# Solving Algebra equation 3x=15

• MHB
• Victor23
In summary: To be clear, "adding 1 to the numerator" does NOT change the "sign" of the equation (ie. it doesn't make the equation "positive" or "negative"). It's just like saying "x+1 is 2". In this example, "adding 1 to the numerator" does change the "value" of x, but that's a different story.
Victor23
My workings ;3x=15

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

3x = 15

x = 15/3

x = 5

Victor23 said:
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.

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."

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".

## 1. How do I solve an algebra equation with one variable?

To solve an algebra equation with one variable, you need to isolate the variable on one side of the equation by using inverse operations. In the equation 3x=15, you can divide both sides by 3 to get x=5.

## 2. What is the order of operations in solving algebra equations?

The order of operations in solving algebra equations is PEMDAS (Parentheses, Exponents, Multiplication and Division from left to right, Addition and Subtraction from left to right). This means you should first simplify any expressions within parentheses, then solve any exponents, followed by multiplication and division, and finally addition and subtraction.

## 3. Can I check my answer to make sure it is correct?

Yes, you can always check your answer by plugging it back into the original equation. In the equation 3x=15, if you substitute x=5, you will get 3(5)=15, which is a true statement.

## 4. How do I know if there are multiple solutions to an algebra equation?

If an algebra equation has one variable and one solution, it will have a unique solution. However, if there are multiple variables or if the equation has an infinite number of solutions, it will have multiple solutions.

## 5. Can I solve an algebra equation without using inverse operations?

No, inverse operations are necessary to solve algebra equations. They allow you to isolate the variable and solve for its value. Without inverse operations, the equation would remain unsolved and the value of the variable would be unknown.

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