For what positive values of p the following pair of linear equations have infinite solutions

Solution:

Given, the pair of linear equations is

4x + 5y = 2

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

We have to find the values of p and q for which the linear pair of equations will have infinitely many solutions.

We know that,

For a pair of linear equations in two variables be a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0,

If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\), then

i) the pair of linear equations is dependent and consistent

ii) the graph will be a pair of coincident lines. Each point on the lines will be a solution and so the pair of equations will have infinitely many solutions.

Here, a1 = 4, b1 = 5, c1 = 2

a2 = (2p + 7q), b2 = (p + 8q), c2 = 2q - p + 1

So, a1/a2 = 4/(2p + 7q)

b1/b2 = 5/(p + 8q)

c1/c2 = 2/(2q - p + 1)

So, 4/(2p + 7q) = 5/(p + 8q) = 2/(2q - p + 1)

Case 1) 4/(2p + 7q) = 5/(p + 8q)

On cross multiplication,

4(p + 8q) = 5(2p + 7q)

4p + 32 q = 10p + 35q

4p - 10p = 35q - 32q

-6p = 3q

Dividing by 3 on both sides,

q = -2p ------------ (1)

Case 2) 4/(2p + 7q) = 2/(2q - p + 1)

On cross multiplication,

4(2q - p + 1) = 2(2p + 7q)

8q - 4p + 4 = 4p + 14q

By grouping,

8q - 14q - 4p - 4p + 4 = 0

-6q - 8p = -4

Now, 8p + 6q = 4 ------------------ (2)

Substitute (1) in (2),

8p + 6(-2p) = 4

8p - 12p = 4

-4p = 4

p = -4/4

p = -1

Put p = -1 in (1),

q = -2(-1)

q = 2

Therefore, the values of p and q are -1 and 2

✦ Try This: For which values of p and q, will the following pair of linear equations have infinitely many solutions? 5x + 4y = 12; (p + 7q) x + (3p + 2q) y = 2q - p + 1

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 3

NCERT Exemplar Class 10 Maths Exercise 3.3 Sample Problem 1

Summary:

For values of p = -1 and q =2, the pair of linear equations 4x + 5y = 2; (2p + 7q) x + (p + 8q) y = 2q - p + 1 has infinitely many solutions.

☛ Related Questions:

This 2019 CBSE class 10 Maths 2 mark question is from Linear Equations. An easy 2 mark question in CBSE class 10 sample question paper. A standard NCERT text book question.

Question 12: For what value of p will the following pair of linear equations have infinitely many solutions: (p - 3)x + 3y = p and px + py = 12

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NCERT Solution to Class 10 Maths

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For two equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Condition for infinitely many solutions => \\frac{a_1}{a_2}) = \\frac{b_1}{b_2}) = \\frac{c_1}{c_2})

The given equations are (p - 3)x + 3y - p = 0 and px + py – 12 = 0
\\frac{a_1}{a_2}) = \\frac{b_1}{b_2}) ⇒ \\frac{p - 3}{p}) = \\frac{3}{p})
p2 – 3p = 3p ⇒ p2 – 6p = 0 p(p – 6) = 0 ⇒ p = 0 or 6

\\frac{b1}{b2}) = \\frac{c1}{c2}) ⇒ \\frac{3}{p}) = \\frac{-p}{-12})

The only value that satisfies both the ratios is +6. The value of p is 6.