Solve simultaneous linear equations: (i) 1/ 2(2x + 3y) + 12/ 7(3x – 2y) = ½

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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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