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## Old school equations with Sal

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# Linear equations 3

## Video transcript

Welcome to my presentation on level three linear, yeah, level three linear
equations. [LAUGH] Okay. So let me, let's, let's make up a problem. Let's say I had X plus 2x plus 3 is equal to, minus 7x minus 5. Well, in all of these linear equations,
the first things that we, the first thing that we try to do is,
get all of our variables on one side of the
equation, and then get all of our concept terms on the other side
of the equation. And then it actually will become a level
one linear equation. So, the first thing we can do is we can
try to simplify each of the sides. Well we, on this, on this left side we
have this X plus 2x. Well, what is X plus 2x? Well that's like saying I have one apple
and now I have two apples. So here I have one X and now I have two
more Xs that I'm adding together. So that's equal 3x, 3x plus 3 is equal to
minus 7x minus 5. Now let's bring the 7x over onto the
left-hand side. And we could do that by adding 7x to both
sides, 7x. This is a review. We, we're adding the opposite. So, it's negative 7x, so we add 7x so
that's why. And we do that, become the right side,
these two will cancel. And the left side, we get 10x plus 3 equals, and on the right side, all we
have left is the negative 5. Almost there, now we're at a level, what
is this, a level two problem. And now we just have to take this 3 and
move it to the other side. And we can do that by subtracting 3 from
both sides. That's a 3 minus 3. The left-hand side, the 3s cancel out,
that's why we subtract it in the first place. And you have 10x equals and then minus 5
minus 3, well that equals minus 8. Now, we just multiply both sides of this
equation by 1 over 10, or the reciprocal of 10, which is the
coefficient on x, times 1 over 10. You could also, some people would say,
well, we're just dividing both side by 10 which is
essential what we're doing. If you divide by 10, that's the same thing
as multiplying by 1 over 10. Well, anyway, the left-hand side, 1 over
10 times 10. Well, that equals 1, so we're just left
with x equals negative 8 over 10. And that can be reduced further. They both share the common factor 2. So you divide by 2. So it's minus 4 over 5. I think that's right, assuming that I
didn't make any careless mistakes. Let's do another problem. Let's say I had 5, that's a 5x minus 3 minus 7x equals x plus 8. And in general if you wanna work this out
before I give you how I do it that now's a good time to
actually pause the video. And you could, you could try to work it
out and then, play it again and, and see what I have to
say about it. But assuming you wanna hear it, let me go
and do it. So let's do the same thing. We, first of all, we can merge these two
Xs on the left-hand side. Remember, you can't add the 5 and the 3
because the 3 is just a constant term while the 5 is 5
times x. But the 5 times x and the negative 7 times
actually can merge. So 5, you just add the coefficient. So, it's 5 and negative 7. So, that becomes negative 2x minus 3 is
equal to x plus 8. Now, if we wanna take this x that's on the
right-hand side and put it over the left-hand side, we can
just subtract x from both sides. The left-hand side becomes minus 3x minus
3 is equal to, these two Xs cancel out, is equal to
8. Now, we can just add 3 to both sides to get rid of that constant term 3 on left
hand-side. These two 3's will cancel out. And you get minus 3x is equal to 11. Now, you just multiply both sides by
negative one-third. And once again, this is just the same thing as dividing both sides by negative
3. And you get x equals negative 11 over 3. Actually let's, let's, just for fun, let's check
this just to see. And the cool thing about algebra is if you
have enough time, you can always make sure you got the
right answer. So we have 5x, so we have 5 times negative
11 over 3. So that's, I'm just, I'm just gonna take this and substitute it back into the
original equation. And you might wanna try that out, too. So you have minus 55 over 3, that's just 5 times negative 11 over 3, that's a 3,
minus 3. And what's 3? Three could also be written as, minus 9
over 3. I'm skipping some steps, but I think you, you know your fractions pretty good
by this point. So that's minus 9 over 3. And then, minus 7x is the same thing, as
plus 77 over 3. Because we have the minus 7 times minus
11, so it's plus 77. And, and the equation is saying that
should equal minus 11 over 3, that's what x is, plus an 8 is nothing
more than 24 over 3. Let's add this up. Minus 55 minus 9, that's minus 64, if I'm
right, yeah, that's minus 64. And then, plus 77 minus 64 plus 77 is 13. So the left-hand side becomes 13 over 3. And on the right-hand side minus 11 plus
24, well that's 13 and we still have over 3. So looks like we got the right solution. It checks out. So the correct answer was minus 11 over 3. Hopefully you're ready by now to, do some
level three problems. The only thing that makes this a little
bit more complicated than level two is you
just have to remember to merge the variables in the
beginning, and, know that you could subtract variables or
constants from both sides. Have fun.