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Question
Solve the following simultaneous equations.
`1/(2(3x + 4y)) + 1/(5(2x - 3y)) = 1/4; 5/(3x + 4y) - 2/(2x - 3y) = (-3)/2`.
Solution
`1/(2(3"x" + 4"y")) + 1/(5(2"x" - 3"y")) = 1/4` .........(I)
`5/((3"x" + 4"y")) - 2/((2"x" - 3"y")) = -3/2` .............(II)
`1/2"m" + 1/5"n" = 1/4` .......(III)
5m − 2n = `-3/2` .......(IV)
10m + 4n = 5
+ 10m − 4n = − 3
∴ 20m = 2
∴ m = `2/20 = 1/10`
∴ `10 xx 1/10 + 4"n" = 5`
∴ 1 + 4n = 5
∴ 4n = 5 − 1 = 4
∴ n = 1
But, `1/((3"x" + 4"y"))` = m and `1/((2"x" - 3"y"))` = n`1/((3"x" + 4"y")) = 1/10`
∴ 3x + 4y = 10 .........(VII)
And `1/((2"x" - 3"y")) = 1`
∴ 2x − 3y = 1 .......(VIII)
Multiplying by 3 in equation (VII),
9x + 12y = 30 .....................(IX)
Multiplying by 4 in equation (VIII),
8x − 12y = 4 .....................(X)
Adding equation (IX) and (X), we get
9x + 12y = 30
+ 8x − 12y = 4
17x = 34
∴ x = `34/17` = 2
∴ x = 2
2x − 3y = 1 .......(VIII)
∴ 2 × 2 − 3y = 1
∴ 4 − 3y = 1
∴ 4 − 1 = 3y
∴ 3y = 3
∴ y = 1
∴ The solution of the given equations is (x, y) = (2, 1).