Are the following pair of linear equations consistent? Justify your answer.
(i) \(-3x - 4y = 12\), \(4y + 3x = 12\)
(ii) \(\dfrac{3}{5}x - y = 12\), \(\dfrac{1}{5}x - 3y = 16\)
(iii) \(2ax + by = a\), \(4ax + 2by - 2a = 0\); \(a,b \ne 0\)
(iv) \(x + 3y = 11\), \(2(2x + 6y) = 22\)
(i) Inconsistent, (ii) Consistent, (iii) Consistent, (iv) Inconsistent
Rule to check: If
\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}\)
the lines are parallel ⇒ inconsistent (no solution).
Otherwise, they are consistent (they meet or coincide).
(i) Put both in standard form.
\(-3x - 4y - 12 = 0\)
\(3x + 4y - 12 = 0\)
Compute the three ratios.
\(\dfrac{a_1}{a_2} = \dfrac{-3}{3} = -1\)
\(\dfrac{b_1}{b_2} = \dfrac{-4}{4} = -1\)
\(\dfrac{c_1}{c_2} = \dfrac{-12}{-12} = 1\)
Here the first two ratios are equal, but the third is different.
So the lines are parallel ⇒ inconsistent.
(ii) Clear fractions.
Multiply both equations by 5.
\(3x - 5y = 60\)
\(x - 15y = 80\)
Compare ratios.
\(\dfrac{a_1}{a_2} = \dfrac{3}{1}\)
\(\dfrac{b_1}{b_2} = \dfrac{-5}{-15} = \dfrac{1}{3}\)
These are not equal, so the lines intersect ⇒ consistent.
(iii) Arrange both equations.
\(2ax + by - a = 0\)
\(4ax + 2by - 2a = 0\)
Notice that the second is exactly twice the first.
Multiply the first by 2 to get the second.
Hence all three ratios are equal, so the lines are coincident.
Therefore, they are consistent with infinitely many solutions.
(iv) Expand the second equation.
\(2(2x + 6y) = 22\)
\(4x + 12y = 22\)
Compare with \(x + 3y = 11\).
Multiply \(x + 3y = 11\) by 4 to match the left side.
\(4x + 12y = 44\)
Left sides match but right sides differ (44 vs 22).
So the ratios of coefficients are equal but constants differ ⇒ parallel.
Hence inconsistent.