Old school equations with Sal
Algebra: Linear Equations 4 Solving linear equations with variable expressions in the denominators of fractions
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- Welcome to the presentation on level four linear equations.
- So, let's start doing some problems.
- So.
- Let's say I had the situation-- let me give me a couple of
- problems-- if I said three over x is equal to, let's just say five.
- So, what we want to do -- this problem's a little unusual from
- everything we've ever seen.
- Because here, instead of having x in the numerator, we actually
- have x in the denominator.
- So, I personally don't like having x's in my denominators,
- so we want to get it outside of the denominator into a
- numerator or at least not in the denominator as
- soon as possible.
- So, one way to get a number out of the denominator is, if we
- were to multiply both sides of this equation by x, you see
- that on the left-hand side of the equation these two
- x's will cancel out.
- And in the right side, you'll just get five times x.
- So this equals -- the two x's cancel out.
- And you get three is equal to fivex.
- Now, we could also write that as fivex is equal to three.
- And then we can think about this two ways.
- We either just multiply both sides by one / five, or you could just
- do that as dividing by five.
- If you multiply both sides by one / five.
- The left-hand side becomes x.
- And the right-hand side, three times one / five, is equal to three / five.
- So what did we do here?
- This is just like, this actually turned into a level
- two problem, or actually a level one problem,
- very quickly.
- All we had to do is multiply both sides of this
- equation by x.
- And we got the x's out of the denominator.
- Let's do another problem.
- Let's have -- let me say, x plus two over x plus one is
- equal to, let's say, seven.
- So, here, instead of having just an x in the denominator,
- we have a whole x plus one in the denominator.
- But we're going to do it the same way.
- 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.
- 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.
- On the left-hand side, the x plus one's cancel out.
- And you're just left with x plus two.
- It's over one, but we can just ignore the one.
- And that equals seven times x plus one.
- And that's the same thing as x plus two.
- And, remember, it's seven times the whole thing, x plus one.
- So we actually have to use the distributive property.
- And that equals sevenx plus seven.
- So now it's turned into a, I think this is a level
- three linear equation.
- And now all we do is, we say well let's get all the x's on
- one side of the equation.
- And let's get all the constant terms, like the two and the seven, on
- the other side of the equation.
- So I'm going to choose to get the x's on the left.
- So let's bring that sevenx onto the left.
- And we can do that by subtracting sevenx from both sides.
- Minus sevenx, plus, it's a minus sevenx.
- The right-hand side, these two sevenx's will cancel out.
- And on the left-hand side we have minus sevenx plus x.
- Well, that's minus six plus two is equal to, and on the
- right all we have left is seven.
- Now we just have to get rid of this two.
- And we can just do that by subtracting two from both sides.
- And we're left with minus six x is equal to six.
- Now it's a level one problem.
- We just have to multiply both sides times the reciprocal
- of the coefficient on the left-hand side.
- And the coefficient's negative six.
- So we multiply both sides of the equation by negative one / six.
- Negative one / six.
- The left-hand side, negative one over six times negative six.
- Well that just equals one.
- So we just get x is equal to five times negative one / six.
- Well, that's negative five / six.
- And we're done.
- And if you wanted to check it, you could just take that x
- equals negative five / six and put it back in the original question
- to confirm that it worked.
- Let's do another one.
- I'm making these up on the fly, so I apologize.
- Let me think.
- three times x plus five is equal to eight times x plus two.
- Well, we do the same thing here.
- Although now we have two expressions we want to get
- out of the denominators.
- We want to get x plus five out and we want to get
- this x plus two out.
- So let's do the x plus five first.
- Well, just like we did before, we multiply both sides of
- this equation by x plus five.
- You can say x plus five over one.
- Times x plus five over one.
- On the left-hand side, they get canceled out.
- So we're left with three is equal to eight times x plus five.
- All of that over x plus two.
- Now, on the top, just to simplify, we once again
- just multiply the eight times the whole expression.
- So it's eightx plus forty over x plus two.
- Now, we want to get rid of this x plus two.
- So we can do it the same way.
- We can multiply both sides of this equation by
- x plus two over one.
- x plus two.
- We could just say we're multiplying both
- sides by x plus two.
- The one is little unnecessary.
- So the left-hand side becomes threex plus six.
- Remember, always distribute three times, because you're
- multiplying it times the whole expression.
- x plus two.
- And on the right-hand side.
- Well, this x plus two and this x plus two will cancel out.
- And we're left with eightx plus forty.
- And this is now a level three problem.
- Well, if we subtract eightx from both sides, minus eightx, plus-- I
- think I'm running out of space.
- Minus eightx.
- Well, on the right-hand side the eightx's cancel out.
- On the left-hand side we have minus fivex plus six is equal
- to, on the right-hand side all we have left is forty.
- Now we can subtract six from both sides of this equation.
- Let me just write out here.
- Minus six plus minus six.
- Now I'm going to, hope I don't lose you guys by
- trying to go up here.
- But if we subtract minus six from both sides, on the left-hand
- side we're just left with minus fivex equals, and on the
- right-hand side we have thirty-four.
- Now it's a level one problem.
- We just multiply both sides times negative one / five.
- Negative one / five.
- On the left-hand side we have x.
- And on the right-hand side we have negative thirty-four / five.
- Unless I made some careless mistakes, I think that's right.
- And I think if you understood what we just did here, you're
- ready to tackle some level four linear equations.
- Have fun.
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