Question

If aq is not equal to bp then the system of equations, ax + by = c and px + qy= r have :

Answer 1

Answer:

the following system of equation has a unique solution.

Step-by-step explanation:

ax+by=c

px+qy=r

so, a1/a2=a/p

b1/b2= b/q

and, c1/c2= c/r

it is given that

aq≠bp

so, a/p≠b/q

this means that a1/a2≠b1/b2

thus the equation has a unique solution.

Answer 2

If aq ≠ bp then system of equations ax + by = c and px + qy = r have unique solution

Correct question : If aq ≠ bp then system of equations ax + by = c and px + qy = r have

Given :

The system of equations ax + by = c and px + qy = r

To find :

The number of solutions when aq ≠ bp

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 ( coincident lines) :

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

Inconsistent :

No solution

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

Solution :

Solution :Step 1 of 2 :

Write down the given system of equations

Here the given system of equations are

ax + by = c - - - - - (1)

px + qy = r - - - - - (2)

Step 2 of 2 :

Find number of solutions when aq ≠ bp

Comparing with the equation

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

a₁ = a , b₁ = b , c₁ = - c and a₂ = p , b₂ = q , c₂ = - r

Now it is given that

$\displaystyle \sf aq \ne bp$

$\displaystyle \sf{ \implies } \frac{a}{p} \ne \frac{b}{q}$

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

So the system of equations ax + by = c and px + qy = r have unique solution

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