Question

For what value of k for which the given system of equations has infinitely many solution (1) 5x+2y = K , 10x+4y =3
Please help me​

Answer 1

Answer:

3/2 is the required value of k .

Step-by-step explanation:

According to the Question

1st equation = 5x+2y=k

➻ 5x+2y-k = 0

where,

a₁ = 5 , b₁ = 2 and c₁ = -k

2nd equation = 10x+4y = 3

➻ 10x+4y-3 = 0

where,

a₂= 10 , b₂ = 4 and c₂ = -3

As we know the condition for Equation which infinitely many solution .

• a₁/a₂ = b₁/b₂ = c₁/c₂

Substitute the value we get

➻ 5/10 = 2/4 = -k/-3

➻ 1/2 = k/3

➻ k = 3/2

• Hence, the given system of equations will have infinitely many solutions if k = 3/2.

Answer 2

QuesTion

• For what value of k for which the given system of equations has infinitely many solution (1) 5x + 2y = K , 10x + 4y = 3.

AnsweR

• The given equations have infinitely many solutions if value of k is 3/2.

Step - By - Step - Explanation

Given that:

• Equations (1) 5x + 2y = K , 10x + 4y = 3.

To Find:

• For what value of k the given equations have infinitely many solutions?

Solution:

• Here, we have two equations (1) 5x + 2y = k or 5x + 2y - k = 0, 10x + 4y = 3 or 10x + 4y - 3 = 0. We know that in case of infinitely many solutions :
• $\pmb{\boxed{\bf{\purple{\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}}}}}$
• 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 coefficient 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 $\bf a_{1}$ = 5, $\bf a_{2}$ = 10, $\bf b_{1}$ = 2, $\bf b_{2}$ = 4, $\bf c_{1}$ = -k and $\bf c_{2}$ = -3.

❏ Finding value of k :

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

❏ Substituting all values :

↦ $\sf {\cancel{\dfrac{5}{10}}} = {\cancel{\dfrac{2}{4}}} = \dfrac{\cancel{-}k}{\cancel{-}3}$

↦ $\sf \dfrac{1}{2} = \dfrac{1}{2} = \dfrac{k}{3}$

↦ $\sf \dfrac{k}{3} = \dfrac{1}{2}$

↦ $\sf k = \dfrac{1}{2}\:\times\:3$

↦ $\sf k = \dfrac{1\:\times\:3}{2}$

➦ $\pmb{\underline{\boxed{\bf{\pink{k = \dfrac{3}{2}}}}}}$

∴ Hence, the value of k for which the given equations have infinitely many solutions is 3/2.

❏ 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 :

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

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