But we can't just multiply the left hand side by x.

And we can still get this expression over here, which we've set to zero.

Since we did it on the left hand side we also have to do it on the right hand side, and this is just seven one, times x plus one over one.

But again, we could walk through this and, and do all the math.

If I think I'm being followed, I walk out with the bag over my right shoulder.

So it's going to be 7 times 10, so 70, plus 7 times 10 plus 5 times 12, plus 60.

That's what the left hand side simplifies to.

So what can we multiply this by, so that we are left with just x .

And let's say we're taking something that we know well sine of t minus pi.

And so you are left with on the left hand side your fourteen x plus four and on your right hand side you just have to multiply eight times negative three x minus two.

You could say that the work is equal to 5 newtons times, I'll just write a dot for times.

The time is 5 times 10 to the second seconds.

But of course, we're doing the chain rule.

4 times 2r is 8r, 4 times negative 13 over 4 is negative 13, is less than and I multiplied by a positive number so I didn't have to worry about swapping the inequality is less than 5 2 times 4 is 10, right?

To get that x plus one out of the denominator, we multiply both sides of this equation times x plus one over one times this side.

And that makes sense because two to the three is two multiplying by itself three times, to the fifth is two multiplying by itself five times, and then we're multiplying the two, so we're going to multiply two eight times.

You can say x plus five over one.

When I take something times 10 to the second power, I'm essentially shifting the decimal point 2 to the right.

And that sum I'm going to define as a times s.

Is it practical?

If it's false, then the flag hit will valuate to zero, 1 hit will be 1.

And then the denominator will have the b.

So in this case, n factorial was 7.

And the denominator would have been 2 times 2 times 2 (that's what 2 to the third power is) 2 times 2 times 2 times 5 times 5

I'm just going to do minus a positive 3 times each of these.

I'm gonna call you back.

Wherever we want to go, we could go arbitrarily we could scale a up by some arbitrary value.

And, of course, the whole time you have that 5 sitting outside.

So just to get the math here, so to get to this number right here, we just multiplied.

That results to 1 σ³ and then 2 multiplied in cancels the minus 1 2 over here very conveniently to with we have all these sigmas in the denominator.

But we can't just multiply one of the terms by x.

Two to the third is two times two times two, right?

This is the delta function shifted to c.

The height is the value of the function, which is x y squared.

So if you divide that side by 5, and then the right hand side by 5.

2. If you multiply 1 by any number, say m, it always equals m.

Oh, actually, I made a mistake. y over x is equal to the natural log of x plus c.

And then to get to a cubed, what did we do?

And then to get to a to the fourth, what did we do?

Thank you.

If you do anything to the left , we need to do the same thing to the right to keep it equal.

I hung my hat on the peg.

What is that going to look like? e to the minus st times f of t?

We can multiply both sides of this equation by x plus two over one. x plus two.

So this is the first step.