Question

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

Answer 1

The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0 ...(i)
And, a(x + y) − b(x − y) = 3a + b − 2
⇒ ax + ay − bx + by = 3a + b − 2
⇒ (a − b)x + (a + b)y − (3a + b − 2) = 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), c₂ = −(3a + b − 2)
For an infinite number of solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{2}{\left(a-b\right)}=\frac{3}{\left(a+b\right)}=\frac{7}{\left(3a+b-2\right)}$
$\Rightarrow \frac{2}{\left(a-b\right)}=\frac{3}{\left(a+b\right)}\mathrm{and}\frac{3}{\left(\mathrm{a}+\mathrm{b}\right)}=\frac{7}{\left(3\mathrm{a}+\mathrm{b}-2\right)}$
⇒ 2(a + b) = 3(a − b) and 3(3a + b − 2) =7(a + b)
⇒ 2a + 2b = 3a − 3b and 9a + 3b − 6 = 7a + 7b
⇒ a = 5b ...(iii)
And, 2a − 4b = 6
or a − 2b = 3 ....(iv)
On substituting a = 5b in (iv), we get:
5b − 2b = 3
⇒ 3b = 3
⇒ b = 1
On substituting b = 1 in (iii), we get:
a = 5
∴ a = 5 and b = 1