Question

Solve : (a+2b)X + (2a-b)y=2; (a -2b)X + (2a +b)y =3 by the cross multiplication method.​

Answer 1

$\huge\frak\red{answer}$

Step-by-step explanation:

SOLUTION

• (a+2b)x+(2a-b)y-2=0

• (a-2b)x+(2a+b)y-3=0

Here,

a1= a+2b, b1= 2a-b, c1= -2

a2= a-2b, b2= 2a+b, c2= -3

By cross- multiplication, we have

$\begin{gathered} = > \frac{x}{ - 3(2a - b) - ( - 2)(2a + b)} = \frac{ - y}{3(a + 2b) - ( - 2)(a - 2b)} = \frac{1}{(a + 2b)(2a + b) - (2a - b)(a - 2b)} \\ \\ = > \frac{x}{ - 6a + 3b + 4a + 2b} = \frac{ - y}{ - 3 - 6b + 2a - 4b} = \frac{1}{2 {a}^{2} + ab + 4ab + 2 {b}^{2} - (2 {a}^{2} - 4ab - ab + 2 {b}^{2}) } \\ \\ = > \frac{x}{ - 2a + 5b} = \frac{ - y}{ - a - 10b} = \frac{1}{2 {a}^{2} + ab + 4ab + 2 {b}^{2} - (2a {}^{2} - 4ab - ab + 2b {}^{2} } \\ \\ = > \frac{x}{ - 2a + 5b} = \frac{ - y}{ - (a + 10b)} = \frac{1}{10ab} \\ \\ = > \frac{x}{ - 2a + 5b} = \frac{y}{a + 10b} = \frac{1}{10ab} \\ = > now \\ = > \frac{x}{ - 2a + 5b} = \frac{1}{10ab} \\ \\ = > x = \frac{5b - 2a}{10ab} \\ and \: \\ = > \frac{y}{a + 10b} = \frac{1}{10ab} \\ \\ = > y = \frac{a + 10b}{10ab} \\ \\ \\ = > hence \: x = \frac{5b - 2a}{10ab} \: and \: y = \frac{a + 10b}{10ab} \end{gathered}$

Hope it helps ✔️

Answer 2

Question:-

Solve: (a + 2b)x + (2a – b)y = 2 ;

(a – 2b)x + (2a + b)y = 3 by the cross multiplication method.

Given:-

(a + 2b)x + (2a – b)y = 2 ; (a – 2b)x + (2a + b)y = 3.

To Solve:-

(a + 2b)x + (2a – b)y = 2 ; (a – 2b)x + (2a + b)y = 3 by the cross multiplication method.

Solution:-

The given system of equation can be written as

• (a + 2b)x + (2a – b)y – 2 = 0.

• (a – 2b)x + (2a + b)y – 3 = 0.

Here,

• a₁ = a + 2b, b₁ = 2a – b, c₁ = – 2.

• a₂ = a – 2b, b₂ = 2a + b, c₂ = – 3.

By cross multiplication, we have

$\sf \large ⇒ \frac{x}{ - 3(2a - b) - ( - 2)(2a + b)} = \frac{ - y}{3(a + 2b) - ( - 2)(a - 2b)} = \frac{1}{(a + 2b)(2a - b) - (2a - b)(a - 2b)}$

$\sf \large⇒ \frac{x}{ - 6a + 3b + 4a + 2b} = \frac{ - y}{ - 3a - 6b + 2a - 4} = \frac{1}{2a {}^{2} + ab + 4ab + 2b {}^{2} - (2a {}^{2} - 4ab - ab + b {}^{2}) }$

$\sf \large⇒ \frac{x}{ - 2a + 5b} = \frac{ - y}{ - a - 10b} = \frac{1}{2a {}^{2} + ab + 4ab + 2b {}^{2} - (2a {}^{2} - 4ab - ab + 2b {}^{2}) }$

$\sf \large⇒ \frac{x}{ - 2a + 5b} = \frac{y}{ - a + 10b} = \frac{1}{10ab}$

$\sf \large⇒ \frac{x}{ - 2a + 5b} = \frac{y}{a + 10b} = \frac{1}{10ab}$

Now,

$\sf \large⇒ \frac{x}{ - 2a + 5b} = \frac{1}{10ab}$

$\sf \large ⇒y = \frac{a + 10b}{10ab}$

And

$\sf \large⇒\frac{y}{a + 10b} = \frac{1}{10ab}$

$\sf \large ⇒y = \frac{a + 10b}{10ab}$

Answer:-

$\sf \large Hence, \: x = \frac{5b - 2a}{ 10ab} ,y = \frac{a + 10b}{10ab} \: is \\ \sf \large the \: solution \: of \: the \: given \: system \: of \: equation.$

Hope you have satisfied. ⚘