Question

Find the value of p and q for which the system of equations represent coincident lines
2x + 3y = 7, (p+q+1)x+(p+2q+2)y=4(p+q)+1

Answer 1

→ p = 3 and q= 2

→ refer to attachments
________________

$hope \: \: it \: \: helps$

Answer 2

Answer:

The required values of p and q are : p = 3 and q = 2

Step-by-step explanation:

Given equations of lines :

2x + 3y = 7

(p+q+1)x + (p+2q+2)y = 4(p+q)+1

Since, the coincident line lies one over the other so the given system of equation has infinitely many solutions.

$\implies\frac{2}{p+q+1}=\frac{3}{p+2q+2}=\frac{7}{4(p+q)+1}\\\\\text{Taking first two, We get}\\\\\frac{2}{p+q+1}=\frac{3}{p+2q+2}\\\\\implies 2p+4q+4=3p+3q+3\\\implies p-q=1........(1)\\\text{Now, taking first and three. We get,}\\\\\frac{2}{p+q+1}=\frac{7}{4(p+q)+1}\\\\\implies 8p+8q+2=7p+7q+7\\\implies p+q=-5........(2)\\\\\text{Now, taking second and three. We get, }\\\\\frac{3}{p+2q+2}=\frac{7}{4(p+q)+1}\\\\\implies 12p+12q+3=7p+14q+14\\\implies 5p-2q=11........(3)$

Now, solve equation (1) and (2) then verify the value of p and q with equation (3)

On solving equation (1) and (2) for p and q :

p = 3 and q = 2

Now, putting these values in equation(3)

LHS : 5p - 2q = 5×3 - 2×2 = 15 - 4 = 11 = RHS

So, The required values of p and q are : p = 3 and q = 2