WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we will be talking about solving systems of equations algebraically.
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In the previous lesson, we talked about solving systems of equations by graphing.
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However, that method has some limitations; therefore, we are going to talk about a few other methods of solving.
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And this is some review from Algebra I; so again, if you need more detail--more review on these concepts--check out our Educator.com Algebra I series.
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First, we are going to review solving by substitution, first jotting down a system of equations:
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2x + 3y = 12, and the second equation in the system is x + 4y = 1.
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In this method, you are going to solve one equation for one variable in terms of the other variable.
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Then, substitute the expression for the variable in the other equation.
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Let's look at what this means: first step: solve one equation for one variable in terms of the other one.
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The easiest thing to do is to find a situation where you have a variable that has a coefficient of 1; and I have that right here.
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So then, I can solve for x in terms of y pretty easily.
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So, x + 4y = 1, so I am going to solve for x in terms of y; this gives me x = -4y + 1.
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The second step is to substitute this expression for the variable in the other equation.
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So, I am going to substitute -4y + 1 for x in the first equation; and that is going to create
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an equation that has only one variable, and then I will be able to solve that.
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It is important that you go back into the other equation to substitute in.
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This is going to give me 2(-4y + 1) + 3y = 12; then, I can just solve for y.
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2 times -4y; that is -8y, plus 2 times 1--that is just 2--plus 3y, equals 12.
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Combine like terms: -8y and 3y gives me -5y, plus 2 equals 12; subtract 2 from both sides to get -5y = 10.
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Dividing both sides by -5, I get y = -2.
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Once I have one variable solved, I can easily solve for the other.
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So, I just go back to either one of these (and this one is easier to work with), and I am going to substitute in...I am going to let y equal -2.
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x + 4(-2) = 1; this is going to give me x + -8 = 1, or x - 8 = 1; adding 8 to both sides, I get x = 9.
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Here, y equals -2 and x equals 9.
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This method of substitution works really well when the coefficient of one of the variables is 1.
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So, use substitution when a variable in one of the equations has a coefficient of 1.
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So again, we looked at this system of equations, saw that this variable had a coefficient of 1,
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then solved for this variable x in terms of the other variable y to get x = -4y + 1.
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And then, I went ahead and substituted this 4x in the other equation; that gave me one equation with one variable, and I can solve for y.
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Once you have y, you can substitute that value into either equation and then solve for x.
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The second method is solving by elimination: in elimination, you add or subtract the two equations to eliminate one of the variables from the resulting equation.
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And this system works well when you have variables, either the two x's or the two y's, that have either the same coefficient or opposite coefficients.
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By "opposite coefficients," I mean the same number with opposite signs, such as 2 and -2 or 3 and -3.
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For example, if you were asked to solve this system, you look and see that there is no variable with a coefficient of 1.
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Therefore, substitution is not the ideal--it is not the easiest way to go.
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But what you see is that you have one variable--you have y--that has the same coefficient.
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So, adding is certainly not going to help me; if I add these together, I will get 6x + 10y = 8.
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What I need to do is subtract, because my goal is to get one variable to drop out.
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So, I am going to subtract the second equation from the first.
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And just to help me keep my signs straight, I rewrite that as adding the opposite; so I am adding -4x, -5y, and -3.
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OK, this is going to give me 2x - 4x is -2x; the y's are going to drop out: 5y - 5y gives me 0, so I can write + 0, but they just drop out.
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And then, 5 minus 3 equals 2; this is -2x = 2; I can easily solve for x, and x equals -1.
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Once I have one variable, I can go to either original equation, and then substitute in order to solve for the other variable.
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I know that x equals -1, so I am going to substitute that up here.
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Add 2 to both sides (that is going to give me 5y = 7), and divide both sides by 5 to get y = 7/5.
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So, solving by elimination, I was able to come up with the solution that x equals -1 and y = 7/5, which will satisfy both of these equations.
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Again, this works well; use elimination when the same variable (meaning both x's or both y's)
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have the same or opposite coefficients (the same coefficient, but opposite signs).
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Sometimes, you look, and you see that there is no variable with a coefficient of 1; and then you say, "OK, I will use elimination."
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But then, you realize that none of the variables have the same or opposite coefficients.
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In that case, you can still use elimination, but you are going to have to take an extra step before you do the elimination.
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And the extra step is to multiply one or both equations so that one of the variables has the same or opposite coefficient in the two new equations.
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So, your goal is to create a situation where one of the variables has the same or opposite coefficients.
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Then, you just use elimination, as we did previously.
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If you had a system of equations, 5x + 4y = 1, and 6x + 3y = 3, you see that neither of these
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has a variable with a coefficient of 1, and neither set of variables has the same or opposite coefficients.
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But if I look at these two, 4y and 3y, the least common multiple is 12; so what I want to do
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is multiply each of these by something, in order to end up with a coefficient of 12 in front of the y.
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Sometimes you will be lucky, and all you will have to do is multiply one of the equations by a number to get the opposite or same coefficient.
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Other times, like this, you are going to have to multiply both.
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So, for the first equation, I am going to multiply by 3; and this is going to give me 15x + 12y = 3.
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The second equation I am going to multiply by 4: this is going to give me...4 times 6x is 24x + 12 y = 12.
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OK, I now see that I have the same coefficient for y; so rewriting this over here, I need to subtract.
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So, I am going to subtract the second equation from the first.
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And to keep everything straight, I like to go about this by adding the opposite to make sure that I keep my signs straight.
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So, I am going to add -24x - 12y, and then that is going to be a -12.
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OK, -24x - 12y - 12: add these together: I get 15 - 24x--that is going to give me -9x; the y's drop out, and then I have 3 - 12; that is -9.
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Divide both sides by -9 to get x = 1.
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From here, I substitute back; I will choose this top equation, 5x + 4y = 1, and I know that x equals 1.
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This is 5(1) + 4y = 1; so that is 5 + 4y = 1, or 4y = -4, so y = -1.
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So, x equals 1; y equals -1.
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Again, this is just the first step in elimination; and you use it in a situation like this,
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where you want to use elimination, but you don't already have a set of variables with the same or opposite coefficients.
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So, you figure out what the least common multiple is, and multiply one or both equations in order to achieve that.
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From there, you just proceed as we did before, solving by elimination.
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There are several different possibilities that can occur when you are solving systems of equations algebraically.
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Usually, in the problems that you will see, what will happen is what just happened previously,
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that I showed you, where you will end up with a value for x and a value for y that satisfy both equations.
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However, there are times when that doesn't occur.
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You can be going along, doing substitution, doing elimination, and things are going fine;
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and then you end up with something like this: c = d.
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Your variables drop out, and you end up with an equation that is saying that a constant is equal to a different constant--for example, 4 = 5.
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Well, that is not true; so when you see this, this is an equation that is never true.
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And what this tells me is that the system of equations is inconsistent.
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So, this is a situation where there is no solution.
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If you start seeing a situation where the constants drop out, and then you see something like 4 = 5
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or 9 = 10--something that is not true--then you know you have an inconsistent system.
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The other possibility is that you could have a system that is dependent, or always true.
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In that case, you are going along; you are doing elimination; you are doing substitution;
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and then you see that you end up with variables dropping out, and a constant that equals a constant--the same constant--for example, 3 = 3.
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Well, that is always true; so if you end up with an equation--the sum or difference of the two equations--
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the system of equations--that is always true, this system is dependent, and it has an infinite number of solutions.
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Recall, when we were solving systems of equations by graphing: this is analogous to the situation where you would end up with the same line.
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So, if you have two equations (a system of equations), you graph both equations,
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and you find out that they are the same line, well, then, there is an infinite number of solutions--all points along those lines.
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Here, no solution would be like parallel lines, where they never intersect; there is no solution to that system.
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OK, the first example is: Solve algebraically for the system of equations.
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As soon as I see that I have variables (or even one variable) with a coefficient of 1, I recognize that substitution would be a really good method to use.
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So, use substitution when a variable has a coefficient of 1.
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So, I am going to solve for x in terms of y: x + y = 5, so x = 5 - y.
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Now, I have solved for x in terms of y; and now I am going to substitute this value into the other equation.
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2x + 3y = 13: I am going to substitute this expression for x, so 2 times (5 - y), plus 3y, equals 13.
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Or, 10 - 2y + 3y = 13; this is 10 + y = 13; then subtract 10 from both sides to get y = 3.
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Once I have this, I can plug y into this simple equation up here: x + y = 5.
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Substitute in; let y equal 3; and I get x = 2.
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And this is very easy to check; you can see that, if x is 2 and y is 3, this equation holds true.
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And you could do the same--plug it back into that first equation, put these values in for x and y, and just verify that these solutions are correct.
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OK, so again, we are solving by substitution because you had a situation where you had a variable (actually, two variables) with a coefficient of 1.
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That is the easiest method to use.
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Here, I am solving this system of equations algebraically; and I don't have any variables with a coefficient of 1, so I am not going to use substitution.
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But if I look; I have variables (y) that have the opposite coefficient, -3 and 3.
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What that means is that, if I add these equations, the y's are going to drop out; then, I can just solve for x.
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Rewrite this right here and add: I am going to add these two.
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2x + 4x: that gives me 6x; the y's drop out: -3y + 3y is 0; and then 0 + 0 is 9.
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I am going to divide both sides by 6; this is going to give me 9/6, and that simplifies to 3/2.
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I have x = 3/2; I am going to pick either one--I will go ahead and pick the top one: 2x -3y = 0.
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And let x equal 3/2; substitute that for x.
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The 2's cancel out; this gives me 3 - 3y = 0, or -3y = -3; if I divide both sides of the equation by -3, I will get y = 1.
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So, the solution is x = 3/2, y = 1; again, I could always check these solutions
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by substituting the values in here and making sure that this equation holds true.
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And this was a perfect setup to use elimination, because I already had variables that had the opposite coefficients.
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I simply added these equations together; the y dropped out, allowing me to solve for x, and then substitute 3/2 in for x.
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OK, Solve algebraically: again, there are no variables with a coefficient of 1, so I am going to go to elimination.
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But these do not have the same or opposite coefficients, and the y's do not have the same or opposite coefficients.
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So, this time I am going to need to use that extra step; I am going to need to use multiplication
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in order to create a situation where I have variables with the same or opposite coefficients.
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And I see that, if I multiply the top equation by 3, I will get 6x; and if I multiply the bottom equation by 2, I will get 6x.
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So, I am going to go ahead and do that--multiply the top equation by 3: that is going to give me 3 times 2x - 3y = -16.
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And this comes out to 3(2x) is 6x; 3 times -3...that is -9y; and 3 times -16 is -48.
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The second equation I am going to multiply by 2.
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So, 2 times 3x, plus 5y, equals 14; OK, 2 times 3x gives me 6x; 2 times 5y is 10y; and 2 times 14 is 28.
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I have the same coefficient here; so what I need to do is subtract.
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I am going to subtract, and again, I am going to convert this so that I am adding the opposite.
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Rewrite the top equation the same, and the bottom as adding the opposite (adding -6x - 10y - 28).
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OK, when I do that, the 6x's drop out, and that is going to give me -19y equals...-48 and -28 comes out to -76.
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Divide both sides by -19, and -76 divided by -19 actually equals 4, so it divided very nicely and evenly; so I ended up with y = 4.
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Now that I have that, I am going to substitute y = 4 into either equation; I am going to choose the top one.
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So, 2x - 3y = 16; let y equal 4; so, 2x - 3(4) = -16, or 2x - 12 = 16.
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Add 12 to both sides to get 2x = -4, and divide both sides by 2 to get x = -2.
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So, the solution is that x equals -2 and y equals 4.
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So, when I looked at this, I saw that I could use elimination, if I got the x's to have the opposite coefficient
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by multiplying the top equation by 3 to give me 6x, and the bottom equation by 2 to give me 6x here, as well.
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So, I did that; and then, these have the same coefficient, so I subtracted and solved for y.
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Once I had a value for y, I substituted that value into one of the equations and then solved for x to get the solution.
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Here, I have 2x + 3y = 8, and -4x - 6y = -16.
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Since there is no variable with a coefficient of 1, I am going to use elimination.
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But again, I am going to have to do a little work to get the opposite coefficients.
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And I see here that all I have to do is multiply the top by 2, and that will give me 4x; that is opposite coefficients.
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I don't have to do anything to the bottom equation.
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So, that is 2 times 2x, plus 3y, equals 8, and this is going to give me 4x + 6y = 16.
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That is the opposite coefficient, so I am going to add this new equation and this equation: + -4x - 6y - 16.
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And you may have already realized that each one of these terms is opposite.
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And so, here is what is going to happen: 4x - 4x is 0; 6y - 6y is 0; 16 - 16 is 0; so I end up with 0 = 0.
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And I didn't make any mistakes--I did everything correctly--but then all my variables went away.
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And what this tells me is that I have a dependent system, and it has an infinite number of solutions.
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If I were to graph this, I would see that I have an infinite number of solutions--that these are intersecting at every point along the line.
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So, we call this a dependent system.
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If I had been going along, and all of the variables dropped out, and then I got something that wasn't true,
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like 4 = 2 or 4 = 0, then I would have a situation where it is an inconsistent system and there are just no solutions.
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But here, this is always true; so I have an infinite number of solutions.
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So today, we covered solving systems of equations algebraically.
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And that concludes today's lesson for Educator.com; I will see you again.