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Question 68
Let k be a constant. The equations $$kx + y = 3$$ and $$4x + ky = 4$$ have a unique solution if and only if
Solution
Two linear equations ax+by= c and dx+ ey = f have a unique solution if $$\frac{a}{d}\ne\ \frac{b}{e}$$
Therefore, $$\frac{k}{4}\ne\ \frac{1}{k}$$ => $$k^2\ne\ 4$$
=> k $$\ne\ $$ |2|
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