How do you solve #sqrt(6x-5)+10=3#?

2 Answers
Mar 12, 2017

Answer:

See the entire solution process below:

Explanation:

First, subtract #color(red)(10)# from each side of the equation to isolate the square root term while keeping the equation balanced:

#sqrt(6x - 5) + 10 - color(red)(10) = 3 - color(red)(10)#

#sqrt(6x - 5) + 0 = -7#

#sqrt(6x - 5) = -7#

Next, square both sides to eliminate the radical while keeping the equation balanced:

#(sqrt(6x - 5))^2 = (-7)^2#

#6x - 5 = 49#

Then, add #color(red)(5)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#6x - 5 + color(red)(5) = 49 + color(red)(5)#

#6x - 0 = 54#

#6x = 54#

Now, divide each side of the equation by #color(red)(6)# to solve for #x# while keeping the equation balanced:

#(6x)/color(red)(6) = 54/color(red)(6)#

#(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) = 9#

#x = 9#

Mar 12, 2017

Answer:

no solution.

Explanation:

#color(blue)"Isolate the square root"# by subtracting 10 from both sides.

#sqrt(6x-5)cancel(+10)cancel(-10)=3-10#

#rArrsqrt(6x-5)=-7#

#color(blue)"square both sides"#

#(sqrt(6x-5))^2=(-7)^2#

#rArr6x-5=49#

add 5 to both sides.

#6xcancel(-5)cancel(+5)=49+5#

#rArr6x=54#

divide both sides by 6

#(cancel(6) x)/cancel(6)=54/6#

#rArrx=9#

#color(blue)"As a check"#

Substitute this value into the left side of the equation and if equal to the right side then it is the solution.

#"left side "=sqrt((6xx9)-5)+10#

#=color(white)(left side)=sqrt49+10#

#=color(white)("left side)=7+10#

#color(white)(xxxxxxxx)=17!=3#

#rArr" there is no solution"#