How to solve for the variable "X"

Basic Tips for Solving Equations

Suppose I start with something like

x = 5

That's no fun, is it? There's nothing to figure out. But I can multiply both sides of the equation by 3 without changing the truth of the equation, right?

3x = 15

And there are a lot of different ways I can write 3:

3 = 1 + 2

= 7 - 4

= -5 + 8

So I can change my equation to look like this:

3x = 15

x(3) = 15

x(7 - 4) = 15

7x - 4x = 15

And since I can add or subtract anything on both sides of an equation without changing the truth of the equation, I can add 4x to both sides:

7x - 4x = 15

7x - 4x + 4x = 15 + 4x

7x = 15 + 4x

And of course, I can do the same thing with any number I want:

7x = 15 + 4x

7x - 5 = 15 + 4x - 5

7x - 5 = 15 - 5 + 4x

7x - 5 = 10 + 4x

Now I have something that looks just like your problems, don't I?

The trick, then, is to reverse the steps that I just did, to get back
to the original, simple, boring equation:

7x - 5 = 10 + 4x

7x - 5 + 5 = 10 + 4x + 5 Add 5 to both sides

7x = 15 + 4x

7x - 4x = 15 + 4x - 4x Subtract 4x from both sides

3x = 15

(3x)/3 = 15/3 Divide both sides by 3

x = 5

It's easy to forget what's really going on when you're solving an algebraic expression for a variable. You're really just trying to rewrite a complicated expression in a simpler way, without doing anything to change the meaning of the expression. It's a little like what you would do to change a complicated sentence like:

"Under a mammal of feline persuasion is located, at this time, in an unspecified location, a rectangle of woven fabric."

into a simpler one like:

"The cat is on the mat."

Algebra is basically tying knots and then untying them again.

With that in mind, let's look at one of your example problems:

3 + 3x = 9 + 2x

The first thing to notice is that we have numbers on both sides. We know that someone put them there by adding a number to both sides of the equation, so we can un-put them there by subtracting a number from both sides:

3 + 3x = 9 + 2x

3 + 3x - 3 = 9 + 2x - 3

(3 - 3) + 3x = (9 - 3) + 2x

3x = 6 + 2x

Now we have multiples of x on both sides. We can do the same kind of thing to fix that:

3x = 6 + 2x

3x - 2x = 6 + 2x - 2x

(3 - 2)x = 6 + (2 - 2)x

x = 6

Can you follow the same kinds of steps to solve the rest of the problems?

I hope this helps.

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