WEBVTT
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A system of two linear equations with the constants a, b, c, d, e, and f
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is of the form ax plus by equals c, and second equation dx plus ey equals f.
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What we are going to do is learn how to solve a
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system of two linear equations by substitution and elimination.
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It is also possible to solve these systems by graphing.
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If we graph one linear equation and we graph a second one,
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one of the possibilities in a rectangular coordinate plane
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is that the two lines intersect at a single point with the coordinates (x, y).
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This system is, because it has one solution, is called consistent.
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Now that is one possibility. A second possibility is that the two lines may be parallel,
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in which case there is no solution because the two lines do not intersect.
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This is called an inconsistent system.
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The third possibility is that the two lines are actually the same line.
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In this case, we have an infinite number of solutions,
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and this system is known as a dependent system.
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This is the background material for a system of two linear equations.
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We will look at our first example.
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Solve by substitution or elimination.
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We have the system 3x minus 6y equals 2, 5x plus 4y equals 1.
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We have a choice as to which method we want to use.
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Now because the coefficients of our x’s and y’s are greater than one,
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substitution is not a good choice. I am going to use elimination,
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and in elimination I need for the coefficients of the x’s to be able to cancel
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or to give a zero when we add them.
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In other words, I would need this to be 15x and this to be negative 15x,
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or if I want to eliminate the y’s I would need
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this term to be negative 12y and this one to be positive 12y.
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The choice is yours as to which of the two variables you are going to eliminate.
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Let’s write our system, and I think I will eliminate the y’s.
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So, to get negative 12 I need to multiply by a positive 2,
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and I will multiply both sides of the equation.
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To get a positive 12 I am going to multiply both sides by positive 3.
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Let’s be very careful as we distribute.
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2 times 3x is 6x, 2 times negative 6y is negative 12y, and 2 times 2 is 4.
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3 times 5x is 15x, 3 times 4y is 12y, 3 times 1 is 3.
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Now we want to add. 6x and 15x is 21x, the y’s are eliminated, 4 and 3 is 7.
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Solving now by dividing both sides by 21 we get 7 divided by 21,
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but we reduce that to lowest terms, so we get x is a positive 1/3.
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Now we need to find y, so we are going to take the value that we found for x,
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substitute it into one of the equations. I am just going to choose the first equation.
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3 times 1/3 minus 6y equals 2. 3 times 1/3 is 1.
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Subtract the 1. Divide by the negative 6.
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We get negative 1/6. So we see that our solution is the ordered pair (1/3, negative 1/6).
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Now if we remember from our earlier discussion graphically what this means
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in this system the lines that are represented here intersect at this one point,
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so it is a consistent system. We will look at a second example.
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Solve by substitution or elimination. 3x plus y equals 10. 12x plus 4y equals 14.
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We have a y that has a coefficient of 1, so this system would be
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very convenient for us to solve by substitution. So let’s try that method.
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y equals, subtracting the 3x, y equals 10 minus 3x.
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Now I am going to take that expression for y and substitute into the second equation.
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So I will have 12x plus 4 times 10 minus 3x equals 14.
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So what has happened is the y has been eliminated by substitution.
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We will distribute the 4, 40 minus 12x equals 14.
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We will combine the like terms. 12x minus 12x is 0. 40 equals 14.
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In this example, not only was the y eliminated here, but now the x’s are eliminated.
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We look at the remaining sentence, which we know is not possible,
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so this system has no solution. Let’s circle that as our final answer.
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Recall graphically, what does this look like?
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Well this line is about here, with a y-intercept at 10 and a negative slope.
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If we solve this equation for y, it has a negative slope and a y-intercept of 5/2.
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Because the slopes are equal, we know that the lines are parallel.
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On my graph they don’t look very parallel, but they are parallel.
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We have an inconsistent system, which gives us no solution.
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We will look at a third example.
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Solve by substitution or elimination. We have the system
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x plus 2y equals 4, 3x plus 6y equals 12.
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Substitution, we could simply solve for x in the first equation
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and then substitute into the second equation.
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If we use elimination, we have several choices.
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I think the choice that I will use is to multiply the first equation by negative 3.
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Now of course you may recognize that
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if you divided the second equation by 3 you would get x plus 2y equals 4.
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So that gives us a hint that if we end up with the same two equations
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what is going to happen with our solution.
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Let’s just take that first equation and multiply the equation by negative 3.
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So that is going to give us negative 3x minus 6y equals negative 12.
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That becomes our first equation.
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Second equation, 3x plus 6y equals positive 12.
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When we add the x’s give us 0. The y’s, negative 6y plus 6y also gives us 0.
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Then on the right side, we have negative 12 plus 12 which is 0.
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This is what we suspected, that we just have a dependent system,
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actually consistent and dependent, and our solution is going to be
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an infinite number of ordered pairs.
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Let’s go ahead and solve for y and see what that graph would look like.
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2y would be negative x plus 4. y will be negative 1/2 x plus 2.
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So quick graph. y- intercept 2, slope of negative 1/2.
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We get that line, which would be the same for both.
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We are going to write our solution as the set of ordered pairs (x, y)
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such that, we have solved for y, y is negative 1/2 x plus 2
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where x is any real number. Some students are always sometimes confused
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because they think all real numbers will work,
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but it is only the ordered pairs (x, y) where x is a real number and then y has this form.
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So for example one solution is (0, 2), our y-intercept.
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If we let x be 2, we would have y is negative 1/2 times 2, that would be,
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oh let’s figure that out. That would be negative 1/2 times 2 plus 2.
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That will give me negative 1 plus 2, or positive 1.
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Those are just two of the solutions to this system of linear equations.
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These three examples should show you how to solve any system
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of two linear equations by substitution or elimination.