English

Using cross-multiplication method, solve the following system of simultaneous linear equations: 4x + 3y = 5, 2x + 5y = –1 - Mathematics

Advertisements
Advertisements

Question

Using cross-multiplication method, solve the following system of simultaneous linear equations:

4x + 3y = 5, 2x + 5y = –1

Sum
Advertisements

Solution

Given system of equations:

4x + 3y = 5

2x + 5y = –1

Step 1: Write equations in form (a1x + b1y + c1 = 0) and (a2x + b2y + c2 = 0) by bringing all terms to one side: 

4x + 3y – 5 = 0 

2x + 5y + 1 = 0

Here, a1 = 4, b1 = 3, c1 = –5 and a2 = 2, b2 = 5, c2 = 1.

Step 2: According to the cross-multiplication method, the variables (x) and (y) are given by the ratio:

`x/(b_1c_2 - b_2c_1) = y/(c_1a_2 - c_2a_1) = 1/(a_1b_2 - a_2b_1)`

Calculate each numerator and denominator:

b1c2 – b2c1 = 3 × 1 – 5 × (–5)

b1c2 – b2c1 = 3 + 25

b1c2 – b2c1 = 28

c1a2 – c2a1 = (–5) × 2 – 1 × 4

c1a2 – c2a1 = –10 – 4

c1a2 – c2a1 = –14

a1b2 – a2b1 = 4 × 5 – 2 × 3

a1b2 – a2b1 = 20 – 6

a1b2 – a2b1 = 14

Step 3: So, `x/28 = y/(-14) = 1/14`

Step 4: From this,

`x = 28/14`

x = 2 

`y = (-14)/(14)`

y = –1

The solution of the system is x = 2, y = –1.

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Simultaneous Linear Equations - Exercise 5C [Page 105]

APPEARS IN

Nootan Mathematics [English] Class 9 ICSE
Chapter 5 Simultaneous Linear Equations
Exercise 5C | Q 3. | Page 105
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×