How do you solve #sqrt(6x-5)+10=3#?
2 Answers
See the entire solution process below:
Explanation:
First, subtract
Next, square both sides to eliminate the radical while keeping the equation balanced:
Then, add
Now, divide each side of the equation by
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"#