Find the value of P and …

The given system of equations may be written as
2x + 3y = 7
So,  2x + 3y - 7 = 0
 and (p + q) x +(2p -q)y - 21 = 0
The given system of equations is of the form
a₁x + b₁ y + c₁ =0
a₂x + b₂y + c₂=0
where,  a₁ = 2, b₁ = 3, c₁ = -7 and a₂ =(p+ q), b₂= (2p -q), c₂= -21
We have, {tex}\frac{a_1}{a_2}=\frac2{p\;+\;q}\;,\frac{b_1}{b_2}=\frac3{2p-\;q}\;\text{ and }\frac{c_1}{c_2}=\frac{-7}{-21}=\frac{\;1}3{/tex}
If {tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }{/tex} then the pair of linear equations has infinitely many solutions.
The given system of equations will have infinite number of solutions, if
{tex}\frac { 2 } { p + q } = \frac { 3 } { 2 p - q } = \frac { 7 } { 21 }{/tex}
{tex}\Rightarrow \quad \frac { 2 } { p + q } = \frac { 3 } { 2 p - q } = \frac { 1 } { 3 }{/tex}
{tex}\Rightarrow \quad \frac { 2 } { p + q } = \frac { 1 } { 3 } \text { and } \frac { 3 } { 2 p - q } = \frac { 1 } { 3 }{/tex}
{tex}\Rightarrow{/tex} p + q = 6 and 2p - q = 9
{tex}\Rightarrow{/tex} (p + q) + (2p - q ) = 6 + 9
{tex}\Rightarrow{/tex} 3p = 15 [On adding]
{tex}\Rightarrow{/tex} p = 5
Putting p = 5 in p + q = 6 or, 2p - q = 9, we get q = 1.
Hence, the given system of equations will have infinitely many solutions, if p = 5 and q = 1.