Question

for which values of p and q does the following pair of linear equation have an infinite number of solution? 2x+3y=7 (p-q)x + (p+q)y=3p+q-2​

Answer 1

$\huge {\underline {\mathtt {\red {☆A}\pink {N}\green {S}\blue {W}\purple {E}\orange {R}}}}$

Given equations :-

$2x + 3y = 9$

$(p + q)x + (2p - q)y = 3(p + q + 1)$

Here,

$\frac{a1}{a2} = \frac{2}{p + q}$

$\frac{b1}{b2} = \frac{3}{2p - q}$

$\frac{c1}{c2} = \frac{9}{3(p + q + 1)}$

For a pair of linear equation to have infinitely many solutions :-

$\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}$

so we need,

$\frac{2}{p + q} = \frac{3}{2p - q} = \frac{9}{3(p + q + 1)}$

or,

$\frac{2}{p + q} = \frac{3}{2p - q}$

$= > 2(2p - q) = 3(p + q)$

$= > 4p - 2q = 3p + 3q$

$= > p = 5q....(1)$

Also,

$\frac{3}{2p - q} = \frac{9}{3(p + q + 1)}$

$= > 9(p + q + 1) = 9(2p - q)$

$= > p + q + 1 = 2p - q$

$= > 2p - p = q + q + 1$

$= > p = 2q + 1$

Substituting (1) we have,

$5q = 2q + 1$

$q = \frac{1}{3}$

Also,

$p = 5q = 5 \binom{1}{3} = \frac{5}{3}$

$p = \frac{5}{3} \: and \: q = \frac{1}{3}$

Answer 2

Given equations :-

2x + 3y = 92x+3y=9

(p + q)x + (2p - q)y = 3(p + q + 1)(p+q)x+(2p−q)y=3(p+q+1)

Here,

\frac{a1}{a2} = \frac{2}{p + q}

a2

a1

=

p+q

2

\frac{b1}{b2} = \frac{3}{2p - q}

b2

b1

=

2p−q

3

\frac{c1}{c2} = \frac{9}{3(p + q + 1)}

c2

c1

=

3(p+q+1)

9

For a pair of linear equation to have infinitely many solutions :-

\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}

a2

a1

=

b2

b1

=

c2

c1

so we need,

\frac{2}{p + q} = \frac{3}{2p - q} = \frac{9}{3(p + q + 1)}

p+q

2

=

2p−q

3

=

3(p+q+1)

9

or,

\frac{2}{p + q} = \frac{3}{2p - q}

p+q

2

=

2p−q

3

= > 2(2p - q) = 3(p + q)=>2(2p−q)=3(p+q)

= > 4p - 2q = 3p + 3q=>4p−2q=3p+3q

= > p = 5q....(1)=>p=5q....(1)

Also,

\frac{3}{2p - q} = \frac{9}{3(p + q + 1)}

2p−q

3

=

3(p+q+1)

9

= > 9(p + q + 1) = 9(2p - q)=>9(p+q+1)=9(2p−q)

= > p + q + 1 = 2p - q=>p+q+1=2p−q

= > 2p - p = q + q + 1=>2p−p=q+q+1

= > p = 2q + 1=>p=2q+1

Substituting (1) we have,

5q = 2q + 15q=2q+1

q = \frac{1}{3}q=

3

1

Also,

p = 5q = 5 \binom{1}{3} = \frac{5}{3}p=5q=5(

3

1

)=

3

5

p = \frac{5}{3} \: and \: q = \frac{1}{3}p=

3

5

andq=

3

1