How do you translate into mathematical expressions and find the number given The product of three more than twice a number and two is fourteen?

1 Answer
Apr 17, 2017

Answer:

#2(2x+3)=14#
#x=2#

Explanation:

Let the number be represented by #x#.

Start by looking for the equals sign. In this case, the word "is" represents the equals sign. Whatever is to the left of the word "is" is the left side of the equation and whatever is to the right of the word "is" is the right side of the equation. Since the right side is just "fourteen", let's put that in and write down what we have so far.
#=14#
It doesn't mean much but we're getting there. Now we need our left side.

Continue by looking as close to the variable as possible. The problem asks for "twice a number", so let's add that to our equation.
#2x=14#

Work outwards. We need "three more than twice a number", so the equation becomes
#2x+3=14#

Finally we're left with the tricky word. When you see "the sum/product/difference/etc. of", we'll need parentheses. In general you should put parentheses around the whole "the sum/product/difference/etc. of" expression and around the things to the right and to the left of the word "and" within that expression, then remove them as necessary.
Our "three more than twice a number" is the left side of the "and" in a "the product of" clause and the right side is "two", so our equation becomes
#((2x+3)times(2))=14#

The parentheses around the entire left side can be removed as well as around just the number #2#, leaving
#2(2x+3)=14#.

Now it's time to solve for #x#. To do that you need to isolate it. Start by dividing both sides by #2#.
#color(green)((color(red)cancelcolor(black)2color(black)((2x+3)))/color(red)cancelcolor(green)2)=color(green)(color(black)14/2#
#2x+3=7#

Then subtract #3# from both sides.
#2xcolor(red)cancelcolor(black)(+3)color(red)cancelcolor(green)(-3)=7color(green)(-3)#
#2x=4#

Finally divide both sides by #2# again.
#color(green)((color(red)cancelcolor(black)2color(black)x)/color(red)cancelcolor(green)2)=color(green)(color(black)4/2)#
#x=2#