Question

If $2x-3y=7\mathrm{and}(\mathrm{a}+\mathrm{b})\mathrm{x}-(\mathrm{a}+\mathrm{b}-3)\mathrm{y}=4\mathrm{a}+\mathrm{b}$ have an infinite number of solutions, then
(a) a = 5, b = 1
(b) a = −5, b = 1
(c) a = 5, b = −1
(d) a = −5, b = −1

Answer 1

The correct option is (d).

The given system of equations can be written as follows:
2x − 3y − 7 = 0 and (a + b)x − (a + b − 3)y − (4a + b) = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁ = −3, c₁ = −7 and a₂ = (a + b), b₂ = −(a + b − 3) and c₂ = −(4a + b)
∴ $\frac{a_1}{a_2}=\frac{2}{\left(a+b\right)},\frac{b_1}{b_2}=\frac{-3}{-\left(a+b-3\right)}=\frac{3}{\left(a+b-3\right)}\mathrm{and}\frac{c_1}{c_2}=\frac{-7}{-\left(4a+b\right)}=\frac{7}{\left(4a+b\right)}$
For an infinite number of solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
∴ $\frac{2}{\left(a+b\right)}=\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}$
Now, we have:
$\frac{2}{\left(a+b\right)}=\frac{3}{\left(a+b-3\right)}\Rightarrow 2a+2b-6=3a+3b$
⇒ a + b + 6 = 0 ...(i)
Again, we have:
$\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}\Rightarrow 12a+3b=7a+7b-21$
⇒ 5a − 4b + 21 = 0 ...(ii)

On multiplying (i) by 4, we get:
4a + 4b + 24 = 0 ...(iii)
On adding (ii) and (iii), we get:
9a = −45 ⇒ a = −5
On substituting a = −5 in (i), we get:
−5 + b + 6 = 0 ⇒ b = −1
∴ ​a = −5 and b = −1