$21x + 6y = k$
$(7x-20)^2 + (6y+10)^2 = 200$
if $k = 1$ , how to explain there are no solutions without solving?
How to show that there are no solutions without solving? [simultaneous]
1
$\begingroup$
systems-of-equations
intuition
quadratics
1 Answers
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Let $x' = 7x, y' = 6y$, the the equations become $3 x'+y'=1$, $(x'-20)^2+(y'+10)^2 = 200$.
The distance from the centre $(20,-10)$ to the line is ${49 \over \sqrt{10}}$ and the radius is $10 \sqrt{2}$.
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0is there another way to do this without graphs? – 2017-02-19
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0How do graphs come in to it? If every point in the line is more than a radius away, the line cannot intersect the circle and so there is no solution. – 2017-02-19