How do you solve #4=7-sqrt(33x-2)#?

Question

1209 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-14T13:56:35

#x=1/3#

Given:#" "4=7-sqrt(33x-2)#

Swap the 4 with the square root (sign changes)

#sqrt(33x-2)=7-4 " "=" "3#

Square both sides

#33x-2=9#

Add 2 to both sides

#33x=11#

Divide both sides by 33

#x=11/33#

#x=(11 -:11)/(33-:11)= 1/3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Check")#

#4=7-sqrt(33x-2)#

consider the RHS only

#7-sqrt(33(11/33)-2)#

#7-sqrt(9)#

#4#

Thus LHS=RHS