Question

2х + Зу = 7
(а - b) х + (a + b) y = За + b – 2​

Answer 1

Given : 2х + Зу = 7

(а - b) х + (a + b) y = За + b – 2​

To find :  Solve

Solution:

2х + Зу = 7  multiply by a + b

=> 2(a + b)x  + 3(a + b)y  = 7a  + 7b   Eq1

(а - b) х + (a + b) y = За + b – 2​    multiply by 3

=> 3(a - b)x + 3(a + b) y  = 9a + 3b - 6   Eq2

Eq2  - Eq1

=>  (a  - 5b) x   =  2a - 4b  - 6

=> x  =  (2a - 4b - 6) / (a - 5b )    

2х + Зу = 7  multiply by a-  b

=>  2(a - b)x  + 3(a - b)y  = 7a - 7b   Eq3

(а - b) х + (a + b) y = За + b – 2​    multiply by 2

=> 2(a - b)x + 2(a + b) y  = 6a + 2b - 4   Eq4  

Eq3 - Eq4

=>  ( a  - 5b)y  =  a  - 9b  + 4

=> y  = (a - 9b + 4)/(a - 5b )    

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Answer 2

Step-by-step explanation:

Given:

$2x + 3y = 7...eq1 \\ \\ (a - b)x + (a + b)y = 3a + b - 2 ...eq2$

To find: Value of x and y

Solution:

Method used: Substitution Method

Step 1: Write eq1 in terms of y

$3y = 7 - 2x...eq3$

Step 2: Multiply eq 2 by 3 and substitute the value of 3y from eq3

$3(a - b)x + (a + b)3y = 3(3a + b - 2 ) \\ \\ 3(a - b)x + (a + b)(7 - 2x) = 9a + 3b - 6 \\ \\ 3ax - 3bx + 7a - 2ax + 7b - 2bx = 9a + 3b - 6 \\ \\ ax - 5bx = 9a - 7a + 3b - 7b - 6 \\ \\ x(a - 5b) = 2a - 4b - 6 \\ \\ x = \frac{2a - 4b - 6}{a - 5b} ...eq4$

Step 3: Put value of x from eq 4 into eq1

$2x + 3y = 7 \\ \\ 2 \left( \frac{2a - 4b - 6}{a - 5b} \right) + 3y = 7 \\ \\ \frac{4a - 8b - 12 }{a - 5b} + 3y = 7 \\ \\ 3y = \frac{7a - 35b - 4a + 8b + 12}{a - 5b} \\ \\ 3y = \frac{3a - 27b + 12}{a - 5b} \\ \\ y = \frac{ \cancel3(a - 9b + 4)}{ \cancel3(a - 5b)} \\ \\ y = \frac{a - 9b + 4}{a - 5b}$

Final answer:

Value of x and y are

$\bold{\pink{ x = \frac{2a - 4b - 6}{a - 5b}}} \\ \\\bold{\green{ y = \frac{a - 9b + 4}{a - 5b}}}$

respectively.

Hope it helps you.

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