Question

draw the graph and find the many solution of 2x+y=1 and 4x +y =2​

Answer 1

SolutioN :

We have, Pair of linear Equation.

• 2x + y = 1.
• 4x + y = 2.

A/Q,

‡ Many Solution,

• When coefficient of a1 upon a2 equai to b1 upon b2 should be equal to c1 upon c2.

Taking Given Equation.

$\tt\mapsto \dfrac{a_1}{a_2} = \dfrac{b_1} {b_2 } = \dfrac{c_1} {c_2 }$

Where as,

• a1 = 2.
• a2 = 4.
• b1 = 1.
• b2 = 1.
• c1 = 1.
• c2 = 2.

A/Q,

$\tt\mapsto \dfrac{2}{4} = \dfrac{1} {1 } = \dfrac{1} {2 }$

$\tt\mapsto \dfrac{ \cancel{2}}{ \cancel{4}} = \dfrac{1} {1 } = \dfrac{1} {2 }$

$\tt\mapsto \dfrac{1}{2} = \dfrac{1} {1 } = \dfrac{1} {2 }$

★ We are absorb ( Many solution not satisfied given Equation )

$\tt\mapsto \dfrac{1}{2} \red{ \neq \dfrac{1} {1 } }= \dfrac{1} {2 }$

★ According to condition ( it have Unique solution )

Note : Graph provide in attachment.

• Red ( 2x + y = 1. )
• Blue ( 4x + y = 2. )

Answer 2

Step-by-step explanation:

We have, Pair of linear Equation.

2x + y = 1.

4x + y = 2.

A/Q,

‡ Many Solution,

When coefficient of a1 upon a2 equai to b1 upon b2 should be equal to c1 upon c2.

Taking Given Equation.

\tt\mapsto \dfrac{a_1}{a_2} = \dfrac{b_1} {b_2 } = \dfrac{c_1} {c_2 }↦a2a1=b2b1=c2c1

Where as,

a1 = 2.

a2 = 4.

b1 = 1.

b2 = 1.

c1 = 1.

c2 = 2.

A/Q,

\tt\mapsto \dfrac{2}{4} = \dfrac{1} {1 } = \dfrac{1} {2 }↦42=11=21

\tt\mapsto \dfrac{ \cancel{2}}{ \cancel{4}} = \dfrac{1} {1 } = \dfrac{1} {2 }↦42=11=21

\tt\mapsto \dfrac{1}{2} = \dfrac{1} {1 } = \dfrac{1} {2 }↦21=11=21

★ We are absorb ( Many solution not satisfied given Equation )

\tt\mapsto \dfrac{1}{2} \red{ \neq \dfrac{1} {1 } }= \dfrac{1} {2 }↦21≠11=21

★ According to condition ( it have Unique solution )