Question

Value of k for which the pair of linear equations 3x + 4y = 7 and 3x + ky =8 has no solution is?


for class 10

please give correct answer ​

Answer 1

Answer:

Value of k is 4

Step-by-step explanation:

3x + 4y = 7

∴ a₁ = 3. b₁ = 4 c₁ = -7

3x + ky = 8

∴ a₂ = 3. b₂ = k c₂ = - 8

Codition for no solution is

a₁/a₂ = b₁ /b₂ ≠ c₁ /c₂

3/3 = 4/k

∴ k = 4

Answer 2

Given :-

A Pair of linear equations :-

3x + 4y = 7 and 3x + ky = 8

To Find :-

For what values of "k" , the Given equations have no solution .

Used Concepts :-

For two Equations ;

${ \sf { a_{1} x + b_{1} y + c_{1} = 0 } }$

${ \sf { a_{2} x + b_{2} y + c_{2} = 0 } }$

The Pair of Equations have no solution if and only ;

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

Solution :-

3x + 4y = 7 can be written as ;

3x + 4y - 7 = 0

3x + ky = 8 can be written as ;

3x + ky - 8 = 0

On Comparing 3x + 4y - 7 = 0 and 3x + ky - 8 = 0 with ${ \sf { a_{1} x + b_{1} y + c_{1} = 0 } }$ and ${ \sf { a_{2} x + b_{2} y + c_{2} = 0 } }$ . We gets ;

• ${ \sf { a_{1} = 3 } }$
• ${ \sf { a_{2} = 3 } }$
• ${ \sf { b_{1} = 4 } }$
• ${ \sf { b_{2} = k } }$
• ${ \sf { c_{1} = - 7 } }$
• ${ \sf { c_{2} = - 8 } }$

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

${ \sf { \dfrac{3}{3} = \dfrac{4}{k} \neq \dfrac{ - 7 }{ - 8 } } }$

${ \sf { 1 = \dfrac{4}{k} \neq \dfrac{7}{8} } }$

${ \sf { 1 = \dfrac{4}{k} } }$

${ \sf { k = 4 } }$

Henceforth , For k = 4 , the given equations have no solution .