Question

2. The value of k for which the system of equations x + y - 5 = 0 and 3x + ky = 7 has no
solution, is
(a) 3
(b) 2
(c ) 8
(d ) 9​

Answer 1

Answer:-

Correct option is (a) 3.

Note:-

• The standard form of linear equation in two variables is ax + by + c = 0 .

• Two lines , a1x + b1y + c1 = 0 and a2 + b2y + c2 = 0 have no solution if , they are parallel lines and the Reaction between co-efficients is :-

$\boxed{\sf \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}}$

• For infinite solutions , condition is

$\boxed{\sf \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}}$

• For unique solution , Condition is ;

$\boxed{\sf \dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}}$

Explanation :-

Here given two equations are ,

• x + y - 5 = 0
• 3x + ky - 7 = 0 .

With respect to Standard form a1x + b1y + c1 = 0 and ax2 + b2y + c2 = 0 ,

$\boxed{\begin{minipage}{6cm} a_1=1 \qquad b_1=1 \qquad c_1=(-5) \\ \\ a_2=3\qquad b_2=k \qquad c_2=(-7)\end{minipage}}$

$\underline{\sf For\: no\: solutions\:}$

$=> \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2}\\\\=> \dfrac{1}{3}=\dfrac{1}{k}\neq\dfrac{-5}{-7}\\\\=> \dfrac{1}{3}=\dfrac{1}{k}\\\\=>\bf k = 3$

$\underline{\sf Also , \:}$

$=> \dfrac{-5}{-7}\neq\dfrac{1}{k}\\\\=> \dfrac{5}{7}\neq\dfrac{1}{k}\\\\\bf=> k \neq \dfrac{7}{5}$

Answer 2

Answer:

Answer:-

Correct option is (a) 3.

Note:-

• The standard form of linear equation in two variables is ax + by + c = 0 .

• Two lines , a1x + b1y + c1 = 0 and a2 + b2y + c2 = 0 have no solution if , they are parallel lines and the Reaction between co-efficients is :-

• For infinite solutions , condition is

• For unique solution , Condition is ;

Explanation :-

Here given two equations are ,

x + y - 5 = 0

3x + ky - 7 = 0 .

With respect to Standard form a1x + b1y + c1 = 0 and ax2 + b2y + c2 = 0 ,

• Step-by-step explanation:

hope you under stood