Question

for what values of k does the following system of equations has a unique solution 5 x - 8 y = 12 , 10 X + k y = - 7​

Answer 1

Answer:

ANSWER

x−Ky=2

3x+2y=−5

a

1

=1 , b

1

=−K

a

2

=3 , b

2

=2

b

1

a

1

=

−K

1

b

2

a

2

=

2

3

b

1

a

1



=

b

2

a

2

→

−K

1



=

2

3

K



=

3

−2

∴ K∈R−(

3

2

)

Answer 2

We have,

$\: \rm{}5x - 8y = 12 \\ \rm{}10x +ky = - 7$

Above Equations are of the form :

$\huge{ \bf{ax + by = c}}$

• For the condition of this question,

We need to have,

$\bf \dfrac{a_1}{b_1} \neq \dfrac{a_2}{b_2} \neq \dfrac{a_3}{b_3}$

So,

$\tt \frac{5}{10} \neq \frac{ - 8}{k} \neq \frac{12}{ - 7}$

Thus,

• k = {0, 1, 2, 3, ...... 8}