Question

☺️☺️50 points________

♣️ for what values of p and q will the following pair of linear equations has infinitely solutions ?

4x 5y = 2 ; (2p 7q)x (p 8q)y = 2q-p1

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Answer 1

Your question is wrong.

It should be >

For what values of p and q will the following pair of linear equations has infinite solutions?

4x + 5y = 2

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

ANSWER

p = -1 and q =2

Step By Step Explanation

The given equations are >>

4x + 5y - 2 = 0

(2p + 7q)x + (p + 8q)y - (2q-p +1) =0

We know that standard form of any pair of a linear equation is of the form>

$a_{1}x + b_{1}y + c_{1} = 0$

$a_{2}x + b_{2}y + c_{2} = 0$

Also for having infinitely many solutions we have ->

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

And on comparing with the standard form of the linear equation we get →

$a_{1} = 4 \: and \: a_{2} = 2p + 7q$

$b_{1} = 5 \: and \: b_{2} = p + 8q$

$c_{1} = -2 \: and \: c_{2} = -(2q - p+1)$

Now we have →

$\dfrac{4 }{2p + 7q } = \dfrac{5} {p+8q} = \dfrac{-2 }{-(2q - p+1) }$

$\implies \dfrac{4 }{2p + 7q } = \dfrac{5} {p+8q} = \dfrac{2}{2q - p+1}$

So,

Either

$\dfrac{4 }{2p + 7q } =\dfrac{5} {p+8q}$

On cross multiplying we get →

$\implies 4(p+8q) =5 (2p + 7q)$

$\implies 4p + 32q= 10p + 35q$

$\implies 35q - 32q + 10p - 4p=0$

$\implies 3q +6p=0 -----(i)$

$\implies 2p + q =0$

or

$\dfrac{5} {p+8q} = \dfrac{2}{2q - p+1}$

On cross multiplying we get→

$\implies 5(2q - p +1) =2(p + 8q)$

$\implies 10q - 5p + 5 = 2p + 16q$

$\implies 5 = 2p + 5p + 16q - 10q$

$\implies 7p + 6q = 5 ---(ii)$

So on multiplying equation (i) by 6 and (ii)by 1, we get→

12p + 6q = 0

7p + 6q = 5 (Substracting )

___________________________

5p = -5

=> p = -1

On putting p = -1 in equation {i}we get→

2(-1)+ q = 0

=> q = 2

Hence, for p = -1 and q =2 the pair of linear equations has infinitely many solutions.