Question

13. The pair of equations x + 2y – 5 = 0 and −3x – 6y + 15 = 0 have:
1 point
unique solution
infinite many solution
no sulution
two solution

Answer 1

Answer :

Infinitely many solutions

Note:

★ A linear equation is two variables represent a straight line .

★ The word consistent is used for the system of equations which consists any solution .

★ The word inconsistent is used for the system of equations which doesn't consists any solution .

★ Solution of a system of equations : It refers to the possibile values of the variable which satisfy all the equations in the given system .

★ A pair of linear equations are said to be consistent if their graph ( Straight line ) either intersect or coincide each other .

★ A pair of linear equations are said to be inconsistent if their graph ( Straight line ) are parallel .

★ If we consider equations of two straight line

ax + by + c = 0 and a'x + b'y + c' = 0 , then ;

• The lines are intersecting if a/a' ≠ b/b' .

→ In this case , unique solution is found .

• The lines are coincident if a/a' = b/b' = c/c' .

→ In this case , infinitely many solutions are found .

• The lines are parallel if a/a' = b/b' ≠ c/c' .

→ In this case , no solution is found .

Solution :

Here ,

The given linear equations are ;

x + 2y - 5 = 0

-3x - 6y + 15 = 0

Clearly , we have ;

a = 1

a' = -3

b = 2

b' = -6

c = -5

c' = 15

Now ,

a/a' = 1/-3 = -⅓

b/b' = 2/-6 = -⅓

c/c' = -5/15 = -⅓

Clearly ,

a/a' = b/b' = c/c' = -⅓

Hence ,

The given pair of lines have infinitely many solutions .

Answer 2

Option (2) infinite many solution.

Solution :-

The given equations are :

• x + 2y - 5 = 0
• -3x - 6y + 15 = 0

From the given equation we have :

$\sf \: \mapsto \: \frac{a_1}{a_2} = \frac{x}{ - 3x} = \frac{1}{ - 3}$

$\mapsto \sf \: \frac{b_1}{b_2} = \frac{2y}{ - 6y} = \frac{1}{ - 3}$

$\mapsto \sf \: \frac{c_1}{c_2} = \frac{ - 5}{15} = -\frac{1}{3}$

Hence,

$\mapsto \: { \boxed { \red{\sf \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} }}}$

Therefore, The given pair of lines have infinite many solutions.