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3x + y - 3 = 0; 2x + (2/3)y = 2, does this pair of linear equations have no solution

Solution:

Given, the pair of linear equations is

3x + y - 3 = 0

2x + (2/3)y = 2

We have to determine whether the pair of equations has a solution or not.

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.

Here, a₁ = 3, b₁ = 1, c₁ = -3

a₂ = 2, b₂ = 2/3, c₂ = -2

So, a₁/a₂ = 3/2

b₁/b₂ = 1/(2/3) = 3/2

c₁/c₂ = -3/-2 = 3/2

3/2 = 3/2 = 3/2

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

Therefore, the pair of equations will has infinitely many solutions.

✦ Try This: Do the pair of linear equations x - 2y = 0; 2x - 2y = 0 have no solution? Justify your answer.

Given, the pair of equations are

x - 2y = 0

2x - 2y = 0

We have to determine whether the pair of equations has a solution or not.

Here, a₁ = 1, b₁ = -2, c₁= 0

a₂ = 2, b₂ = -2, c₂ = 0

So, a₁/a₂ = 1/2

b₁/b₂ = -2/-2 = 1

c₁/c₂ = 0

1/2 ≠ 2

\(\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\)

We know that,

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

If \(\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\), then the graph will be a pair of lines intersecting at a unique point, which is the solution of the pair of equations.

Therefore, the pair of equations has a unique solution

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

NCERT Exemplar Class 10 Maths Exercise 3.2 Problem 1 (iii)

3x + y - 3 = 0; 2x + (2/3)y = 2, does this pair of linear equations have no solution

Summary:

The pair of linear equations 3x + y - 3 = 0; 2x + (2/3)y = 2 has infinitely many solutions.

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