For what value of p does the pair of equation has unique solution?

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Pair Of Linear Equations In Two Variables

Question

For which values of p does the pair of equations given below has unique solution?

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Solution: For a pair of equations to have unique solutions,

Here,
So,

So, above pair equations has a unique solutions for all valuesof .

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Given, pair of linear equations is

2x + 3y – 5 = 0

px – 6y – 8 = 0

On comparing with ax + by + c = 0 we get

Here, a1 = 2, b1 = 3, c1 = – 5;

And a2 = p, b2 = – 6, c2 = – 8;

a1 /a2 = 2/p

b1 /b2 = – 3/6 = – ½

c1 /c2 = 5/8

Since, the pair of linear equations has a unique solution.

a1/a2 ≠ b1/b2

so 2/p ≠ – ½

p ≠ – 4

Hence, the pair of linear equations has a unique solution for all values of p except – 4.

Solution:

In the above equation a1= 4, a2 = 2, b1 = p and b2 = 2.

If the solution of a pair of linear equations is unique, then

a1/a2 ≠ b1/b2

4/2 ≠ p/ 2

Thus, the pair of linear equations has a unique solution for all values of p except 4

☛ Check: NCERT Solutions for Class 10 Maths Chapter 3

For which values of p does the pair of equations given below has unique solution? 4x + py + 8 = 0, 2x + 2y + 2 = 0

Summary:

The pair of linear equations has a unique solution for all the values of p except 4

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Solution : For a pair of equations to have unique solutions,
`a_1/a_2 != b_1/b_2`
Here, `a_1 = 4,a_2=2,b_1=p and b_2 = 2`
So,
`4/2!=p/2=>p!=4`
So, above pair equations has a unique solutions for all valuesof `p` `except` `4`.

\(\frac{a1}{a2}\) ≠ \(\frac{b1}{b2}\) is the condition for the given pair of equations to have unique solution.

\(\frac{4}{2}\) ≠ \(\frac{p}{2}\)

p ≠ 4

Therefore, for all real values of p except 4, the given pair of equations will have a unique solution.

Example 15 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables (Term 1)

Last updated at Dec. 18, 2020 by

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Example 15 For which values of p does the pair of equations given below has unique solution? 4x + py + 8 =0 2x + 2y + 2 = 0 4x + py + 8 = 0 2x + 2y + 2 = 0 4x + py + 8 = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = 4 , b1 = p , c1 = 8 2x + 2y + 2 = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = 2 , b2 = 2 , c2 = 2 So, a1 = 4 , b1 = p , c1 = 8 & a2 = 2 , b2 = 2 , c2 = 2 Since equations has unique number of solutions 𝒂𝟏/𝒂𝟐 ≠ 𝒃𝟏/𝒃𝟐 4/2 ≠ 𝑝/2 𝑝/2 "≠" 4/2 p "≠" 𝟒 So, for all values of p except 4, We will have a unique solution for the given set of questions

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