2(3x−v)=5uv⇒2(3xuv−vuv)=5uvuv2(3v−1u)=5→(i)and2(u+3v)=5uv⇒2(uuv+3vuv)=5uvuv⇒2(1v+3u)=5→(ii)Let1v=x,1u=yfromequation(i)and(ii)⇒2(3x−y)=5⇒3x−y=52→(ii)and⇒2(x+3y)=5⇒x+y=52addingequation(iii)and(iv)3x−y=52×1x+3y=52×33x−y=5/23x–––+––9y=15−2−10y=−5,y=510=12=1uu=2Putting=12inequation(iii)⇒3x−12=52⇒3x=3x=1=1v∴v=1andu=2Ans.