Math, asked by StarlightPhoenix, 1 month ago

The graphical representation of a pair of equations 4x + 3y -1 = 5 and 12x + 9y = 15 will be ?

a) parallel lines
b) coincident lines
c) interesting lines
d) perpendicular lines​

Answers

Answered by amitnrw
7

Given : pair of equations 4x + 3y -1 = 5 and 12x + 9y = 15

To Find : graphical representation of a pair of equations  will be

a) parallel lines

b) coincident lines

c) interesting lines

d) perpendicular lines​

Solution:

Pair of linear equations

a₁x  +  b₁y + c₁  =  0

a₂x  +  b₂y + c₂  =  0

Consistent

if a₁/a₂ ≠ b₁/b₂   (unique solution  and lines intersects each others)

  a₁/a₂ = b₁/b₂ = c₁/c₂   (infinite solutions and line coincide each other )

Inconsistent

if  a₁/a₂ = b₁/b₂ ≠  c₁/c₂  ( No solution , lines are parallel to each other)

4x + 3y -1 = 5

=> 4x + 3y  = 6

12x + 9y = 15

4/12 = 3/9 ≠ 6/15

1/3 = 1/3  ≠ 2/5

Hence No solution  as  lines are parallel to each other

parallel lines

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Answered by FιҽɾყPԋσҽɳιx
22

 \sf \: Question

The graphical representation of a pair of equations 4x + 3y -1 = 5 and 12x + 9y = 15 will be ?

a) parallel lines

b) coincident lines

c) interesting lines

d) perpendicular lines

 \large \sf \: Solution -

 \sf \: 4x + 3y = 6  \:  -  -  - (1)

 \sf \: 12x + 9y = 15  \: -  -  -  - (2)

  \sf\dfrac{ \cancel4 {}^{1} }{ \cancel12_3}  =  \dfrac{ \cancel3 {}^{1} }{ \cancel9 _5} \ne \:   \dfrac{ \cancel6 {}^{2} }{5_2}

\sf \dfrac{1}{3}  =  \dfrac{1}{3} \ne \dfrac{2}{5}

{\rule{500pt}{5pt}}

Additional informatioN

 \sf \: case  \: \:1 = when \:  \dfrac{a1}{a2} \ne \dfrac{b1}{b2}

The lines are intersecting and solution is unique

\sf \: case  \: \:2 = when \:  \dfrac{a1}{a2}  =  \dfrac{b1}{b2}  =  \dfrac{c1}{c2}

The lines are coincident and there are infinitely many solution.

\sf \: case  \: \:3 = when \:  \dfrac{a1}{a2}  =  \dfrac{b1}{b2} \ne   \dfrac{c1}{c2}

Then lines are parallel and no solution

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