Math, asked by rehandigra0211, 7 months ago

If the following system equations have infinitely many solutions then value of k will be:2x+3y=5;4x+ky=10​

Answers

Answered by TheMoonlìghtPhoenix
41

Step-by-step explanation:

ANSWER:-

We know that:-

  • If it contains infinite solutions, then the equation stands as:-

 \dfrac{a1}{a2}  =  \dfrac{b1}{b2}  =  \dfrac{c1}{c2}

In this we have :-

  • a1 as 2
  • a2 as 4
  • b1 as 3
  • b2 as k
  • c1 as 5
  • c2 as 10

We need to place the equations.

If I consider first 2, it stands:-

 \dfrac{2}{4}  =  \dfrac{3}{k}

2k = 12

 \boxed{k = 6}

Now, let us verify our answer with second pair.

 \dfrac{3}{k}  =  \dfrac{5}{10}

5k = 30

 \boxed{k = 6}

So, K = 6 is the answer.

More to know:-

If no solutions, then:-

 \dfrac{a1}{a2}  =  \dfrac{b1}{b2}  \neq \dfrac{c1}{c2}


amitkumar44481: Perfect :-)
TheMoonlìghtPhoenix: Thanks!
Answered by Anonymous
12

Given, 2x+3y=5, 4x+ky=10

where

a_1 = 2, b_1 = 3,c_1 = 5 \: in \: 2x + 3y = 5 \: and \\ a_2 = 4,b_2 = k ,c_2 = 10 \: in \: 4x + ky = 10

as \: we \: know,

 \frac{a_1}{a_2}  =  \frac{b_1}{b_2}  =  \frac{c_1}{c_1} [ ∵given \: set \: has \: infinite \: solutions ]

 =  >  \frac{2}{4}  =  \frac{3}{k}  =  \frac{5}{10}

 =  >  \frac{ \cancel 2}{ \cancel 4}  =  \frac{3}{k}  =  \frac{ \cancel 5}{ \cancel10}

 =  >  \frac{1}{2}  =  \frac{k}{2}   = \frac{1}{2}

 =  >  \frac{3}{k}  =  \frac{1}{2}

cross MULTIPLY this,

 =  > k = 3 \times 2

k = 6

hence, \boxed{k=6}

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