Do the following pair of linear equations have no solution? Justify your answer.
(i) \(2x + 4y = 3\), \(12y + 6x = 6\)
(ii) \(x = 2y\), \(y = 2x\)
(iii) \(3x + y - 3 = 0\), \(\dfrac{2}{3}x + \dfrac{1}{2}y = 2\)
(i) Yes, (ii) No, (iii) No
How to decide: First write each equation in the form \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\).
If \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\) but \(\dfrac{c_1}{c_2}\) is different, the lines are parallel.
Parallel lines never meet, so there is no solution.
(i) Convert both to standard form.
\(2x + 4y - 3 = 0\)
\(6x + 12y - 6 = 0\)
Compute ratios one by one.
\(\dfrac{a_1}{a_2} = \dfrac{2}{6} = \dfrac{1}{3}\)
\(\dfrac{b_1}{b_2} = \dfrac{4}{12} = \dfrac{1}{3}\)
\(\dfrac{c_1}{c_2} = \dfrac{-3}{-6} = \dfrac{1}{2}\)
The first two ratios are equal, but the third is different.
Therefore, the lines are parallel ⇒ no solution.
(ii) Write in standard form.
\(x - 2y = 0\)
\(-2x + y = 0\)
Here \(\dfrac{a_1}{a_2} = \dfrac{1}{-2}\) and \(\dfrac{b_1}{b_2} = \dfrac{-2}{1}\).
These are not equal, so the lines intersect at one point (unique solution).
In fact, using the original form: from \(x = 2y\) and \(y = 2x\), substitute to get \(x = 2(2x)\).
That is \(x = 4x\) ⇒ \(3x = 0\) ⇒ \(x = 0\), and hence \(y = 0\).
So there is a unique solution, not “no solution”.
(iii) Standard form first.
\(3x + y - 3 = 0\)
\(\dfrac{2}{3}x + \dfrac{1}{2}y - 2 = 0\)
Compare ratios.
\(\dfrac{a_1}{a_2} = \dfrac{3}{\,2/3\,} = \dfrac{9}{2}\)
\(\dfrac{b_1}{b_2} = \dfrac{1}{\,1/2\,} = 2\)
Since these are unequal, the lines meet at one point (unique solution).
Therefore, the answer is “No” to “no solution”.