Question

Solve the pairs of equation: 6 + 3 = 6, 2 + 4 = 5
​

Answer 1

Answer:

Solve the following pairs of equations by reducing them to a pair of linear equations:

(i)

2x

1

+

3y

1

=2;

3x

1

+

2y

1

=

6

13

(ii)

x

2

+

y

3

=2;

x

4

−

y

9

=−1

(iii)

x

4

+3y=14;

x

3

−4y=23

(iv)

x−1

5

+

y−2

1

=2;

x−1

6

−

y−2

3

=1

(v)

xy

7x−2y

=5;

xy

8x+7y

=15

(vi) 6x+3y=6xy;2x+4y=5xy

(vii)

x+y

10

+

x−y

2

=4;

x+y

15

−

x−y

5

=−2

(viii)

3x+y

1

+

3x−y

1

=

4

3

;

2(3x+y)

1

−

2(3x−y)

1

=

8

−1

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ANSWER

(i)

2x

1

+

3y

1

=2

3x

1

+

2y

1

=

6

13

Lets assume

x

1

=p and

y

1

=q, both equations will become

2

p

+

3

q

=2

⇒3p+2q=12 ... (A)

3

p

+

2

q

=

6

13

⇒2p+3q=13 ...(B)

Multiply (A) by 3 and (B) by 2, we get

9p+6q=36

4p+6q=26

Subtracting these both, we get

5p=10

⇒p=2

⇒x=

p

1

=

2

1

∴q=

6

26−8

=3

⇒y=

q

1

=

3

1

(ii)

x

2

+

y

3

=2

x

4

−

y

9

=−1

Lets assume

x

1

=p and

y

1

=q, both equations will become

2p+3q=2 ...(A)

4p−9q=−1 ...(B)

Multiply (A) by 3, we get

6p+9q=6

4p−9q=−1

Adding these both, we get

10p=5

⇒p=

2

1

⇒ x=

p

2

1

=4

∴q=

3

1

⇒y=

q

2

1

=9

(iii)

x

4

+3y=14

x

3

−4y=23

Lets assume

x

1

=p, both equations become

4p+3y=14 ...(A)

3p−4y=23 ...(B)

Multiply (A) by 4 and (B) by 3, we get

16p+12y=56

9p−12y=69

Adding these both, we get

25p=125

⇒p=5

⇒x=

p

1

=

5

1

⇒y=−2

(iv)

x−1

5

+

y−2

1

=2

x−1

6

−

y−2

3

=1

Lets assume

x−1

1

=p and

y−2

1

=q, both equations become

5p+q=2

6p−3q=1

Multiply (A) by 3, we get

15p+3q=6

6p−3q=1

Adding both of these, we get

21p=7

⇒p=

3

1

⇒ x=4

∴q=

3

1

⇒ y=5

(v)

xy

7x−2y

=5

⇒

y

7

−

x

2

=5

xy

8x+7y

=15

⇒

y

8

+

x

7

=15

Let t=

x

1

and r=

y

1

equation will be

7r−2t=5;8r+7t=15

On solving we get

t=1r=1

∴x=1andy=1

(vi)

Dividing both side by xy then we have

y

6

+

x

3

=6;

y

2

+

x

4

=5

Let t=

x

1

andr=

y

1

equation will be

6r+3t=6;2r+4t=5 on solving we get

t=1r=

2

1

∴x=1andy=2

(vii)

Let t=

x+y

1

andr=

x−y

1

then equation will be

10t+2r=4;15t−5r=−2 on solving we have

t=

5

1

andr=1

∴x+y=5andx−y=1

Hence, x=4,y=1

(viii)

Let t=

3x+y

1

andr=

3x−y

1

then equation will be

t+r=

4

3

;

2

t

−

2

r

=

8

−1

On solving we have

t=

4

1

andr=

2

1

∴3x+y=4and3x−y=2

Hence, x=1,y=1