Solve for #x#: #y= sqrt( (4x+1)/(3x-3))# ?

1 Answer
Jim G.
Nov 22, 2017

#x=(1+3y^2)/(3y^2-4)#

Explanation:

#"note that "sqrtaxxsqrta=(sqrta)^2=a#

#y=sqrt((4x+1)/(3x-3))#

#color(blue)"squaring both sides"#

#y^2=(sqrt((4x+1)/(3x-3)))^2#

#rArry^2=(4x+1)/(3x-3)#

#rArry^2(3x-3)=4x+1larrcolor(blue)"cross-multiplying"#

#rArr3xy^2-3y^2=4x+1#

#rArr3xy^2-4x=1+3y^2larrcolor(blue)"collect terms in x"#

#rArrx(3y^2-4)=1+3y^2larrcolor(blue)"factorising"#

#rArrx=(1+3y^2)/(3y^2-4)to(y!=+-4/3)#

#color(blue)"As a check"#

#"let x = 2"#

#"then " y=sqrt(9/3)=sqrt3#

#"substitute into the expression for x we should get 2"#

#x=(1+3(sqrt3)^2)/(3(sqrt(3)^2-4))=(1+9)/(9-4)=10/5=2#

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