Question

pair of the equations 2x+3y=9 and 4x+6y=18 has​

Answer 1

Answer:

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Answer 2

The pair of equations 2x + 3y = 9 and 4x + 6y = 18 have infinite number of solutions

Given :

The pair of equations 2x + 3y = 9 and 4x + 6y = 18

To find :

The number of solutions of the pair of equations

Concept :

For the given two linear equations

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

Consistent :

One of the Below two condition is satisfied

1. Unique solution :

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

2. Infinite number of solutions :

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

Inconsistent :

No solution

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

Solution :

Step 1 of 2 :

Write down the given pair of equations

Here the given pair of linear equations are

2x + 3y - 9 = 0 - - - - - (1)

4x + 6y - 18 = 0 - - - - - (2)

Step 2 of 2 :

Find the number of solutions

2x + 3y - 9 = 0 - - - - - (1)

4x + 6y - 18 = 0 - - - - - (2)

Comparing with the equations

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we get

a₁ = 2 , b₁ = 3 , c₁ = - 9 and a₂ = 4 , b₂ = 6 , c₂ = - 18

Now we have ,

$\displaystyle \sf\frac{a_1}{a_2} = \frac{2}{ 4} = \frac{1}{2}$

$\displaystyle \sf \frac{b_1}{b_2} = \frac{3}{ 6} = \frac{1}{2}$

$\displaystyle \sf \frac{c_1}{c_2} = \frac{ - 9}{ - 18} = \frac{1}{2}$

Thus we get ,

$\displaystyle \sf\frac{a_1}{a_2} = \frac{b_1}{b_2} = \: \frac{c_1}{c_2} = \frac{1}{2}$

Hence the pair of equations 2x + 3y = 9 and 4x + 6y = 18 have infinite number of solutions

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