#### Solve simultaneous linear equations:

(i) 1/ 2(2x + 3y) + 12/ 7(3x – 2y) = ½

7/ (2x + 3y) + 4/ (3x – 2y) = 2

(ii) 1/ 2(x + 2y) + 5/ 3(3x – 2y) = – 3/2

5/ 4(x + 2y) – 3/ 5 (3x – 2y) = 61/60.

**Solution:**

(i) 1/ 2(2x + 3y) + 12/ 7(3x – 2y) = ½

7/ (2x + 3y) + 4/ (3x – 2y) = 2

Consider 2x + 3y = a and 3x – 2y = b

We can write it as

1/2a + 12/7b = ½

7/a + 4/b = 2

Now multiply equation (1) by 7 and (2) by ½

7/2a + 12/b = 7/2

7/2a + 2/b = 1

Subtracting both the equations

10/b = 5/2

So we get

b = (10 × 2)/ 5 = 4

Substitute the value of b in equation (2)

7/a + 4/4 = 2

7/a + 1 = 2

So we get

7/a = 2 – 1 = 1

a = 7

Here

2x + 3y = 7 ….. (3)

3x – 2y = 4 ….. (4)

Multiply equation (3) by 2 and (4) by 3

4x + 6y = 14

9x – 6y = 12

So we get

13x = 26

x = 26/13 = 2

Substitute the value of x in (3)

2 × 2 + 3y = 7

By further calculation

4 + 3y = 7

So we get

3y = 7 – 4 = 3

y = 3/3 = 1

Therefore, x = 2 and y = 1.

(ii) 1/ 2(x + 2y) + 5/ 3(3x – 2y) = – 3/2

5/ 4(x + 2y) – 3/ 5 (3x – 2y) = 61/60

Consider x + 2y = a and 3x – 2y = b

1/2a + 5/3b = – 3/2 …. (1)

5/4a – 3/5b = 61/60 ….. (2)

Now multiply equation (1) by 5/2 and (2) by (1)

5/4a + 25/6b = – 15/4

5/4a – 3/5b = 61/60

Subtracting both the equations

25/6b + 3/5b = – 15/4 – 61/60

Taking LCM

(125 + 18)/ 30b = (-225 – 61)/ 60

By further calculation

143/30b = – 286/60

By cross multiplication

30b × (-286) = 60 × 143

So we get

b = (60 × 143)/ (30 × -286) = – 1

Substitute the value of b in equation (1)

1/2a + 5/ (3 × – 1) = -3/2

By further calculation

1/2a – 5/4 = – 3/2

We can write it as

1/2a = – 3/2 + 5/3

Taking LCM

1/2a = (-9+ 10)/ 6 = 1/6

So we get

a = 6/2 = 3

Here

x + 2y = 3 ….. (3)

3x – 2y = – 1 ….. (4)

Adding both the equations

4x = 2

x = 2/4 = ½

Substitute the value of x in equation (3)

½ + 2y = 3

By further calculation

2y = 3 – ½

Taking LCM

2y = 5/2

y = 5/ (2 × 2) = 5/4

Therefore, x = ½ and y = 5/4.

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