Question

the pair of equations 3x-5y=7 and -6x+10y=7 has
a) unique solution
b)infinitely many solutions
c)no solution
d)two solutions

Answer 1

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Answer 2

The pair of equations 3x - 5y = 7 and - 6x + 10y = 7 have no solution

Given :

The pair of equations 3x - 5y = 7 and - 6x + 10y = 7

To find :

The pair of equations 3x - 5y = 7 and - 6x + 10y = 7 has

a) unique solution

b) infinitely many solutions

c) no solution

d) two solutions

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

3x - 5y = 7 - - - - (1)

- 6x + 10y = 7 - - - - - (2)

Step 2 of 2 :

Find the number of solutions

The given pair of equations can be rewritten as

3x - 5y - 7 = 0 - - - - - (1)

- 6x + 10y - 7 = 0 - - - - - (2)

Comparing with the equations

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

a₁ = 3 , b₁ = - 5 , c₁ = - 7 and a₂ = - 6 , b₂ = 10 , c₂ = - 7

Now we have ,

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

$\displaystyle \sf \frac{b_1}{b_2} = \frac{-5}{ 10} = - \frac{1}{2}$

$\displaystyle \sf \frac{c_1}{c_2} = \frac{ - 7}{ - 7} = 1$

Thus we get ,

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

So the pair of equations 3x - 5y = 7 and - 6x + 10y = 7 have no solution

Hence the correct option is c) no solution

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