(a+2b)x + (2a-b)y= 2,

(a-2b)x +(2a+b)y = 3

Solve the eq. by the method of elimination by equating coefficients.

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(a+2b)x + (2a-b)y= 2,
(a-2b)x +(2a+b)y = 3
Solve the eq by the method of elimination by equating
coefficients

Ankush Jain · Accepted Answer

Given, equation are : $(a + 2 b) x + (2 a - b) y = 2 (1) (a - 2 b) x + (2 a + b) y = 3 (2)$ Multiply (1) by (2a + b) and (2) by (2a â€“ b) and subtract them $(a + 2 b) (2 a + b) x + (2 a - b) (2 a + b) y - (a - 2 b) (2 a - b) x - (2 a + b) (2 a - b) y = 2 (2 a + b) - 3 (2 a - b) \Rightarrow (a + 2 b) (2 a + b) x - (a - 2 b) (2 a - b) x = 4 a + 2 b - 6 a + 3 b \Rightarrow (2 a^{2} + a b + 4 a b + 2 b^{2}) x - (2 a^{2} - 4 a b - a b + 2 b^{2}) x = 5 b - 2 a \Rightarrow x [2 a^{2} + 2 b^{2} + 5 a b - 2 a^{2} + 5 a b - 2 b^{2}] = 5 b - 2 a \Rightarrow x = \frac{5 b - 2 a}{10 a b}$ Now, on putting value of x in (1), we get $(a + 2 b) (\frac{5 b - 2 a}{10 a b}) + (2 a - b) y = 2 \Rightarrow \frac{5 a b - 2 a^{2} + 10 b^{2} - 4 a b}{10 a b} + (2 a - b) y = 2 \Rightarrow (2 a - b) y = 2 - (\frac{10 b^{2} - 2 a^{2} + a b}{10 a b}) \Rightarrow (2 a - b) y = \frac{20 a b - 10 b^{2} + 2 a^{2} - a b}{10 a b} \Rightarrow y = \frac{19 a b - 10 b^{2} + 2 a^{2}}{10 (2 a - b) a b}$

(a+2b)x + (2a-b)y= 2, (a-2b)x +(2a+b)y = 3 Solve the eq. by the method of elimination by equating coefficients.

(a+2b)x + (2a-b)y= 2,

(a-2b)x +(2a+b)y = 3

Solve the eq. by the method of elimination by equating coefficients.