Question

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$$\begin{gathered} 2x-3y=7, \\ (a+b)x-(a+b-3)y=4a+b. \end{gathered}$$

Answer 1

The given system of equations:
2x − 3y = 7
⇒ 2x − 3y − 7 = 0 ....(i)
And, (a + b)x − (a + b − 3)y = 4a + b
⇒ (a + b)x − (a + b − 3)y − (4a + b) = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁= −3, c₁ = −7 and a₂ = (a + b), b₂ = −(a + b − 3), c₂ = −(4a + b)
For an infinite number of solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{2}{a+b}=\frac{-3}{-\left(a+b-3\right)}=\frac{-7}{-\left(4a+b\right)}$
$\Rightarrow \frac{2}{a+b}=\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}$
$\Rightarrow \frac{2}{a+b}=\frac{7}{\left(4a+b\right)}\mathrm{and}\frac{3}{\left(\mathrm{a}+\mathrm{b}-3\right)}=\frac{7}{\left(4\mathrm{a}+\mathrm{b}\right)}$
⇒ 2(4a + b) = 7(a + b) and 3(4a + b) = 7(a + b − 3)
⇒ 8a + 2b = 7a + 7b and 12a + 3b = 7a + 7b − 21
⇒ a = 5b ....(iii)
And, 5a = 4b − 21 ....(iv)
On substituting a = 5b in (iv), we get:
25b = 4b − 21
⇒ 21b = −21
⇒ b = −1
On substituting b = −1 in (iii), we get:
a = 5(−1) = −5
∴ a = −5 and b = −1