# Solving Algebra equation 3x=15

#### Victor23

##### New member
My workings ;3x=15

Any help is appreciated, thanks
= 5x, as 15 divide by 3x=5x

Staff member
3x = 15

x = 15/3

x = 5

#### HallsofIvy

##### Well-known member
MHB Math Helper
My workings ;3x=15

Any help is appreciated, thanks
= 5x, as 15 divide by 3x=5x
This makes no sense! In the first place, what does it mean to divide by an equation?
In the second place, 3x is NOT equal to 5x unless x= 0. There is nothing here that says either x= 0 or 3x= 5x.

To "solve for x" means to get x by itself: "x= something".

In the original equation, 3x= 15, x is not "by itself" because it is multiplied by 3. To get x "by itself" we need to "undo" that- and we undo "multiply by 3" by dividing by 3. And, of course, anything we do to one side of the equation we must do to the other.

Rather than "15 divided by 3x= 5x" you should have said "15 divided by 3 is 5".
Dividing both sides of 3x= 15 by 3 gives
(3x)/3= 15/3

x= 5.

#### mrtwhs

##### Active member
HallsofIvy has given the correct advice for solving this equation ... but ... I'm going to take an epsilon amount of issue with one phrase ... "anything we do to one side of the equation we must do to the other". I've said this myself in Algebra classes but I paid for it once. I had a student solve an equation in the following way:

$$\displaystyle \dfrac{x-1}{6}=\dfrac{1}{3}$$

$$\displaystyle \dfrac{x}{6}=\dfrac{2}{3}$$

$$\displaystyle x = \dfrac{2}{3} \cdot 6 = 4$$

When I pressed him on what he did, he said "in the second step I added one to the numerator of both sides of the equation and anything I do to one side of an equation must be done to the other side of the equation ... which is what you said ... so it must be correct."

#### HallsofIvy

##### Well-known member
MHB Math Helper
Thank you! I would argue that "adding 1 to the numerator" is "doing something" to the numerator, not to the "left side of the equation".