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Solve the following simultaneous equations. 12(3x+4y)+15(2x-3y)=14;53x+4y-22x-3y=-32. - Algebra

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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`.

Sum

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)

Substituting `1/((3"x" + 4"y"))` = m and `1/((2"x" - 3"y"))` = n in equation (I) and (II),

`1/2"m" + 1/5"n" = 1/4` .......(III)

5m − 2n = `-3/2` .......(IV)

Multiply equation (III) by 20,
∴ 10m + 4n = 5 .......(V)
 
Multiply equation (IV) by 2,
∴ 10m − 4n = − 3 ..................(VI)
 
Adding equation (I) and (II),

    10m + 4n = 5 
+ 10m − 4n = − 3
∴ 20m = 2

∴ m = `2/20 = 1/10`

Substituting m = `1/10` in equation (V),

∴ `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
 
∴ Substituting the values of m and n in the above equation, we get

`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

 Substituting x = 2 in equation (VIII), we get

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).

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Chapter 1: Linear Equations in Two Variables - Problem Set 1 [Page 28]

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Balbharati Algebra (Mathematics 1) [English] 10 Standard SSC Maharashtra State Board
Chapter 1 Linear Equations in Two Variables
Problem Set 1 | Q 6.5 | Page 28
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