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U6.1 Systems of Equations Intervention Assignment

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U6.1 Systems of Equations Intervention Assignment

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Write each problem and show all work on notebook paper.  When done, complete the U6.1 Systems of Equations Intervention Quiz in Blackboard.  You will enter answers to problems on this assignment.

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1) Watch the videos to review solving systems of linear equations.  You may also find it helpful to review your notes.  2) Complete the U6.1 Systems of Equations Intervention Assignment. 3) Then complete the U6.1 Systems of Equations Quiz in Blackboard by March 26.  4) Turn in completed assignment to Mrs. Mosher on Monday, March 27. 

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1) Watch the videos to review solving systems of linear equations.  You may also find it helpful to review your notes.  2) Complete the U6.1 Systems of Equations Intervention Assignment. 3) Then complete the U6.1 Systems of Equations Quiz in Blackboard by March 26.  4) Turn in completed assignment to Mrs. Mosher on Monday, March 27. 

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Let's say I have the equation y is equal to x plus 3. And I want to graph all of the sets, all of the coordinates x comma y that satisfy this equation right there. And we've done this many times before. So we draw our axis, our axes. That's my y-axis. This is my x-axis. And this is already in mx plus b form, or slope-intercept form. The y-intercept here is y is equal to 3, and the slope here is 1. So this line is going to look like this. We intersect at 0 comma 3-- 1, 2, 3. At 0 comma 3. And we have a slope of 1, so every 1 we go to the right, we go up 1. So the line will look something like that. It's a good enough approximation. So the line will look like this. And remember, when I'm drawing a line, every point on this line is a solution to this equation. Or it represents a pair of x and y that satisfy this equation. So maybe when you take x is equal to 5, you go to the line, and you're going to see, gee, when x is equal to 5 on that line, y is equal to 8 is a solution. And it's going to sit on the line. So this represents the solution set to this equation, all of the coordinates that satisfy y is equal to x plus 3. Now let's say we have another equation. Let's say we have an equation y is equal to negative x plus 3. And we want to graph all of the x and y pairs that satisfy this equation. Well, we can do the same thing. This has a y-intercept also at 3, right there. But its slope is negative 1. So it's going to look something like this. Every time you move to the right 1, you're going to move down 1. Or if you move to the right a bunch, you're going to move down that same bunch. So that's what this equation will look like. Every point on this line represents a x and y pair that will satisfy this equation. Now, what if I were to ask you, is there an x and y pair that satisfies both of these equations? Is there a point or coordinate that satisfies both equations? Well, think about it. Everything that satisfies this first equation is on this green line right here, and everything that satisfies this purple equation is on the purple line right there. So what satisfies both? Well, if there's a point that's on both lines, or essentially, a point of intersection of the lines. So in this situation, this point is on both lines. And that's actually the y-intercept. So the point 0, 3 is on both of these lines. So that coordinate pair, or that x, y pair, must satisfy both equations. And you can try it out. When x is 0 here, 0 plus 3 is equal to 3. When x is 0 here, 0 plus 3 is equal to 3. It satisfies both of these equations. So what we just did, in a graphical way, is solve a system of equations. Let me write that down. And all that means is we have several equations. Each of them constrain our x's and y's. So in this case, the first one is y is equal to x plus 3, and then the second one is y is equal to negative x plus 3. This constrained it to a line in the xy plane, this constrained our solution set to another line in the xy plane. And if we want to know the x's and y's that satisfy both of these, it's going to be the intersection of those lines. So one way to solve these systems of equations is to graph both lines, both equations, and then look at their intersection. And that will be the solution to both of these equations. In the next few videos, we're going to see other ways to solve it, that are maybe more mathematical and less graphical. But I really want you to understand the graphical nature of solving systems of equations. Let's do another one. Let's say we have y is equal to 3x minus 6. That's one of our equations. And let's say the other equation is y is equal to negative x plus 6. And just like the last video, let's graph both of these. I'll try to do it as precisely as I can. There you go. Let me draw some. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. And then 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. I should have just copied and pasted some graph paper here, but I think this'll do the job. So let's graph this purple equation here. Y-intercept is negative 6, so we have-- let me do another [? slash-- ?] 1, 2, 3, 4, 5, 6. So that's y is equal to negative 6. And then the slope is 3. So every time you move 1, you go up 3. You moved to the right 1, your run is 1, your rise is 1, 2, 3. That's 3, right? 1, 2, 3. So the equation, the line will look like this. And it looks like I intersect at the point 2 comma 0, which is right. 3 times 2 is 6, minus 6 is 0. So our line will look something like that right there. That's that line there. What about this line? Our y-intercept is plus 6. 1, 2, 3, 4, 5, 6. And our slope is negative 1. So every time we go 1 to the right, we go down 1. And so this will intersect at-- well, when y is equal to 0, x is equal to 6. 1, 2, 3, 4, 5, 6. So right over there. So this line will look like that. The graph, I want to get it as exact as possible. And so we're going to ask ourselves the same question. What is an x, y pair that satisfies both of these equations? Well, you look at it here, it's going to be this point. This point lies on both lines. And let's see if we can figure out what that point is. Just eyeballing the graph here, it looks like we're at 1, 2, 3 comma 1, 2, 3. It looks like this is the same point right there, that this is the point 3 comma 3. I'm doing it just on inspecting my hand-drawn graphs, so maybe it's not the exact-- let's check this answer. Let's see if x is equal to 3, y equals 3 definitely satisfies both these equations. So if we check it into the first equation, you get 3 is equal to 3 times 3, minus 6. This is 9 minus 6, which is indeed 3. So 3 comma 3 satisfies the top equation. And let's see if it satisfies the bottom equation. You get 3 is equal to negative 3 plus 6, and negative 3 plus 6 is indeed 3. So even with our hand-drawn graph, we were able to inspect it and see that, yes, we were able to come up with the point 3 comma 3, and that does satisfy both of these equations. So we were able to solve this system of equations. When we say system of equations, we just mean many equations that have many unknowns. They don't have to be, but they tend to have more than one unknown. And you use each equation as a constraint on your variables, and you try to find the intersection of the equations to find a solution to all of them. In the next few videos, we'll see more algebraic ways of solving these than drawing their two graphs and trying to find their intersection points.

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Systems of equations with graphing (video) | Khan Academy

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Let's say I have the equation y is equal to x plus 3. And I want to graph all of the sets, all of the coordinates x comma y that satisfy this equation right there. And we've done this many times before. So we draw our axis, our axes.

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Solve the system of linear equations by graphing, and they give us two equations here. 5x plus 3y is equal to 7, and 3x minus 2y is equal to 8. When they say, "Solve the system of linear equations," they're really just saying find an x and a y that satisfies both of these equations. And when they say to do it by graphing, we're essentially going to graph this first equation. Remember, the graph is really just depicting all of the x's and y's that satisfy this first equation, and then we graph the second equation that's depicting all of the x's and y's that satisfy that one. So if we were looking for an x and a y that satisfies both, that point needs to be on both equations or it has to be on both graphs. So it'll be the intersection of the two graphs. So let's try to see if we can do that. So let's focus on this first equation, and I want to graph it. So I have 5x plus 3y is equal to 7. There's a couple of ways we could graph this. We could put this in slope-intercept form, or we could just pick some points here. You just really need two points to graph a line. So let me just set some points over here. Let's say x and y. When x is equal to 0, what does y need to be equal to? So when x is equal to 0, we have-- let me do it over here-- we have 5 times 0, plus 3 times y, is equal to 7. That's just 0 over there. So you have 3y is equal to 7. Divide both sides by 3, you get y is equal to 7/3, which is the same thing as 2 and 1/3 if we want to write it as a mixed number. Now let's set y equal to 0. So if we set y as equal to 0, we get 5x, plus 3 times 0, is equal to 7. Or in this part right over here, it just becomes 0. So we have 5x is equal to 7. Divide both sides by 5, and we get x equals 7/5, which is the same thing as 1 and 2/5. So let's graph both of these points, and then we should be able to graph this line, or at least a pretty good approximation of that line. So we have the point, 0, 2 and 1/3. So that's that point right over there. So I'll call it 0, 7/3 right over there, and then we have the point, 7/5, 0, or 1 and 2/5, 0. So 1 and 2/5. 2/5 is a little less than a half. So 1 and 2/5, 0. So our line is going to look something like this. I just have to connect the dots. It's always hard to draw the straight line. I'll draw it as a dotted line. So it would look something like this. Normally, when you have to solve a system of equations by graphing, they normally give you a little bit cleaner numbers, but we'll try our best and see if we can see where these two lines intersect. So now, let's worry about the second one right over there. So we have 3x minus 2y is equal to 8. So I'll do the same thing. So 3x minus 2y is equal to 8. We'll just look at the x- and the y-intercepts. So first, the y-intercept. When x is equal to 0, this whole thing boils down to 3 times 0 minus 2y is equal to 8. That's just 0. So you have negative 2y is equal to 8. Divide both sides by negative 2, we get y is equal to negative 4. So the y-intercept is 0, negative 4. Right over here, and we mark it 0, negative 4. And then, let's set y equal to 0. So when y is equal to 0, this term right over here just becomes 0. So we get 3x minus 2 times 0, so that's just 0. So 3x is equal to 8. Divide both sides by 3, you get x is equal to 8/3. And 8/3 is the same thing as 2 and 2/3. So it puts it right about there. That's the point, 0, 8/3. Now let me try my best to graph it. Connect these two dots. So let me do my very best. I drew a dotted line there. It goes something like that. And just eyeballing it, it looks like these two lines intersect right over there. I'm hoping that this will give us a clean answer. And this is the point 2, negative 1. So that is the point, 2, negative 1. Right? The x value here is 2. y value is negative 1. Now, that's what we got just by eyeballing it, and clearly these are hand-drawn graphs. Not very precise. Let's verify or let's see if 2, negative 1 does satisfy both of these equations, if it's an x and y value that satisfies both and lies on both the graphs. So if you put 2, negative 1 in this first equation, you get 5 times 2, plus 3 times negative 1, and we're going to see if that is equal to 7. So this is 10 plus negative 3. So that's the same thing as 10 minus 3. Does that equal 7? And yeah, it does. 10 minus 3 is equal to 7. So 2, negative 1 is definitely on that graph or definitely satisfies that equation. And then, let's do it with the other one. So if we do 2, negative 1, you have 3 times 2, minus 2 times negative 1. And we're testing to see if that is equal to 8. So 3 times 2. 3 times 2 is 6. And then 2 times negative 1 is negative 2. But then, we're subtracting that, so it's 6 plus 2. 6 plus 2 is equal to 8. And it definitely does equal 8. So we have the coordinate or we have the point, 2, negative 1 satisfying both equations. So we've solved the system of linear equations by graphing.

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Systems of equations with graphing: 5x+3y=7 & 3x-2y=8 (video) | Khan Academy

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Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

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Systems of equations with substitution: -3x-4y=-2 & y=2x-5

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Learn to solve the system of equations -3x - 4y = -2 and y = 2x - 5 using substitution.

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Systems of equations with elimination: 3t+4g=6 & -6t+g=6

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Sal solves the system of equations 3t + 4g = 6 and -6t + g = 6 using elimination.

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Number of solutions to a system of equations algebraically (video) | Khan Academy

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Sal solves several examples where he reasons about the number of solutions of systems of equations using algebraic reasoning.

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Systems of equations with elimination: apples and oranges

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Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.