Math, asked by mandriyaaadil3, 23 days ago

Solve the pair of equations px + qy = 2pq and qx + py = p2 + q2 by the cross multiplication method.​

Answers

Answered by 8123456
0

(i)px+qy=p−q;qx−py=p+q

(ii)ax+by=c;bx+ay=1+c

Medium

Solution

verified

Verified by Toppr

(i) px+qy=p−q…(i)

   qx−py=p+q…(ii)

Multiplying (i) by p and (ii) by q we get (iii) and (iv) respectively as

p  

2

x+pqy=p  

2

−pq…(iii)

q  

2

x−pqy=pq+q  

2

…(iv)

Adding (iii) and (iv) we get,

(p  

2

+q  

2

)x=(p  

2

+q  

2

)

⇒x=1

Substituting value of x in (i), we get

p+qy=p−q

qy=−q

y=−1

Hence, x=1,y=−1

(ii) ax+by=c…(i)

    bx+ay=1+c…(ii)

Multiplying (i) by b and (ii) by a we get (iii) and (iv) respectively as

abx+b  

2

y=bc…(iii)

abx+a  

2

y=a+ac…(iv)

Subtracting (iv) from (iii), we get

(b  

2

−a  

2

)y=(b−a)c−a

y=  

(b  

2

−a  

2

)

(b−a)c−a

 

Substituting value of y in (i), we get

ax+b×  

(b  

2

−a  

2

)

(b−a)c−a

=c

ax=c−  

(b  

2

−a  

2

)

(b−a)bc−ab

 

ax=  

(b  

2

−a  

2

)

c(b  

2

−a  

2

)−b  

2

c+abc+ab

 

x=  

a(b  

2

−a  

2

)

cb  

2

−ca  

2

−b  

2

c+abc+ab

 

x=  

a(b  

2

−a  

2

)

ab+abc−ca  

2

 

 

x=  

(b  

2

−a  

2

)

b−bc−ca

 

x=  

(b  

2

−a  

2

)

b+c(b−a

Class 10

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>>Pair of Linear Equations in Two Variables

>>Reducing a Pair of Equations to Linear Form

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Solve the following pair of linear equations:

(i)px+qy=p−q;qx−py=p+q

(ii)ax+by=c;bx+ay=1+c

Medium

Solution

verified

Verified by Toppr

(i) px+qy=p−q…(i)

   qx−py=p+q…(ii)

Multiplying (i) by p and (ii) by q we get (iii) and (iv) respectively as

p  

2

x+pqy=p  

2

−pq…(iii)

q  

2

x−pqy=pq+q  

2

…(iv)

Adding (iii) and (iv) we get,

(p  

2

+q  

2

)x=(p  

2

+q  

2

)

⇒x=1

Substituting value of x in (i), we get

p+qy=p−q

qy=−q

y=−1

Hence, x=1,y=−1

(ii) ax+by=c…(i)

    bx+ay=1+c…(ii)

Multiplying (i) by b and (ii) by a we get (iii) and (iv) respectively as

abx+b  

2

y=bc…(iii)

abx+a  

2

y=a+ac…(iv)

Subtracting (iv) from (iii), we get

(b  

2

−a  

2

)y=(b−a)c−a

y=  

(b  

2

−a  

2

)

(b−a)c−a

 

Substituting value of y in (i), we get

ax+b×  

(b  

2

−a  

2

)

(b−a)c−a

=c

ax=c−  

(b  

2

−a  

2

)

(b−a)bc−ab

 

ax=  

(b  

2

−a  

2

)

c(b  

2

−a  

2

)−b  

2

c+abc+ab

 

x=  

a(b  

2

−a  

2

)

cb  

2

−ca  

2

−b  

2

c+abc+ab

 

x=  

a(b  

2

−a  

2

)

ab+abc−ca  

2

 

 

x=  

(b  

2

−a  

2

)

b−bc−ca

 

x=  

(b  

2

−a  

2

)

b+c(b−a

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