2x + 3y = 7 and 2px + py = 28 - qy, if the pair of equations have infinitely many solutions. Find the value(s) of p and q for the pair of equations
Solution:
Given, the pair of linear equations are
2x + 3y = 7
2px + py = 28 - qy
We have to determine the values of p and q for which the pair of linear equations 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 equation 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.
The equation 2px + py = 28 - qy can be written as
2px + py + qy = 28
2px + (p + q)y = 28
Here, a₁ = 2, b₁ = 3, c₁ = 7
a₂ = 2p, b₂ = p + q, c₂ = 28
So, a₁/a₂ = 2/2p = 1/p
b₁/b₂ = 3/(p + q)
c₁/c₂ = 7/28 = 1/4
For infinitely many solutions,
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
1/p = 3/(p + q) = 1/4
Case 1) 1/p = 3/(p + q)
p + q = 3p
q = 3p - p
q = 2p ----------------- (1)
Case 2) 3/(p + q) = 1/4
3(4) = p + q
p + q = 12 ------------- (2)
Substitute (1) in (2),
p + 2p = 12
3p = 12
p = 12/3
p = 4
Put p = 4 in (1),
q = 2(4)
q = 8
Therefore, for the values of p = 4 and q = 8, the pair of linear equations have infinitely many solutions.
✦ Try This: For which value(s) of λ, do the pair of linear equations λx + 8y = λ and 9x + λy = 11 have infinitely many solutions
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 3
NCERT Exemplar Class 10 Maths Exercise 3.3 Problem 4 (v)
2x + 3y = 7 and 2px + py = 28 - qy, if the pair of equations have infinitely many solutions. Find the value(s) of p and q for the pair of equations
Summary:
For the values of p = 4 and q = 8, the pair of linear equations x + 2y = 1; (a - b)x + (a + b)y = a + b - 2 have infinitely many solutions.
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