How do you solve # 8x-1=63#?

3 Answers
Jun 19, 2018

#x=8#

Explanation:

#"isolate "8x" by adding 1 to both sides of the equation"#

#8x=63+1=64#

#"divide both sides by 8"#

#(cancel(8) x)/cancel(8)=64/8rArrx=8#

#color(blue)"As a check"#

#(8xx8)-1=64-1=63larr"correct"#

Jun 19, 2018

x = 8

Explanation:

#8x = 63 + 1#
#8x = 64#
#x = 64/8#
#x = 8#

Jun 19, 2018

Using Addition/Subtraction, 'move' all the constants to one side of the equation, and then divide through by #x#'s coefficient to find #x=8#

Explanation:

Our goal is to get #x# by itself. We will do this by applying factors to both sides of the equation.

The first step is to get all constants to the opposite side of the equation as #x#. For this scenario, that means eliminating the '-1' from the left hand side (LHS) and doing the same operation to the right hand side (RHS). How will we eliminate the '-1'? By adding 1 to both sides. This preserves equality for both sides.

#8x-1color(red)(+1)=63color(red)(+1)#

#8xcolor(red)(+0)=63color(red)(+1)#

#8x=64#

Next, we need to eliminate #x#'s coefficient. You would normally do this by multiplying or dividing both sides of the equation. For our purposes, the coefficient is 8, so we will divide both sides by 8 to determine what #x# is equal to:

#(8x)/color(red)(8)=64/color(red)(8)#

#8/color(red)(8)x=64/color(red)(8)#

#color(red)(1)x=64/color(red)(8)#

#color(green)(x=8)#