Question

For which the values of p and q will the following pair of linear equations have infinitely many solutions 4x+5y=2, (2p+7q)x+(p+8q)y=2q-p+1

Answer 1

Heya user!!!!!!
If the equations have many solution, then a1/a2=b1/b2=c1/c2.....
So, 4/(2p+7q)=5/(p+8q)=2/(2q-p+1)
So,
Taking the first two equations, we get :-
10p+35q=4p+32q
=>6p+3q=0
=>2p+q=0
Multiplying the above equation by 6(we did it because further we will require this).
12p+6q=0
Now, we take the second and third and equate them, we will get :-
2p+16q=10q-5p+5
=>7p+6q=5
Equating both the equations, we get :-
p=-1,q=2

Answer 2

The value of p = -1 and q = 2

Given:

4x+5y=2

(2p+7q)x + (p+8q)y = 2q-p+1

Solution:

If the equations have many solution, then $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \ldots$.

From the equation,

$a_{1}=4$

$b_{1}=5$

$c_{1}=2$

$a_{2}=(2 p+7 q)$

$b_{2}=(p+8 q)$

$c_{2}=2 q-p+1$

Thus,

$\Rightarrow \frac{4}{2 p+7 q}=\frac{5}{p+8 q}=\frac{2}{2 q-p+1}$

So,

Taking the first two equations, we get it,

10p+35q=4p+32q

\Rightarrow 6p+3q=0

\Rightarrow 2p+q=0

Above the equation is multiplied by 6,

$12p+6q=0 \rightarrow (1)$

Now, we take the second and third and equate them, we will get it answer:-

2p+16q = 10q-5p+5

$\Rightarrow 7p+6q=5 \rightarrow (2)$

On solving, equation (1) and (2), we get,

p = -1; q = 2