Question

3x+2y-8=0 and 6x+4y-24=0​

Answer 1

Given :- Solve 3x+2y-8=0 and 6x+4y-24=0

Answer :-

→ 3x - 2y - 8 = 0

→ 3x - 2y = 8 -------- Eqn.(1)

and,

→ 6x + 4y - 24 = 0

→ 6x + 4y = 24 ---------- Eqn.(2)

multiply Eqn.(1) by 2 and adding both,

→ 2(3x - 2y) + (6x + 4y) = 2*8 + 24

→ 6x + 6x - 4y + 4y = 16 + 24

→ 12x = 40

→ x = (40/12) = (10/3) .

putting value of x in Eqn.(1),

→ 3(10/3) - 2y = 8

→ 10 - 2y = 8

→ 2y = 10 - 8

→ 2y = 2

→ y = 1 .

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Answer 2

SOLUTION

COMPLETE QUESTION

How many solutions have the pair of linear equations 3x+2y-8=0 and 6x+4y-24=0

CONCEPT TO BE IMPLEMENTED

A pair of Straight Lines

$\displaystyle \sf{ a_1x+b_1y+c_1=0 \: \: and \: \: \: a_2x+b_2y+c_2=0}$

is said to have no solution if

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \: \frac{c_1}{c_2}}$

EVALUATION

Here the given pair of lines are

3x+2y-8=0 and 6x+4y-24=0

Comparing with the lines

$\displaystyle \sf{ a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0}$

We have

$\displaystyle \sf{ a_1 = 3 \ , \: b_1 = 2 , c_1= - 8 \: and \: \: a_2 = 6 \: , \: b_2 = 4\: , \: \: c_2= - 24 \:} \:$

Now

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{3}{6} \ = \: \frac{1}{2}}$

$\displaystyle \sf{ \: \frac{b_1}{b_2} = \frac{2}{ 4} = \frac{1}{2} }$

$\displaystyle \sf{ \frac{c_1}{c_2} = \frac{ - 8}{ - 24} = \frac{1}{3} }$

$\displaystyle \sf{ \therefore \: \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \: \frac{c_1}{c_2}}$

So the given lines have no solution

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