Question

If the system of equations 2x+3y=7,(a+b)x+(2a-b)y=21 has infinitely many solutions, then find the values of a and b.

Answer 1

There are two lines

Line 1:- 2x+3y-7=0 ; p₁ = 2 , q₁ = 3 , r₁ = -7

Line 2 :- (a+b)x +(2a-b)y-21= 0 ; p₂ = a+b, q₂=2a-b and r₂= -21

Now, As lines have infinitely many solutions both are coincident lines. And the condition for coincident lines is,

p₁/p₂ = q₁/q₂ = r₁/r₂

$\frac{2}{a + b} = \frac{3}{2a - b} = \frac{ - 7}{ - 21}$

Therefore, we get two equations

$\frac{2}{a + b} = \frac{ - 7}{ - 21} \: \: and \: \: \frac{3}{2a - b} = \frac{ - 7}{ - 21}$

Let equation 1 will be,

$\frac{2}{a + b} = \frac{ - 7}{ - 21}$

Let equation 2 will be,

$\frac{3}{2a - b} = \frac{ - 7}{ - 21}$

Now, In equation 1,

$\frac{2}{a + b} = \frac{ - 7}{ - 21}$

$\frac{2}{a + b} = \frac{ 1}{ 3}$

$( 3)2 = (a + b)$

$a + b = 6$

$a = 6 - b$

Now by putting this value of "a" in equation 2

$\frac{3}{2a - b} = \frac{ - 7}{ - 21}$

$\frac{3}{2(6 - b) - b} = \frac{ 1}{ 3}$

$3 \times 3 = 2(6 - b) - b$

$9 = 12 - 2b - b$

$9 = 12 - 3b$

$3b = 12 - 9$

$3b = 3$

$b = 1$

Putting value of "b" in equation of "a"

$a = 6 - b$

$a = 6 - 1$

$a = 5$

Answer:- a=5 and b=1