WEBVTT
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This time around we're going to take a look at systems of equations using the
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substitution method. That is we're going to solve one equation for either x or
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y whatever proves most convenient and replace it or substitute it into the
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other. Taking a look at our two equations in the system we need to be clever
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about which variable is easier to get by itself. We want to work smarter not
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harder. If we have a variable with a coefficient of 1 that's a lot better
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for us because we don't have to do any dividing to get it by itself. Well our
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y variables here are kind of complex. We have a coefficient of 2 and a coefficient
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of 5 but looking at our x inspecting our x variables we have a good situation
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here because in our second equation we have a coefficient of 1 that's just one
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x. Now if we just subtract the 5 y we will have x by itself and then we'll be in
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good shape because again the whole strategy here is once you get either x or
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y by itself you can replace it into the other equation. Moving around long we're
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going to keep our top equation 3x minus 2y equals 8 but we're going to solve the
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bottom one for x. To get x by itself as you can see we need to subtract negative
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5y or rather I'm sorry we have to subtract 5y which is going to give us a
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negative 5y on the other side so if you solve this for x you get x equals
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negative 5y minus 3. I think we can all handle that much. So now we have our
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top equation which is 3x minus 2y equals 8 and we have our new one that we just
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solve for x. x equals negative 5y minus 3. Now since this is called substitution
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we substitute our new found value of x into the other equation. So right here we
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have found x by itself what is x? x is negative 5y minus 3. So instead of seeing x
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here you're going to replace x with what we just found negative 5y minus 3. In
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other words instead of 3 times x we're going to substitute it x. Instead of 3
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times x we're going to do 3 times negative 5y minus 3. Here we go. We have 3
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times what x just became. 3 times negative 5y minus 3 minus the 2y from the
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original problem equals 8. Now we have to use distributive property. 3 times
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negative 5y is negative 15y. 3 times negative 3 is negative 9. Re-copy the
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minus 2y equals 8. Moving right along we're doing pretty well here trying to
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solve for y. We need to collect like terms. I can see that we have a negative 15y
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and a negative 2y. Those combine to a negative 17y. Re-copy your minus 9 equals
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8. Now in this process we have to get y by itself. So we want to get rid of minus
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9. We add 9 to both sides and now we have negative 17y equals 17. We don't
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want negative 17 times y to get rid of it. We divide both sides by negative 17.
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Therefore we have y equals negative 1. That's part of our solution. The y
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coordinate is negative 1 but we also need the x because remember systems are
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looking for an intersection point. So we have our y. Now we need to find our
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x. The easiest thing to do here is to go back up to the beginning and we had an
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equation for x. We have x equals negative 5y minus 3. Well we just found out a
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second ago that our y value is negative 1. If the y value is negative 1 I can
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replace it in there. Negative 5 times negative 1 is positive 5. Take away the
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3. Positive 5, take away 3 is 2. So now x equals 2. x equals 2. So when you have y
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equals negative 1, x equals 2. Writing that as an ordered pair we have 2,
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1, negative 1. That is the final and correct answer. 2, negative 1.