Solving Equations with the Distributive Property Solving Equations with the Distributive Property
Solving Equations with the Distributive Property
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- We have the equation negative 9 minus this whole expression,
- 9x minus 6-- this whole thing is being subtracted from
- negative 9-- is equal to 3 times this whole
- expression, 4x plus 6.
- Now, a good place to start is to just get rid of these
- And the best way to get rid of these parentheses is to kind
- of multiply them out.
- This has a negative 1-- you just see a minus here, but
- it's just really the same thing as having a negative 1--
- times this quantity.
- And here you have a 3 times this quantity.
- So let's multiply it out using the distributive property.
- So the left-hand side of our equation, we have
- our negative 9.
- And then we want to multiply the negative 1 times each of
- these terms. So negative 1 times 9x is negative 9x, and
- then negative 1 times negative 6 is plus 6, or positive 6.
- And then that is going to be equal to-- let's distribute
- the 3-- 3 times 4x is 12x.
- And then 3 times 6 is 18.
- Now what we want to do, let's combine our constant
- terms, if we can.
- We have a negative 9 and a 6 here, on this side, we've
- combined all of our like terms. We can't combine a 12x
- and an 18, so let's combine this.
- So let's combine the negative 9 and the 6, our two constant
- terms on the left-hand side of the equation.
- So we're going to have this negative 9x.
- So we're going to have negative 9x plus-- let's see,
- we have a negative 9 and then plus 6-- so negative 9 plus 6
- is negative 3.
- So we're going to have a negative 9x, and then we have
- a negative 3, so minus 3 right here.
- That's the negative 9 plus the 6, and that is
- equal to 12x plus 18.
- Now, we want to group all the x terms on one side of the
- equation, and all of the constant terms-- the negative
- 3 and the positive 18 on the other side-- I like to always
- have my x terms on the left-hand side, if I can.
- You don't have to have them on the left, so let's do that.
- So if I want all my x terms on the left, I have to get rid of
- this 12x from the right.
- And the best way to do that is to subtract 12x from both
- sides of the equation.
- So let me subtract 12x from the right, and subtract 12x
- from the left.
- Now, on the left-hand side, I have negative 9x minus 12x.
- So negative 9 minus 12, that's negative 21.
- Negative 21x minus 3 is equal to-- 12x minus 12x, well,
- that's just nothing.
- That's 0.
- So I could write a 0 here, but I don't
- have to write anything.
- That was the whole point of subtracting the 12x from the
- left-hand side.
- And that is going to be equal to-- so on the right-hand
- side, we just are left with an 18.
- We are just left with that 18 here.
- These guys canceled out.
- Now, let's get rid of this negative 3 from
- the left-hand side.
- So on the left-hand side, we only have x terms, and on the
- right-hand side, we only have constant terms. So the best
- way to cancel out a negative 3 is to add 3.
- So it cancels out to 0.
- So we're going to add 3 to the left, let's
- add 3 to the right.
- And we get-- the left-hand side of the equation, we have
- the negative 21x, no other x term to add or subtract here,
- so we have negative 21x.
- The negative 3 and the plus 3, or the positive 3, cancel
- out-- that was the whole point-- equals--
- what's 18 plus 3?
- 18 plus 3 is 21.
- So now we have negative 21x is equal to 21.
- And we want to solve for x.
- So if you have something times x, and you just want it to be
- an x, let's divide by that something.
- And in this case, that something is negative 21.
- So let's divide both sides of this equation by negative 21.
- Divide both sides by negative 21.
- The left-hand side, negative 21 divided by negative 21,
- you're just left with an x.
- That was the whole point behind
- dividing by negative 21.
- And we get x is equal to-- what's 21
- divided by negative 21?
- Well, that's just negative 1.
- You have the positive version divided by the negative
- version of itself, so it's just negative 1.
- So that is our answer.
- Now let's verify that this actually works for that
- original equation.
- So let's substitute negative 1 into that original equation.
- So we have negative 9-- I'll do it over here; I'll do it in
- a different color than we've been using-- we have negative
- 9 minus-- that 1 wasn't there originally, it's there
- implicitly-- minus 9 times negative 1.
- 9 times-- I'll put negative 1 in parentheses-- minus 6 is
- equal to-- well, actually, let me just solve for the
- left-hand side when I substitute a negative 1 there.
- So the left-hand side becomes negative 9, minus 9 times
- negative 1 is negative 9, minus 6.
- And so this is negative 9 minus-- in parentheses--
- negative 9 minus 6 is negative 15.
- So this is equal to negative 15.
- And so we get negative 9-- let me make sure I did that--
- negative 9 minus 6, yep, negative 15.
- So negative 9 minus negative 15, that's the same thing as
- negative 9 plus 15, which is 6.
- So that's what we get on the left-hand side of the equation
- when we substitute x is equal to negative 1.
- We get that it equals 6.
- So let's see what happens when we substitute negative 1 to
- the right-hand side of the equation.
- I'll do it in green.
- We get 3 times 4 times negative 1 plus 6.
- So that is 3 times negative 4 plus 6.
- Negative 4 plus 6 is 2.
- So it's 3 times 2, which is also 6.
- So when x is equal to negative 1, you substitute here, the
- left-hand side becomes 6, and the right-hand side becomes 6.
- So this definitely works out.
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