How do you solve #(3w+8)/2=25#?

2 Answers
Jul 21, 2016

w = 14

Explanation:

#(3w + 8)/2 = 25#

#3w + 8 = 50#

#3w = 50 - 8 = 42#

Hence #w = 42/3 = 14#

Jul 21, 2016

#w=42/3 = 14#

Explanation:

#color(magenta)("With practice you will be able to answer this question type in 1, 2 or 3 lines")#

Gradually you manipulate the equation until you have just a single #w# and it is on 1 side of = and everything else on the other side.

For this equation there are three basic principles that are the #ul("seed principles behind the shortcut methods")#

#color(brown)("Principle 1")#
To maintain the 'truth' of the equation (some people use the word balance): what you do to one side of the equation you do to the other.

#color(brown)("Principle 2")#
To 'get rid' of an add or subtract change it to 0

#color(brown)("Principle 3")#
To get rid of a multiply or divide change it to 1

Note that LHS is left hand side and that RHS is right hand side.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:#" "color(brown)((3w+8)/2=25)#

Using principle 3 to 'get rid' of the denominator of 2 on the LHS
Multiply both sides by #color(blue)(2)#

#color(brown)((color(blue)(2))/2xx(3w+8)=color(blue)(2xx)25)color(green)(" " ->" " 3w+8=50)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using principle 2 to 'get rid of the 8 on the LHS

Subtract #color(blue)(8)# from both sides

#color(brown)(3w+8color(blue)(-8)=50color(blue)(-8)color(green)(" "->" "3w=42)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using principle 3 to 'get rid of the 3 on the LHS

Divide both sides by #color(blue)(3)# to 'get rid' of the 3 on the LHS

#color(brown)(3/(color(blue)(3)) w=42/(color(blue)(3))color(green)(" "->" "w=42/3)#