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(3/5)x - y = (1/2); (1/5)x - 3y = (1/6), are these pair of linear equations consistent?

Solution:

Given, the pair of equations is

(3/5)x - y = (1/2)

(1/5)x - 3y = (1/6)

We have to determine if the pair of linear equations is consistent.

We know that,

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

If \(\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\), then the pair of linear equations is consistent

Here, a₁ = 3/5, b₁ = -1, c₁ = 1/2

a₂ = 1/5, b₂ = -3, c₂ = 1/6

So, a₁/a₂ = (3/5)/(1/5) = 3

b₁/b₂ = -1/-3 = 1/3

c₁/c₂ = (1/2)/(1/6) = 3

3 ≠ 1/3

Therefore, the pair of equations is consistent and has unique solutions.

✦ Try This: Determine the pair of equations 2x - 16y = 8 and 4x - 28y = 16 is consistent.

Given the pair of equations are

2x - 14y = 8

4x - 28y = 16

We have to determine if the pair of equations are consistent.

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 the pair of equations is consistent.

Here, a₁ = 2, b₁ = -14, c₁ = -8

a₂ = 4, b₂ = -28, c₂ = -16

So, a₁/a₂ = 2/4 = 1/2

b₁/b₂ = -14/-28 = 1/2

c₁/c₂ = 8/16 = 1/2

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

Therefore, the given pair of equations has infinitely many solutions

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

NCERT Exemplar Class 10 Maths Exercise 3.2 Problem 3 (ii)

(3/5)x - y = (1/2); (1/5)x - 3y = (1/6), , are these pair of linear equations consistent

Summary:

The pair of linear equations (3/5)x - y = (1/2); (1/5)x - 3y = (1/6) is consistent.

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