Question

The value of a for which the given pair of equations (2a - 1)x + (a - 1)y - (2a + 1) = 0 ; 6x + 2 y-1 =0 have no solutions is :
a = 1
a = 2
a = -1
a = -2​

Answer 1

Given that:

• (2a - 1)x + (a - 1)y - (2a + 1) = 0
• 6x + 2y - 1 = 0

To Find:

• For what value of a the equation have no solutions.

We know that:

For no solutions,

• a₁/a₂ = b₁/b₂ ≠ c₁/c₂

We have:

• a₁ = (2a - 1)
• a₂ = 6
• b₁ = (a - 1)
• b₂ = 2
• c₁ = - (2a + 1)
• c₂ = - 1

Finding the value of a:

↣ a₁/a₂ = b₁/b₂

↣ (2a - 1)/6 = (a - 1)/2

Cross multiplication.

↣ 2(2a - 1) = 6(a - 1)

↣ 4a - 2 = 6a - 6

↣ 4a - 6a = - 6 + 2

↣ - 2a = - 4

↣ a = - 4/- 2

↣ a = 2

Verifying the equation:

↣ a₁/a₂ = b₁/b₂ ≠ c₁/c₂

↣ (2a - 1)/6 = (a - 1)/2 ≠ - (2a + 1)/- 1

Putting the value of a.

↣ (2 × 2 - 1)/6 = (2 - 1)/2 ≠ - (2 × 2 + 1)/- 1

↣ (4 - 1)/6 = (2 - 1)/2 ≠ - (4 + 1)/- 1

↣ (3)/6 = (1/2 ≠ - (5)/- 1

↣ 1/2 = 1/2 ≠ 5/1

Hence, verified

So,

• The value of a = 2

Answer 2

Question:

The value of a for which the given pair of equations (2a - 1)x + (a - 1)y - (2a + 1) = 0 ; 6x + 2y - 1 = 0 have no solutions is :

• a = 1
• a = 2
• a = -1
• a = -2

Answer:

• Second option a = 2 is correct.

Step - By - step Explanation:

Given:

• Two equations ; (2a - 1)x + (a - 1)y - (2a + 1) = 0 and 6x + 2y - 1 = 0

To Find:

• Value of a for which given pair of equations have no solution.

Solution:

• Here, we have two equations (2a - 1)x + (a - 1)y - (2a + 1) = 0 and 6x + 2y - 1 = 0, we know that in case of no solution ::
• $\pmb{\boxed{\bf{\purple{\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}}}}}$
• Where, $\bf a_{1}$ is coefficient of x in first equation, $\bf a_{2}$ is coefficient of x in second equation, $\bf b_{1}$ is cofficient of y in first equation, $\bf b_{2}$ is coefficient of y in second equation, $\bf c_{1}$ is constant term in first equation and $\bf c_{2}$ is constant term in second equation.
• We have, $\bf a_{1}$ = (2a - 1), $\bf a_{2}$ = 6, $\bf b_{1}$ = (a - 1), $\bf b_{2}$ = 2, $\bf c_{1}$ = -(2a + 1) and $\bf c_{2}$ = -1.

❏ Finding value of a :

⇒ $\sf \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$

❏ Putting all known values :

⇒ $\sf \dfrac{2a - 1}{6} = \dfrac{a - 1}{2} \neq \dfrac{-(2a + 1)}{-1}$

⇒ $\sf \dfrac{2a - 1}{6} = \dfrac{a - 1}{2}$

★ By cross multiplication :

⇒ $\sf 2(2a - 1) = 6(a - 1)$

⇒ $\sf 4a - 2 = 6a - 6$

⇒ $\sf 6a - 4a = -2 + 6$

⇒ $\sf 2a = 4$

⇒ $\sf a = {\cancel{\dfrac{4}{2}}}$

➠ $\pmb{\blue{\underline{\boxed{\bf{\pink{a = 2}}}}}}$

∴ Hence, value of a for which the given pair of equations have no solution is 2. So, second option a = 2 is correct.

Verification:

↦ $\sf \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$

↦ $\sf \dfrac{2a - 1}{6} = \dfrac{a - 1}{2} \neq \dfrac{2a + 1}{-1}$

❏ Putting value of a in above eqⁿ :

↦ $\sf \dfrac{2(2) - 1}{6} = \dfrac{2 - 1}{2} \neq \dfrac{-[2(2) + 1]}{-1}$

↦ $\sf \dfrac{4 - 1}{6} = \dfrac{1}{2} \neq \dfrac{-(4 + 1)}{-1}$

↦ $\sf {\cancel{\dfrac{3}{6}}} = \dfrac{1}{2} \neq \dfrac{\cancel{-}5}{\cancel{-}1}$

➦ $\pmb{\blue{\underline{\boxed{\bf{\pink{\dfrac{1}{2} = \dfrac{1}{2} \neq \dfrac{5}{1}}}}}}}$

∴ Hence, Verified!

So, our ans i.e, a = 2 is correct.

★ Know More :

$\clubsuit$ In case of unique solution :

• $\pmb{\boxed{\bf{\red{\dfrac{a_{1}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}}}}}}$

$\clubsuit$ In case of infinitely many solutions :

• $\pmb{\boxed{\bf{\blue{\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}}}}}$

$\clubsuit$ In case of no solution :

• $\pmb{\boxed{\bf{\green{\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}}}}}\qquad\Bigg\{\pmb{\sf{Used\:above}}\Bigg\}$

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