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## Algebra 1

### Course: Algebra 1 > Unit 2

Lesson 1: Linear equations with variables on both sides- Why we do the same thing to both sides: Variable on both sides
- Intro to equations with variables on both sides
- Equations with variables on both sides: 20-7x=6x-6
- Equations with variables on both sides
- Equation with variables on both sides: fractions
- Equations with variables on both sides: decimals & fractions
- Equation with the variable in the denominator

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# Equations with variables on both sides: 20-7x=6x-6

Solving equations like 20 - 7x = 6x - 6 with the variable on both sides involves a few steps! First, we add or subtract terms from both sides to separate constants and variables to different sides of the equation. Then, we simplify to isolate the variable. Finally, we check our answer by plugging it back into the original equation. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- What would you do in an equation that has a subtraction sign and a negative sign

Example d+x-(-x)=x+d+x(31 votes)- Also here's a helpful tool:

(-)*(-)=(+)

(-)/(-)= (+)

(+)*(-)=(-)

(+)*(+)=(+)

(-)+(+)=(-)

(-)*(-)*(-)=(-)

(+)*(+)*(+)=(+)

Hope this helps :)(91 votes)

- Hi Kim Seidel, I have seen that you answer hundreds of peoples questions and I just want to say thank you. You have helped so many people unrecognized. I'm guessing you work for Khan Academy? Thanks again.👍😁

P.S.

I wasn't sure you would see this in the Tips & Thanks;I know when you see this I might have forgotten about this "question" But thanks if you do respond.(52 votes)- Thanks for the kind words!

Believe it or not, I don't work for KhanAcademy. I'm just a KA user like you who likes to help other people learn math. I'm glad I was able to help you.

FYI - There are quite a few of KA users besides me that frequently answer questions and provide great help.(49 votes)

- how do you know what side to subtract first like in this equation 5x+2=2x+14?

My textbook says to subtract 2x first but how do you know its 2x instead of 5x, 2, or 14?(11 votes)- In equations like your sample,
**we can choose any of the terms to begin with, and still arrive at the same answer**.

★Starting with subtracting 2x will keep the values in the calculations positive, because 2x < 5x.

…**5x + 2 = 2x + 14**

subtract 2x from both sides

=

5x -2x + 2 = 2x + 14 -2x

=**3x + 2 = 14**

subtract 2 from both sides

=

3x + 2 -2 = 14 -2

=**3x = 12**

divide both sides by 3

=**3x/3 = 12/3**←yay! answer!

=

x = 4

★Starting with 5x will cause mixed positive and negative integer math as 5 > 2.

…**5x + 2 = 2x + 14**

subtract 5x from both sides

=

5x -5x + 2 = 2x + 14 -5x

=**2 = 14 -3x**

subtract 14 from both sides

=**2 -14 = 14 -3x -14**

=

-12 = -3x

divide both sides by -3

=

-12/-3 = -3x/-3*

negative ÷ negative = positive

=

4 = x ←yay! same answer!

★Starting with subtracting 2, will also keep the calculations positive, since 2 < 14.

…**5x + 2 = 2x + 14**

subtract 2 from both sides

=

5x + 2 -2 = 2x + 14 -2

=**5x = 2x + 12**

subtract 2x from both sides

=

5x -2x = 2x + 12 -2x

=**3x = 12**

divide both sides by 3

=**3x/3 = 12/3**←yay! same answer!

=

x = 4

★Starting with subtracting 14 will also cause mixed positive and negative integer calculations, as 14 > 2.

…**5x + 2 = 2x + 14**

subtract 14 from both sides

=

5x + 2 -14 = 2x + 14 -14

=**5x -12 = 2x**

subtract 5x from both sides

=**5x -5x -12 = 2x -5x**

=

-12 = -3x

divide both sides by -3

=

-12/-3 = -3x/-3

negative ÷ negative = positive

=**4 = x**←yay! same answer!

So in this case, beginning with any of the terms will arrive at the same answer, the main lesson is to use Opposite Operations to isolate the variable.

(ㆁωㆁ) Hope this helps someone.(27 votes)

- I don't understand no matter how many times I do it is there someone there who can explain SLOWER so, I can understand

pls hlp(18 votes)- I can help you. Can you give me a specific problem to help you with? (I know I'm two years late. Sry) Try re-watching the video from0:12(1 vote)

- Why wouldn't you add the 20 instead of subtracting it in the beginning?(5 votes)
- Sal is moving the 20 across the equals to the other side of the equation. To move items, we always use the opposite operation. In the equation, the 20 is being added to -7x. The opposite of addition is subtracting which is why Sal is subtracting the 20 from both sides.

If we take your approach and add 20 to both sides, look at what happens:

20+20-7x = 6x-6+20

40-7x = 6x + 14

Nothing has moved across the equals. We've only made the numbers larger than when we started.

Hope this helps.(20 votes)

- What happens if we end up with:

-5x = 5x

If I divide by -5 to get x, I end up with:

x = -x

But I think that's not the correct answer, is it?(7 votes)- instead of dividing try adding/subtracting. Or once you get to -x = x add/ subtract. The answer winds up being x = 0

Still, you want to make sure that is the correct result, just in case there was some error getting to -5x = 5x(7 votes)

- How would you explain how to do this in a few words ?(6 votes)
- Isolate the variable, then see what the variable equals.(7 votes)

- how would you do,

10p - 3 = 2 (12 + 4p) -7?

I come up with 3 1/6 but know that that is incorrect.(1 vote)- Yes, it's incorrect.

This is how to solve it:

10p-3 = 2(12+4p)-7

10p-3 = 24+8p-7 ---> Multiply

10p-3 = 8p+17 ---> Combine like terms

-8p -8p

2p-3 = 17

+3 +3

2p = 20

/2 /2

p = 10

One of the most helpful things to remember when you're solving equations with variables is that you have to isolate the variable (as in figure out how to get the variable on its own). Another important thing is that whatever you do to one side of the equation, you have to do to the other.(7 votes)

- So for class we had this question

-5(x+9)=-5

and so what i did next was

-5x-45=-5

-5x-45+5=-5+5

But my teacher said that you can't add 5, and that you were supposed to add 45. HOW DO YOU TELL which number you're supposed to add?? im so confused, plzz help mee :((4 votes)- For this kind of problem, the goal is to get x on one side, by itself. To do this, we would need to do the opposite to whatever is on the same side as the x.

With the example, if we added 5, it would look like this:

-5x - 45 = -5

-5x - 45 + 5 = -5 + 5

Simplifying, we get: -5x - 40 = 0

Adding 5 to both sides doesn't help the x stay on one side, instead it just makes the opposite side become zero.

Whereas if we added 45, here is what it would look like:

-5x - 45 = -5

-5x - 45 + 45 = -5 + 45

-5x = 40

Then, all we would need to do is divide by -5.

x = -8

As getting x on one side is the goal of this type of problem, we needed to add 45 instead of 5.

In general, we want to do the opposite to*what is happening on the same side as the x.*

Say we have this:

4x + 2 = 6

We would want to subtract 2, as that would help us get closer to having only x on one side.(6 votes)

- i did -6 instead of -20, but it ended up with the same answer(2 votes)
- That's ok. The properties of equality used to solve equations are very flexible. There is more than one sequence of steps that will take you to the correct answer.(9 votes)

## Video transcript

We have the equation 20 minus
7 times x is equal to 6 times x minus 6. And we need to solve for x. So the way I like to do these
is we just like to separate the constant terms, which are
the 20 and the negative 6 on one side of the equation. I'll put them on the
right-hand side. And then we'll put all the x
terms, the negative 7x and the 6x, we'll put it all on
the left-hand side. So to get the 20 out of the way
from the left-hand side, let's subtract it. Let's subtract it from
the left-hand side. But this is an equation,
anything you do to the left-hand side, you also have to
do to the right-hand side. If that is equal to that, in
order for them to still be equal, anything I do to the
left-hand side I have to do to the right-hand side. So I subtracted 20 from the
left, let me also subtract 20 from the right. And so the left-hand side
of the equation, 20 minus 20 is just 0. That was the whole point,
they cancel out. Don't have to write it down. And then I have a negative 7x,
it just gets carried down. And then that is equal to the right-hand side of the equation. I have a 6x. I'm not adding or subtracting
anything to that. But then I have a negative
6 minus 20. So if I'm already 6 below 0 on
the number line, and I go another 20 below that, that's
at negative 26. Now, the next thing we want to
do is we want to get all the x terms on the left-hand side. So we don't want this 6x here,
so maybe we subtract 6x from both sides. So let's subtract 6x from the
right, subtract 6x from the left, and what do we get? The left-hand side, negative
7x minus 6x, that's negative 13x. Right? Negative 7 of something minus
another 6 of that something is going to be negative 13
of that something. And that is going to be
equal to 6x minus 6x. That cancels out. That was the whole point by
subtracting negative 6x. And then we have just a negative
26, or minus 26, depending on how you want to
view it, so negative 13x is equal to negative 26. Now, our whole goal, just to
remember, is to isolate the x. We have a negative 13
times the x here. So the best way to isolate it is
if we have something times x, if we divide by
that something, we'll isolate the x. So let's divide by
negative 13. Now, you know by now, anything
you do to the left-hand side of an equation, you have to
do to the right-hand side. So we're going to have to
divide both sides of the equation by negative 13. Now, what does the left-hand
side become? Negative 13 times x divided
by negative 13, that's just going to be x. You multiply something times x,
divide it by the something, you're just going to
be left with an x. So the left-hand side
just becomes an x. x is equal to negative 26
divided by negative 13. Well, that's just positive
2, right? A negative divided by a negative
is a positive. 26 divided by 13 is 2. And that is our answer. That is our answer. Now let's verify that
it really works. That's the fun thing
about algebra. You can always make sure that
you got the right answer. So let's substitute it back into
the original equation. So we have 20 minus 7 times x--
x is 2-- minus 7 times 2 is equal to 6 times x--
we've solved for x, it is 2-- minus 6. So let's verify that this
left-hand side really does equal this right-hand side. So the left-hand side simplifies
to 20 minus 7 times 2, which is 14. 20 minus 14 is 6. That's what the left-hand
side simplifies to. The right-hand side, we have 6
times 2, which is 12 minus 6. 12 minus 6 is 6. So they are, indeed, equal, and
we did, indeed, get the right answer.