ML Aggarwal Simultaneous Linear Equations Exe-5.3 Class 9 ICSE Maths Solutions

ML Aggarwal Simultaneous Linear Equations Exe-5.3 Class 9 ICSE Maths  Solutions. Step by step solutions  of  Simultaneous Linear Equations problems as council prescribe guideline. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Simultaneous Linear Equations Exe-5.3 Class 9 ICSE Maths Solutions

Board	ICSE
Subject	Maths
Class	9th
Chapter-5	Simultaneous Linear Equations
Topics	Solution of Exe-5.3 Questions
Academic Session	2024-2025

Solution of Exe-5.3 Questions

ML Aggarwal Simultaneous Linear Equations Exe-5.3 Class 9 ICSE Maths  Solutions

Question 1. Solve the following systems of simultaneous linear equations by cross-multiplication method:

(i) 3x + 2y = 4

8x + 5y = 9

(ii) 3x – 7y + 10 = 0

y – 2x = 3.

Answer :

(i) 3x + 2y = 4

8x + 5y = 9

We can write it as

3x + 2y – 4 = 0

8x + 4y – 9 = 0

By cross multiplication method

x/ (-18 + 20) = y/ (-32 + 27) = 1/ (15 – 16)

By further calculation

x/2 = y/-5 = 1/-1

So we get

x/2 = – 1

x = – 2

y = – 5 (-1) = 5

Therefore, x = – 2 and y = 5.

(ii) 3x – 7y + 10 = 0

y – 2x = 3

We can write it as

3x – 7y + 10 = 0

y – 2x – 3 = 0

By cross multiplication method

x/ (21 – 10) = y/ (-20 + 9) = 1/ (3 – 14)

By further calculation

x/11 = y/-11 = 1/-11

So we get

x/11 = 1/-11

x = – 1

Similarly

y/-11 = 1/ -11

y = 1

Therefore, x = – 1 and y = 1.

Question 2. Solve the following pairs of linear equations by cross-multiplication method:

(i) x – y = a + b

ax + by = a² – b²

(ii) 2bx + ay = 2ab

bx – ay = 4ab.

Answer :

(i) x – y = a + b

ax + by = a² – b²

We can write it as

x – y – (a + b) = 0

ax + by – (a² – b²) = 0

By cross multiplication method

x/ [a² – b² + b (a + b)] = y/ [- a (a + b) + a² – b²] = 1/ (b + a)

By further calculation

x/ (a² – b² + ab + b²) = y/ (-a² – ab + a² – b²) = 1/ (a + b)

So we get

x/ [a (a + b)] = y/ [-b (a + b)] = 1/ (a + b)

x = a (a + b)/ (a + b) = a

y = [-b (a + b)]/ (a + b) = – b

Therefore, x = a and y = – b.

(ii) 2bx + ay = 2ab

bx – ay = 4ab

We can write it as

2bx + ay – 2an = 0

bx – ay – 4ab = 0

By cross multiplication method

x/ (- 4a²b – 2a²b) = y/ (-2ab² + 8ab²) = 1/ (-2ab – ab)

By further calculation

x/ -6a²b = y/6ab²= 1/-3ab

So we get

x = -6a²b/ -3ab = 2a

y = 6ab²/-3ab = – 2b

Therefore, x = 2a and b = – 2b.

—  : End of ML Aggarwal Simultaneous Linear Equations Exe-5.3 Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

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