Question

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

Answer 1

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:-

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

Answer 2

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, $\boxed{k=6}$