Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 12x + 13y = 2; 13x + 12y = 136
(ii) 2√x + 3√y = 2; 4√x - 9√y = -1
(iii) 4x + 3y = 14; 3x - 4y = 23
(iv) 5(x - 1) + (y - 2) = 2; 6(x - 1) - 3(y - 2) = 1
(v) 7(x - 2)/yxy = 5; 8x + 7(y - 2)/xy = 15
(vi) 6x + 3y = 6xy; 2x + 4y = 5xy
(vii) 10x + y + 2x - y = 4; 15x + y - 5x - y = -2
(viii) 13x + y + 13x - y = 34; 1/2(3x + y) - 1/2(3x - y) = -1/8

(i)12x+13y=213x+12y=136Lets assume1x=pand1y=q, both equations will becomep2+q3=2⇒3p+2q=12... (A)p3+q2=136⇒2p+3q=13.. (B)Multiply (A) by3and (B) by2, we get9p+6q=364p+6q=26On Subtracting these both, we get9p+6q−(4p+6q)=36−2;6⇒5p=10⇒p=2⇒x=1p=12∴4×12+6q=26q=26−2;6=4⇒y=1q=14(ii)2√x+3√y=24√x−9√y=−1;Lets assume1√x=pand1√y=q,both equations will become2p+3q=2.. (A)4p−9q=−1;.. (B)Multiply (A) by3, we get6p+9q=64p−9q=−1;Adding these both, we get6p+9q+4p−9q=6+(−1;)10p=5⇒p=12⇒ x=1p2=4[∵1√x=p,(1√x)2=p2,1x=p2,x=1p2]∴4×12−9q=−1;⇒−9q=−3⇒q=13⇒y=1q2=9(iii)4x+3y=14 3x−4y=23Lets assume1x=p, both equations become4p+3y=14.. (A)3p−4y=23.. (B)Multiply (A) by4and (B) by3, we get16p+12y=569p−1;2y=69Adding these both, we get25p=125⇒p=5⇒x=1p=15⇒3×5−4y=23⇒−4y=23−1;5=8⇒y=−2;(iv)5x−1;+1y−2;=26x−1;−3y−2;=1Lets assume1x−1;=pand1y−2;=q, both equations become5p+q=26p−3q=1Multiply (A) by3, we get15p+3q=66p−3q=1Adding both of these, we get21p=7⇒p=13⇒1x−1;=13⇒x−1;=3⇒ x=4⇒6×13−3q=1⇒−3q=−1;∴q=13⇒1y−2;=13⇒y−2;=3⇒ y=5(v)7x−2;yxy=5⇒7y−2;x=58x+7yxy=15⇒ 8y+7x=15Lett=1xandr=1yequation will be7r−2;t=5(A);8r+7t=15. (B)on multiplying (A) by7and (B) by2, then adding these, we get⇒(49r−1;4t)+(16r+14)=35+30⇒65r=65On solving we gett=1r=1∴x=1andy=1(vi)6x+3y=6xy;2x+4y=5xyDividing both side byxythen we have6y+3x=6;2y+4x=5Lett=1xandr=1yequation will be6r+3t=6(A);2r+4t=5. (B)On multiplying (B) by3and then subtracting (B) from (A), we get⇒6r+12t−(6r+3t)=15−6;⇒t=1on solving we gett=1,r=12∴x=1andy=2(vii)10x+y+2x−y=4;15x+y−5x−y=−2;Lett=1x+yandr=1x−ythen equation will be10t+2r=4(A);15t−5r=−2;(B)On multiplying (A) by5and (B) by2, and then adding, we get⇒50t+10r+30t−1;0r=20−4⇒80t=16on solving we havet=15andr=1∴x+y=5andx−y=1Hence,x=4,y=1(viii)13x+y+13x−y=34;12(3x+y)−1;2(3x−y)=−1;8Lett=13x+yandr=13x−ythen equation will bet+r=34;t2−r2=−1;84t+4r=3(A)4t−4r=−1;(B)on adding (A) and (B), we get4t+4r+4t−4r=3−1;8t=2On solving we havet=14andr=12∴3x+y=4and3x−y=2Hence,x=1,y=1