Question

please answer it's very important for me.....​

Answer 1

ANSWER:

• k = (-15) then the given equation have no solution.

GIVEN:

• kx- 5y-2 =0. .....(i)
• 6x+2y-1= 0. ......(ii)

TO FIND:

• Value of 'k' for which the given equation have no solution.

SOLUTION:

Formula

$for \: no \: soution \: \\ \frac{a}{p} = \frac{b}{q} \neq \frac{c}{r}$

Here. a = k b= (-5). c = (-2). from eq(i)

p= 6. q= 2. c= (-1). from eq(ii)

Putting these values in the formulae we get;

$\frac{k}{6} = \frac{( - 5)}{2} \neq \: \frac{( - 2)}{( - 1)} \\ when \: \implies \: \frac{k}{6} = \frac{( - 5)}{2} \: \\ \implies \: k = \frac{( - 5)}{2} \times 6 \\ \implies \: k = ( - 15) \\ when \: \implies \: \frac{k}{6} \neq \: \frac{2}{1} \\ \implies \: k \: \neq \: 12$

So , k = (-15) then the given equation have no solution.

NOTE:

some more formulas:

• For unique solution
• $\implies \: \frac{a}{p} \neq \frac{b}{q}$
• For infinite solution
• $\frac{a}{p} = \frac{b}{q} = \frac{c}{r}$
• For no solution
• $\frac{a}{p} = \frac{b}{q} \neq \frac{c}{r}$

Answer 2

AnsWer :

-15.

Given :

• General Equation of pair of liner equation in two variable, i.e. a1/a2 = b1/ b2 = c1/c2.
• We have Equation which is, kx -5y = 2. and 6x + 2y = 1.
• Following equation has no solution, it means a1/a2 = b1/ b2 not equal to c1/c2.

Solution :

We have equation,

$\tt \implies kx - 5y = 2. - - - (1)\\ \tt \implies6x + 2y = 1. - - - (2)$

Where as,

• a1 = k , b1 = -5 and c =2.
• a2 = 6 , b2 = 2 and c = 1.

Now, we know [ What condition required for no solution ]

$\implies \tt \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Putting the value given above,

$\implies \tt \frac{k}{6} = \frac{ - 5}{2} \neq \frac{2}{1}$

$\implies \tt \frac{k}{6} = \frac{ - 5}{2}$

$\implies \tt k = \frac{ - 5 \times 6}{2}$

$\implies \tt k = - 15.$

Therefore, the value of k is -15.

$\rule{200}3$

Some information :

• For many Solution,

a1/a2 = b1/ b2 = c1/c2.

• For No Solution,

a1/a2 = b1/ b2 not equal to c1/c2.

• Unique Solution,

a1/a2 not equal to b1/ b2.