How do you solve #12v+14+10v=80#?

2 Answers
Aug 19, 2017

Answer:

See a solution process below:

Explanation:

First, group and combine like terms on the left side of the equation:

#12v + 10v + 14 = 80#

#(12 + 10)v + 14 = 80#

#22v + 14 = 80#

Next subtract #color(red)(14)# from each side of the equation to isolate the #v# term while keeping the equation balanced:

#22v + 14 - color(red)(14) = 80 - color(red)(14)#

#22v + 0 = 66#

#22v = 66#

Now, divide each side of the equation by #color(red)(22)# to solve for #v# while keeping the equation balanced:

#(22v)/color(red)(22) = 66/color(red)(22)#

#(color(red)(cancel(color(black)(22)))v)/cancel(color(red)(22)) = 3#

#v = 3#

Aug 19, 2017

Answer:

A lot of explanation is given as I wish you to fully understand the process.

#38/11color(white)("n") # or #color(white)("n") 3.45bar(45)color(white)("nn") larr" Precise answers"#

Thar bar over the #bar(45)# means that they repeat for ever.

#3.46# is not precise

Explanation:

The objective is to have just one #v# on the left of the equals and just numbers on the right.

To explain the principle using fruit.

Suppose you have two boxes. One containing 12 apples and the other containing 10 apples. You them empty one box into the other. Putting them together (adding). Consequently, that one box now has 12apples + 10apples =22 apples

Ok! Suppose that instead of using the word apples we just use the its first letter. We than have #12a+10a=22a#

#"==========================="#

Using the above concept.

Given: #color(white)(..)12v+14+10v=80#

Re grouping gives

#(12v+10v)+14=80#
The brackets have no purpose other than to emphasize the grouping.

#(22v)+14=80#

#22v+14=80#

#"..................................................................."#
Now we put all the numbers on the right.

Subtract #color(red)(14)# from both sides.

Note that the shortcut is to move it to the other side and give it the sign of the revers process. On the left it is add so on the right it is subtract

#color(green)(22v+14=80color(white)(mm)->color(white)(mm)22v+14color(red)(-14)=80color(red)(-14))#

#color(white)("bbbnnnnnnnnnnn") -> color(white)("bbcc")22v+0=76#

#"..................................................................."#
Now we get the #v# on its own

divide both sides by #color(red)(22)#.

Note that the short cut is to move it to the other side and give it the sign of the reverse process. On the left it is multiply. On the right it is divide.

#color(green)(22v=76color(white)("ccccccccccc")->color(white)("cccc")22/(color(red)(22))v=76/(color(red)(22))#

But #22/22=1# giving #1xxv# which is the same as just #v# giving:

#v=76/22#

#v=(76-:2)/(22-:2) = 38/11 larr" Exact value"#
#"================================="#

#color(blue)("General comment")#

There is no indication in the question that we may give an imprecise answer. The decimal solution would be

#3.45454545............"repeating for ever"#

so if you rounded this it would not be precise.

Precise answers would be:

#38/11# or #3.45bar(45)#