$

and so $$x = \frac{2}{3}$$ using the first equation.

Adding the fifth, sixth and two times the fourth gives us

$$w_1 x^2 + w_2 x^2 = \frac{1}{4}$$

and so $$x = \frac{\pm 1}{\sqrt{2}}$$

a contradiction.

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Since we didn't use any inequalities, you can allow the variables to be complex too. – Aryabhata May 9 '13 at 6:29

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