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## Algebra (all content)

### Course: Algebra (all content)ย >ย Unit 2

Lesson 12: Old school equations with Sal# Linear equations 2

solving equations of the form AX+B=C. Created by Sal Khan.

## Want to join the conversation?

- how do you do 7x-4=24(13 votes)
- When you have your equation 7x-4 = 24, you can take the number to left side and place it in the right. However, you need to change the operation

7x-4 = 24

7x=24+4

7x=28.

Now, you need to divide each side by the coefficient (if there is). The coefficient here is 7, so you nee to divide each side by 7

7x=28

7x/7=28/7

x=4.

That is how you solve a linear equation. I hope this helps!(2 votes)

- I am lost how he got ten minus nine over twelve(16 votes)
- In order to subtract 3/4 from 5/6 he needs to find a common denominator for the two fractions. (Preferably the smallest possible.) The smallest number that is a multiple of both 4 and 6 is 12. Make two equivalent fractions by multiplying each fraction by the number that gives 12 in the denominator. For 5/6 it's 2, and for 3/4 it'3. This gives 10/12 - 9/12. Whatever the numerator turns out to be the denominator in the answer will be 12, so Sal can write 10-9/12.(20 votes)

- do we have to follow all those steps is their a shorter and/or faster way to do it(8 votes)
- how does Sal check his work at5:14(2 votes)
- You should always get in the habit of checking your work. If you can find and fix errors, you get better grades.(6 votes)

- I am a bit confused I can't figure out how to take away both sides if that makes sense how you have to do the same to both sides

thank you if you can help!(3 votes)- It's okay... Everybody learns!

Say we have this equation...

9(4x-7) + 19 = 12x

we can use the distributive property to multiply:

36x - 63 + 19 = 12x

Let's simplify and clean things up:

36x - 44 = 12x

Now for your question... We want to make the 12x scooch onto the left hand side, and when doing so, we want the constant on the right. 36x stays right where it is. Essentially, what Mr. Sal was doing was (in this example) is he subtracted 12x from both sides and left it as

36x - 44 - 12x = 0

because 0 fills all the random spaces (0's just nice like that).

Now we can move -44 to its rightful place and fill that annoying gap.

36x - 12x = 44

Simplify to clean it up:

24x = 44

x = 44/24

x = 1 5/6

Hope this helped cgr6507!

-Ella :)(2 votes)

- Why does this video not change the pos/neg sign of the reciprocal? I did everything the same except I multiplied 1/12 by 2/1 leaving me with positive 1/6.(3 votes)
- Why didn't the -3 and the positive 3 cancel out in the beginning of the first problem?(2 votes)
- why r linear equation hard?(2 votes)
- I have a question: how would you simplify 6x-2y+10y-11x? I'm not sure if the numbers would be negative or positive when you combine them? Thank you:)(1 vote)
- Use the commutative property to change it to 6x-11x-2y+10y. Combine the like terms to get -5x+8y. Whether it is positive or negative depends on whether the s and y are negative or positive.(2 votes)

- is there any more ways to do the problem 2x + 3= -15, to get -9? thanks!(1 vote)
- The value of x would have to change. In the case of 2x + 3= -9, x would be equal to -6.(2 votes)

## Video transcript

Welcome to level two
linear equations. Let's do a problem. 2x plus 3 is equal to minus 15. Throw the minus in there to
make it a little bit tougher. So the first thing we
want to do whenever we do any linear
equation, is we want to get all of the
variable terms on one hand side of the equation and
all the constant terms on the other side. And it doesn't really
matter, although I tend to get my variables on the
left hand side of the equation. Well, my variables are
already on the left hand side of the equation
but I have this plus 3 that I somehow want to
move to the right hand side of the equation. And the way I can-- you can put
it in quotes, move the 3 is I can subtract 3 from both
sides of this equation. And look at that carefully as
to why you think that works. Because if I subtract 3
from the left hand side, clearly this negative 3 that I'm
subtracting and the original 3 will cancel out and become 0. and as long as I do whatever
I do on the left hand side, as long as I do it on
the right hand side as well, because whatever you do
on one side of the equal side, you have to do to
the other side, then I'm making a
valid operation. So this will simplify to 2x,
because the 3's cancel out. They become just 0. Equals minus 15 minus 3. Well, that's minus 18. And now, we're just at
a level one problem, and you can just
multiply both sides of this equation
times the reciprocal on the coefficient of 2x. I mean, some people
would just say that we're dividing by
2, which is essentially what we're doing. I like to always go
with the reciprocal, because if this
2 was a fraction, it's easier to think
about it that way. But either way, you either
multiply by the reciprocal, or divide by the number. It's the same thing. So 1/2 times 2x. Well, that's just 1x. So you get x equals,
and then minus 18/2. And minus 18/2, well,
that just equals minus 9. Let's do another problem. And actually, well, if we wanted
to check it, we could say, well, the original problem
was 2x plus 3 equals minus 15. So we could say 2
times minus 9 plus 3. 2 times minus 9 is
minus 18 plus 3. Well, that's equal
to minus 15, which is equal to what the
original equation said, so we know that's right. That's the neat
thing about algebra. You can always check your work. Let's do another problem. I'm going to put some
fractions in this time, just to show you that it
can get a little bit hairy. So let's say I had minus 1/2x
plus 3/4 is equal to 5/6. So we'll do the same thing. First, we just want to get
this 3/4 out of the left hand side of the equation,
and actually, if you want to try
working this out yourself, you might want to
pause the video and then play it once you're
ready to see how I do it. Anyway, let me move forward
assuming you haven't paused it. If we want to get rid
of this 3/4, all we do is we subtract 3/4 from
both sides of this equation. Minus 3/4. Well, the left hand side,
the two 3/4 will just cancel. We get minus 1/2x equals, and
then on the right hand side, we just have to do this
fraction addition or fraction subtraction. So the least common
multiple of 6 and 4 is 12. So this becomes 5/6 6 is
10/12 minus 3/4 is 9/12, so we get minus 1/2x
is equal to 1/12. Hopefully, I didn't make
a mistake over here. And if that step confused
you, I went a little fast, you might just want
to review the adding and subtraction of fractions. So going back to where we were. So now all we have to do
is, well, the coefficient on the x term is minus 1/2,
and this is now a level one problem. So to solve for x,
we just multiply both sides by the reciprocal
of this minus 1/2x, and that's minus 2/1 times
minus 1/2x on that side, and then that's times minus 2/1. The left hand side, and
you're used to this by now, simplifies to x. The right hand side
becomes minus 2/12, and we could simplify
that further to minus 1/6. Well, let's check that just
to make sure we got it right. So let's try to
remember that minus 1/6. So the original
problem was minus 1/2x, so here we can substitute
the minus 1/6, plus 3/4. I just wrote only the left hand
side of the original problem. So minus 1/2 times
minus 1/6, well, that's positive 1/12 plus 3/4. Well, that's the
same thing as 12, the 1 stays the same, plus 9. 1 plus 9 is 10 over 12. And that is equal
to 5/6, which is what our original problem was. Our original problem was this. This stuff I wrote later. So it's 5/6, so the
problem checks out. So hopefully, you're now
ready to try some level two problems on your own. I might add some other
example problems. But the only extra step here
relative to level one problems is you'll have this
constant term that you need to add or subtract from
both sides of this equation, and you'll essentially turn
it into a level one problem. Have fun.