Which of the following pairs of linear equations is/are consistent?
A. \(x+y=5,\ 2x+2y=10\)
B. \(2x+y-6=0,\ 4x-2y-4=0\)
C. \(x-y=8,\ 3x-3y=16\)
D. \(x-2y=0,\ 3x+4y-20=0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A, B and D only
A pair \(a_{1}x+b_{1}y=c_{1}\), \(a_{2}x+b_{2}y=c_{2}\) is consistent when it has at least one solution. Comparing the ratios of the coefficients decides it:
\(\frac{a_{1}}{a_{2}}\ne\frac{b_{1}}{b_{2}}\): the lines cross, one solution (consistent)
All three ratios equal: the lines coincide, infinitely many solutions (consistent)
\(\frac{a_{1}}{a_{2}}\) equal to \(\frac{b_{1}}{b_{2}}\) but not to \(\frac{c_{1}}{c_{2}}\): parallel lines, no solution (inconsistent)
A. x + y = 5, 2x + 2y = 10 Correct. The second equation is the first doubled (every ratio is \(\frac{1}{2}\)), so the lines coincide and every point of the line is a solution.
B. 2x + y − 6 = 0, 4x − 2y − 4 = 0 Correct. The x-ratio \(\frac{2}{4}\) differs from the y-ratio \(\frac{1}{-2}\), so the lines cross at one point, (2, 2).
C. x − y = 8, 3x − 3y = 16 Incorrect. The x- and y-ratios are both \(\frac{1}{3}\), but the constants give \(\frac{8}{16}\); the lines are parallel and never meet, so the pair is inconsistent.
D. x − 2y = 0, 3x + 4y − 20 = 0 Correct. The x-ratio \(\frac{1}{3}\) differs from the y-ratio \(\frac{-2}{4}\), so the lines cross at one point, (4, 2).