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# Find the value(s) of $ p $ and $ q $ for the following pair of equations:

$ 2 x+3 y=7 $ and $ 2 p x+p y=28-q y $,

if the pair of equations have infinitely many solutions.

Given:

The given system of equations is:

\( 2 x+3 y=7 \) and \( 2 p x+p y=28-q y \),

To do:

We have to find the values of \( p \) and \( q \) for which the given system of equations has infinitely many solutions.

Solution:

The given system of equations can be written as:

$2x + 3y -7=0$

$2px +(p+q)y -28=0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

The condition for which the above system of equations has infinitely many solutions is

$\frac{a_{1}}{a_{2}} =\frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Comparing the given system of equations with the standard form of equations, we have,

$a_1=2, b_1=3, c_1=-7$ and $a_2=2p, b_2=p+q, c_2=-28$

Therefore,

$\frac{2}{2p}=\frac{3}{p+q}=\frac{-7}{-28}$

$\frac{1}{p}=\frac{1}{4}$

$p=4$

$\frac{3}{p+q}=\frac{1}{4}$

$4\times 3=1(p+q)$

$p+q=12$

$4+q=12$

$q=12-4=8$

The values of $p$ and $q$ for which the given system of equations has infinitely many solutions are $4$ and $8$ respectively.

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