Which one of the following pairs of linear equations has/ have infini…

The correct answer is Option (3) → (C), (D) and (A) only

A pair of linear equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) has infinitely many solutions when one equation is a multiple of the other, that is, when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\); then the two lines coincide. If \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), the lines cross at one point and there is a unique solution.

(A) \(9x+3y+12=0\), and \(18x+6y+24=0\) Correct.

• The three ratios are equal: \(\frac{9}{18}=\frac{3}{6}=\frac{12}{24}=\frac{1}{2}\)

• The second equation is the first multiplied by 2

(B) \(2x-3y=13\), and \(7x-2y=20\) Incorrect.

• The x- and y-ratios differ: \(\frac{2}{7}\ne\frac{-3}{-2}\), so the lines cross at one point

• There is a unique solution (\(x=2,\ y=-3\))

(C) \(x+y=5,\ 2x+2y=10\) Correct.

• The three ratios are equal: \(\frac{1}{2}=\frac{1}{2}=\frac{5}{10}\)

• The second equation is the first multiplied by 2

(D) \(3x+y-3=0\) and \(2x+\frac{2}{3}y=2\) Correct.

Written as \(2x+\frac{2}{3}y-2=0\), the ratios are:

• x-coefficients: \(\frac{3}{2}\)

• y-coefficients: \(1\div\frac{2}{3}=\frac{3}{2}\)

• Constant terms: \(\frac{-3}{-2}=\frac{3}{2}\)

• All equal: the first equation is the second multiplied by \(\frac{3}{2}\)