Question

How many solutions does the pair equations x+y=1 and x+y=-5 have? faf Unique (c) Infinitely mony (b) No solution (d) Can't decide

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm \: x + y = 1 - - - - (1)$

and

$\rm \: x + y = - 5 - - - - (2)$

Comparing the given two equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get  

a₁ = 1

b₁ = 1

c₁ = 1

a₂ = 1

b₂ = 1

c₂ = - 5

Now, Consider

$\rm \: \dfrac{a_1}{a_2} = \dfrac{1}{1} = 1$

$\rm \: \dfrac{b_1}{b_2} = \dfrac{1}{1} = 1$

$\rm \: \dfrac{c_1}{c_2} = - \dfrac{1}{5}$

From this, we concluded that

$\rm \: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$

This implies, System of pair of linear equation is inconsistent having no solution.

It means, Given pair of linear equations have no solution.

So, Option (b) is correct.

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Basic Concept Used

The pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 have

1. Unique solution iff

$\rm \: \dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$

2. Infinitely many solutions iff

$\rm \: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

3. No solution iff

$\rm \: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$

Answer 2

Answer:

${\implies}{ \boxed{\mathbb{\blue{given \: that}}}} \\ x + y = 1 \\ x + y = - 5 \\ to \: find \: \\{ \implies}{ \boxed{\mathbb{\red{solution }}}} \\ \: does \: this \: pair \: of \: equation \: have \\ compared \: the \: given \: equation \: by \: \\ a^{1} x + b^{1} y + c^{1} = 0 \: and \\ a^{2} x + b^{2} y + c^{2}{ }{ {\mathbb{\pink{(where \: {?}^{2} is \: not \: square)}}}}we \: get \: \\ a {}^{1} = 1 \\ b {}^{1} = 1 \: and \: c {}^{1} = - 1 \\ a {}^{2} = 1 \\ b {}^{2} = 1 \\ c {}^{2} = 5 \\ by \: using \: \frac{a {}^{1} }{a {}^{2} } = \frac{b {}^{1} }{b {}^{2} } is \: not \: equal \: to \: \frac{c {}^{1} }{c {}^{2} } we \: get \: \\ \frac{1}{1} = \frac{1}{1} is \: not \: equal \: to \frac{ - 1}{5} \\{ \boxed{\mathbb{\red{therefore \: it \: has \: no \: solution}}}}$

Step-by-step explanation: