SOLVE: 1) (a-2b)x + (2a-b)y = 2, (a-2b)x + (2a+b)y = 3.

2) bx+ay=a+b, ax(1/a-b - 1/a+b)+b(1/b-a - 1/b+a) = 2a/a+b.

3) (a-b)x+(a+b)y=2a2-b2, (a+b)(x+y)=4ab.

4) a2x+b2y=c2, b2x+a2y=d2.

5) x(a-b+ab/a-b)=y(a+b-ab/a+b), x+y=2a2.

Question

SOLVE: 1) (a-2b)x + (2a-b)y = 2, (a-2b)x + (2a+b)y = 3
2) bx+ay=a+b, ax(1/a-b - 1/a+b)+b(1/b-a - 1/b+a) = 2a/a+b
3) (a-b)x+(a+b)y=2a2-b2, (a+b)(x+y)=4ab
4) a2x+b2y=c2, b2x+a2y=d2
5) x(a-b+ab/a-b)=y(a+b-ab/a+b), x+y=2a2

Shridhar Kaushik · Accepted Answer

(1) $(a - 2 b) x + (2 a - b) y = 2 (1) (a - 2 b) x + (2 a + b) y = 3 (2)$ Subtracting (2) from (1) , we get â€“ $y (2 a - b - 2 a - b) = - 1 \Rightarrow - 2 b y = - 1 \Rightarrow y = \frac{1}{2 b}$ âˆ´ From (1) , we get â€“ $(a - 2 b) x + \frac{2 a - b}{2 b} = 2 \Rightarrow (a - 2 b) x = 2 - \frac{(2 a - b)}{2 b} \Rightarrow (a - 2 b) x = \frac{4 b - 2 a + b}{2 b} \Rightarrow x = \frac{5 b - 2 a}{2 b (a - 2 b)} \Rightarrow x = \frac{5 b - 2 a}{2 (a - 2 b) b} ∴ x = \frac{5 b - 2 a}{2 (a - 2 b) b} and y = \frac{1}{2 b}$ Q3 $(a - b) x + (a + b) y = 2 a^{2} - b^{2} (1) (a + b) x + (a + b) y = 4 a b (2)$ Subtracting (2) from (1), we get â€“ $(a - b - a - b) x = 2 a^{2} - b^{2} - 4 a b \Rightarrow - 2 b x = 2 a^{2} - b^{2} - 4 a b \Rightarrow x = \frac{b^{2} + 4 a b - 2 a^{2}}{2 b}$ Again, multiplying (1) by (a + b) and (2) by (a â€“ b) and then subtracting (2) from (1), we get â€“ $(a - b) (a + b) x + (a + b)^{2} y = (2 a^{2} - b^{2}) (a + b) (a - b) (a + b) x + (a + b) (a - b) y = 4 a b (a - b) \underset{̲}{- - -} {(a + b)^{2} - (a + b) (a - b)} y = (a + b) (2 a^{2} - b^{2}) - 4 a b (a - b)$ $\Rightarrow (a + b) {a + b - a + b} y = (a + b) (2 a^{2} - b^{2}) - 4 a b (a - b) \Rightarrow 2 b (a + b) y = (a + b) (2 a^{2} - b^{2}) - 4 a b (a - b) \Rightarrow y = \frac{(a + b) (2 a^{2} - b^{2}) - 4 a b (a - b)}{2 b (a - b)}$ Q4 $a^{2} x + b^{2} y = c^{2} (1) b^{2} x + a^{2} y = d^{2} (2)$ Multiplying (1) by b2 and (2) by a2 and then subtracting (2) from (1), we get â€“ $\underset{̲}{a^{2} b^{2} x + b^{4} y = c^{2} b^{2} a^{2} b^{2} x + a^{4} y = d^{2} a^{2} - - -} (b^{4} - a^{4}) y = (c^{2} b^{2} - d^{2} a^{2})$ $\Rightarrow y = (\frac{c^{2} b^{2} - d^{2} a^{2}}{b^{4} - a^{4}})$ Again, multiplying (1) by a2 and (2) by b2, and then subtracting (2) from (1), we get â€“ $\underset{̲}{a^{4} x + a^{2} b^{2} y = c^{2} a^{2} b^{4} x + a^{2} b^{2} y = d^{2} b^{2} - - -} (a^{4} - b^{4}) x = c^{2} a^{2} - d^{2} b^{2} \Rightarrow x = \frac{c^{2} a^{2} - d^{2} b^{2}}{(a^{4} - b^{4})}$ Similarly, you can find the values of x and y for other equations If you face any problem, then please do get back to us

SOLVE: 1) (a-2b)x + (2a-b)y = 2, (a-2b)x + (2a+b)y = 3. 2) bx+ay=a+b, ax(1/a-b - 1/a+b)+b(1/b-a - 1/b+a) = 2a/a+b. 3) (a-b)x+(a+b)y=2a2-b2, (a+b)(x+y)=4ab. 4) a2x+b2y=c2, b2x+a2y=d2. 5) x(a-b+ab/a-b)=y(a+b-ab/a+b), x+y=2a2.

SOLVE: 1) (a-2b)x + (2a-b)y = 2, (a-2b)x + (2a+b)y = 3.

2) bx+ay=a+b, ax(1/a-b - 1/a+b)+b(1/b-a - 1/b+a) = 2a/a+b.

3) (a-b)x+(a+b)y=2a2-b2, (a+b)(x+y)=4ab.

4) a2x+b2y=c2, b2x+a2y=d2.

5) x(a-b+ab/a-b)=y(a+b-ab/a+b), x+y=2a2.