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})
p2 = 36 => p = ± 6
The only value that satisfies both the ratios is +6. The value of p is 6.