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Let's solve a few more systems of equations using

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elimination, but in these it won't be kind of a one-step

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elimination.

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We're going to have to massage the equations a little bit in

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order to prepare them for elimination.

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So let's say that we have an equation, 5x minus 10y is

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equal to 15.

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And we have another equation, 3x minus 2y is equal to 3.

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And I said we want to do this using elimination.

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Once again, we could use substitution, we could graph

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both of these lines and figure out where they intersect.

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But we're going to use elimination.

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But the first thing you might say, hey, Sal, you know, with

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elimination, you were subtracting the left-hand side

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of one equation from another, or adding the two, and then

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adding the two right-hand sides.

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And I could do that, because it was essentially adding the

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same thing to both sides of the equation.

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But here, it's not obvious that that

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would be of any help.

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If we added these two left-hand sides, you would get

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8x minus 12y.

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That wouldn't eliminate any variables.

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And on the right-hand side, you would just

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be left with a number.

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And if you subtracted, that wouldn't

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eliminate any variables.

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So how is elimination going to help here?

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And the answer is, we can multiply both of these

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equations in such a way that maybe we can get one of these

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terms to cancel out with one of the others.

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And you could really pick which term you

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want to cancel out.

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Let's say we want to cancel out the y terms. So I'll just

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rewrite this 5x minus 10y here.

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5x minus 10y is equal to 15.

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Now, is there anything that I can multiply this green

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equation by so that this negative 2y term becomes a

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term that will cancel out with the negative 10y?

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So I essentially want to make this negative 2y into a

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positive 10y.

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Right?

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Because if this is a positive 10y, it'll cancel out when I

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add the left-hand sides of this equation.

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So what can I multiply this equation by?

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Well, if I multiply it by negative 5, negative 5 times

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negative 2 right here would be positive 10.

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So let's do that.

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Let's multiply this equation times negative 5.

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So you multiply the left-hand side by negative 5, and

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multiply the right-hand side by negative 5.

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And what do you get?

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Remember, we're not fundamentally

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changing the equation.

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We're not changing the information in the equation.

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We're doing the same thing to both sides of it.

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So the left-hand side of the equation becomes negative 5

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times 3x is negative 15x.

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And then negative 5 times negative 2y is plus 10y, is

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equal to 3 times negative 5 is negative 15.

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And now, we're ready to do our elimination.

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If we add this to the left-hand side of the yellow

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equation, and we add the negative 15 to the right-hand

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side of the yellow equation, we are adding the same thing

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to both sides of the equation.

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Because this is equal to that.

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So let's do that.

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So 5x minus 15y-- we have this little negative sign there, we

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don't want to lose that-- that's negative 10x.

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The y's cancel out.

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Negative 10y plus 10y, that's 0y.

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That was the whole point behind multiplying this by

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negative 5.

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Is going to be equal to-- 15 minus 15 is 0.

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So negative 10x is equal to 0.

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Divide both sides by negative 10, and you get

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x is equal to 0.

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And now we can substitute back into either of these equations

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to figure out what y must be equal to.

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Let's substitute into the top equation.

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So we get 5 times 0, minus 10y, is equal to 15.

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Or negative 10y is equal to 15.

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Let me write that.

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Negative 10y is equal to 15.

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Divide both sides by negative 10.

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And we are left with y is equal to

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15/10, is negative 3/2.

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So if you were to graph it, the point of intersection

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would be the point 0, negative 3/2.

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And you can verify that it also satisfies this equation.

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The original equation over here was 3x minus

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2y is equal to 3.

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3 times 0, which is 0, minus 2 times negative 3/2 is, this is

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0, this is positive 3.

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Right?

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These cancel out, these become positive.

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Plus positive 3 is equal to 3.

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So this does indeed satisfy both equations.

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Let's do another one of these where we have to multiply, and

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to massage the equations, and then we can eliminate one of

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the variables.

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Let's do another one.

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Let's say we have 5x plus 7y is equal to 15.

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And we have 7-- let me do another color-- 7x minus 3y is

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equal to 5.

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Now once again, if you just added or subtracted both the

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left-hand sides, you're not going to

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eliminate any variables.

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These aren't in any way kind of have the same coefficient

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or the negative of their coefficient.

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So let's pick a variable to eliminate.

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Let's say we want to eliminate the x's this time.

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And you could literally pick on one of the

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variables or another.

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It doesn't matter.

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You can say let's eliminate the y's first. But I'm going

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to choose to eliminate the x's first. And so what I need to

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do is massage one or both of these equations in a way that

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these guys have the same coefficients, or their

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coefficients are the negatives of each other, so that when I

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add the left-hand sides, they're going to eliminate

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each other.

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Now, there's nothing obvious-- I can multiply this by a

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fraction to make it equal to negative 5.

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Or I can multiply this by a fraction to make it equal to

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negative 7.

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But even a more fun thing to do is I can try to get both of

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them to be their least common multiple.

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I could get both of these to 35.

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And the way I can do it is by multiplying by each other.

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So I can multiply this top equation by 7.

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And I'm picking 7 so that this becomes a 35.

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And I can multiply this bottom equation by negative 5.

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And the reason why I'm doing that is so this becomes a

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negative 35.

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Remember, my point is I want to eliminate the x's.

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So if I make this a 35, and if I make this a negative 35,

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then I'm going to be all set.

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I can add the left-hand and the right-hand

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sides of the equations.

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So this top equation, when you multiply it by 7, it becomes--

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let me scroll up a little bit-- we multiply it by 7, it

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becomes 35x plus 49y is equal to-- let's see, this is 70

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plus 35 is equal to 105.

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Right?

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15 and 70, plus 35, is equal to 105.

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That's what the top equation becomes.

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This bottom equation becomes negative 5 times 7x, is

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negative 35x, negative 5 times negative 3y is plus 15y.

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The negatives cancel out.

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And then 5-- this isn't a minus 5-- this is times

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negative 5.

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5 times negative 5 is equal to negative 25.

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Now, we can start with this top equation and add the same

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thing to both sides, where that same thing is negative

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25, which is also equal to this expression.

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So let's add the left-hand sides and

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the right-hand sides.

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Because we're really adding the same thing to both sides

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of the equation.

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So the left-hand side, the x's cancel out.

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35x minus 35x.

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That was the whole point.

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They cancel out, and on the y's, you get 49y plus 15y,

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that is 64y.

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64y is equal to 105 minus 25 is equal to 80.

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Divide both sides by 64, and you get y is equal to 80/64.

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And let's see, if you divide the numerator and the

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denominator by 8-- actually you could probably do 16.

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16 would be better.

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But let's do 8 first, just because we

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know our 8 times tables.

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So that becomes 10/8, and then you can divide this by 2, and

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you get 5/4.

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If you divided just straight up by 16, you would've gone

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straight to 5/4.

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So y is equal to 5/4.

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Let's figure out what x is.

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So we can substitute either into one of these equations,

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or into one of the original equations.

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Let's substitute into the second of the original

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equations, where we had 7x minus 3y is equal to 5.

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That was the original version of the second equation that we

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later transformed into this.

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So we get 7x minus 3 times y, times 5/4, is equal to 5.

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Or 7x minus 15/4 is equal to 5.

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Let's add 15/4-- Oh, sorry, I didn't do that right.

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This would be 7x minus 3 times 4-- Oh, sorry, that was right.

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What am I doing?

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3 times is 15/4.

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Is equal to 5.

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Let's add 15/4 to both sides.

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And what do we get?

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The left-hand side just becomes a 7x.

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These guys cancel out.

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And that's going to be equal to 5, is the

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same thing as 20/4.

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20/4 plus 15/4.

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Or we get that-- let me scroll down a little bit-- 7x is

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equal to 35/4.

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We can multiply both sides by 1/7, or we could divide both

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sides by 7, same thing.

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Let's multiply both sides by 1/7.

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The same thing as dividing by 7.

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So these cancel out and you're left with x is equal to--

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Here, if you divide 35 by 7, you get 5.

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You divide 7 by 7, you get 1.

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So x is equal to 5/4 as well.

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So the point of intersection of this right here is both x

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and y are going to be equal to 5/4.

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So if you looked at it as a graph, it'd be 5/4 comma 5/4.

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And let's verify that this satisfies the top equation.

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And if you take 5 times 5/4, plus 7 times

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5/4, what do you get?

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It should be equal to 15.

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So this is equal to 25/4, plus-- what is this?

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This is plus 35/4.

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Which is equal to 60/4, which is indeed equal to 15.

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So it does definitely satisfy that top equation.

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And you could check out this bottom equation for yourself,

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but it should, because we actually used this bottom

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equation to figure out that x is equal to 5/4.

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