Question

If the system of equations 2x + 3y = 7 and 2px + (p + q)y = 28 has infinitely many solutions, then​

Answer 1

Refer to the attachment.

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Answer 2

EXERCISES

1. Pair of Linear Equations in Two Variables - Exercise 3.1

2. Pair of Linear Equations in Two Variables - Exercise 3.2

3. Pair of Linear Equations in Two Variables - Exercise 3.3

4. Pair of Linear Equations in Two Variables - Exercise 3.4

2x+ 3y= 7 and 2p + py = 28 – qy, if the pair of equations have infinitely many solutions. Find p and q.

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Given pair of linear equations is

2x + 3y = 7

2px + py = 28 – qy

or 2px + (p + q)y – 28 = 0

On comparing with ax + by + c = 0,

We get,

Here, a1 = 2, b1 = 3, c1 = – 7;

And a2 = 2p, b2 = (p + q), c2 = – 28;

a1/a2 = 2/2p

b1/b2 = 3/ (p+q)

c1/c2 = ¼

Since, the pair of equations has infinitely many solutions i.e., both lines are coincident.

a1/a2 = b1/b2 = c1/c2

1/p = 3/(p+q) = ¼

Taking first and third parts, we get

p = 4

Again, taking last two parts, we get

3/(p+q) = ¼

p + q = 12

Since p = 4

So, q = 8

Here, we see that the values of p = 4 and q = 8 satisfies all three parts.

Hence, the pair of equations has infinitely many solutions for all values of p = 4 and q = 8.