Question

If the pair of equations $2x+3y=11\mathrm{and}(\mathrm{m}+\mathrm{n})\mathrm{x}+(2\mathrm{m}-\mathrm{n})\mathrm{y}=33$ has infinitely many solutions, then

(a) m = −1, n = 5
(b) m = 1, n = 5
(c) m = 5, n = 1
(d) m = 1, n = −1

Answer 1

The correct option is (c).

The given system of equations can be written as follows:
2x + 3y − 11 = 0 and (m + n)x + (2m − n)y − 33 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁ = 3, c₁ =−11 and a₂ = (m + n), b₂ = (2m − n) and c₂ = −33
∴ $\frac{a_1}{a_2}=\frac{2}{\left(m+n\right)},\frac{b_1}{b_2}=\frac{3}{\left(2m-n\right)}\mathrm{and}\frac{c_1}{c_2}=\frac{-11}{-33}=\frac{1}{3}$
For an infinite number of solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
∴ $\frac{2}{\left(m+n\right)}=\frac{3}{\left(2m-n\right)}=\frac{1}{3}$
⇒$\frac{2}{(m+n)}=\frac{1}{3}\mathrm{and}\frac{3}{(2m-n)}=\frac{1}{3}$
⇒ (m + n) = 6 ....(i)
And, (2m − n) = 9 ...(ii)
On adding (i) and (ii) we get:
3m = 15 ⇒ m = 5
On substituting m = 5 in (i) we get:
5 + n = 6 ⇒ n = (6 − 5) = 1
∴​ m = 5 and n = 1