Solving Problems Using Systems Of Equations

Solving Problems Using Systems Of Equations-21
And it's probably not obvious, even though it's sitting right in front of your face. So we know that 3 times x, 3 times 7 over 2-- I'm just substituting the x value we figured out into this top equation-- 3 times 7 over 2, plus 4y is equal to 2.5. Divide both sides by 4, and you get y is equal to negative 2. Is there something we could add to both sides of this equation that'll help us eliminate one of the variables? Well, what if we just added this equation to that equation? And we could substitute this back into either of these two equations. So the solution to this equation is x is equal to 7/2, y is equal to negative 2. Or let me put it this way, is there something we could add or subtract to both sides of this equation that will help us eliminate one of the variables? So if we did that we would be subtracting the same thing from both sides of the equation. You would get Ax plus By, plus D is equal to C plus D. Anything you do to one side of the equation, you have to do to the other side. This second equation is telling me that explicitly. Peter also buys 3 candy bars, but can only afford 1 additional Fruit Roll-Up. What is the cost of each candy bar and each Fruit Roll-Up?

On the right-hand side, you're adding 25.5 to the equation.

Aren't you adding two different things to both sides of the equation?

And you're probably saying, Sal, hold on, how can you just add two equations like that? When I looked at these two equations, I said, oh, I have a 4y, I have a negative 4y. And that's going to be equal to 2.5 plus 25.5 is 28. So let's verify that it also satisfies this bottom equation.

And remember, when you're doing any equation, if I have any equation of the form-- well, really, any equation-- Ax plus By is equal to C, if I want to do something to this equation, I just have to add the same thing to both sides of the equation. If we were to add the left-hand side, 3x plus 5x is 8x. If you just add these two together, they are going to cancel out. 5 times 7/2 is 35 over 2 minus 4 times negative 2, so minus negative 8. Now let's see if we can use our newly found skills to tackle a word problem, our newly found skills in elimination. Nadia buys 3 candy bars, so the cost of 3 candy bars is going to be 3x.

Let's look at an example: Find the ordered pair for which Thus, the solution to the system is .

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We figured out, using elimination, that the cost of a candy bar is equal to [[

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables.

We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.

A system of linear equations is where all of the variables are to the power 1.

What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out.

So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation.

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The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.A system of linear equations is where all of the variables are to the power 1.What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is $1.79 minus $0.35. And 3 goes into $1.44, I think it goes-- well, 3 goes into $1.44, it goes into 1 zero times. If we subtract 0.35 from both sides, what do we get? Matrices can also be used to solve systems of linear equations.In fact, they provide a way to make much broader statements about systems of linear equations.

]].48, and that the cost of a Fruit Roll-Up is equal to [[

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables.

We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.

A system of linear equations is where all of the variables are to the power 1.

What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out.

So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation.

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The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.A system of linear equations is where all of the variables are to the power 1.What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is $1.79 minus $0.35. And 3 goes into $1.44, I think it goes-- well, 3 goes into $1.44, it goes into 1 zero times. If we subtract 0.35 from both sides, what do we get? Matrices can also be used to solve systems of linear equations.In fact, they provide a way to make much broader statements about systems of linear equations.

]].35.A system of linear equations is where all of the variables are to the power 1.What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables.

We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.

A system of linear equations is where all of the variables are to the power 1.

What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out.

So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation.

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The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.A system of linear equations is where all of the variables are to the power 1.What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is $1.79 minus $0.35. And 3 goes into $1.44, I think it goes-- well, 3 goes into $1.44, it goes into 1 zero times. If we subtract 0.35 from both sides, what do we get? Matrices can also be used to solve systems of linear equations.In fact, they provide a way to make much broader statements about systems of linear equations.

.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables.

We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.

A system of linear equations is where all of the variables are to the power 1.

What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out.

So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation.

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The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.A system of linear equations is where all of the variables are to the power 1.What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is $1.79 minus $0.35. And 3 goes into $1.44, I think it goes-- well, 3 goes into $1.44, it goes into 1 zero times. If we subtract 0.35 from both sides, what do we get? Matrices can also be used to solve systems of linear equations.In fact, they provide a way to make much broader statements about systems of linear equations.

.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables.

We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.

A system of linear equations is where all of the variables are to the power 1.

What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out.

So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation.

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The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.A system of linear equations is where all of the variables are to the power 1.What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is $1.79 minus $0.35. And 3 goes into $1.44, I think it goes-- well, 3 goes into $1.44, it goes into 1 zero times. If we subtract 0.35 from both sides, what do we get? Matrices can also be used to solve systems of linear equations.In fact, they provide a way to make much broader statements about systems of linear equations.

.79 minus [[

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables.

We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.

A system of linear equations is where all of the variables are to the power 1.

What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out.

So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation.

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The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.A system of linear equations is where all of the variables are to the power 1.What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is $1.79 minus $0.35. And 3 goes into $1.44, I think it goes-- well, 3 goes into $1.44, it goes into 1 zero times. If we subtract 0.35 from both sides, what do we get? Matrices can also be used to solve systems of linear equations.In fact, they provide a way to make much broader statements about systems of linear equations.

]].35. And 3 goes into

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables.

We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.

A system of linear equations is where all of the variables are to the power 1.

What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out.

So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation.

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The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.A system of linear equations is where all of the variables are to the power 1.What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is $1.79 minus $0.35. And 3 goes into $1.44, I think it goes-- well, 3 goes into $1.44, it goes into 1 zero times. If we subtract 0.35 from both sides, what do we get? Matrices can also be used to solve systems of linear equations.In fact, they provide a way to make much broader statements about systems of linear equations.

.44, I think it goes-- well, 3 goes into

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables.

We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.

A system of linear equations is where all of the variables are to the power 1.

What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out.

So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation.

||

The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.A system of linear equations is where all of the variables are to the power 1.What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? This would be the coordinate of their intersection. Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. And you could try it out on both of these equations right here. And remember, by doing that, I would be subtracting the same thing from both sides of the equation. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is $1.79 minus $0.35. And 3 goes into $1.44, I think it goes-- well, 3 goes into $1.44, it goes into 1 zero times. If we subtract 0.35 from both sides, what do we get? Matrices can also be used to solve systems of linear equations.In fact, they provide a way to make much broader statements about systems of linear equations.

.44, it goes into 1 zero times. If we subtract 0.35 from both sides, what do we get? Matrices can also be used to solve systems of linear equations.In fact, they provide a way to make much broader statements about systems of linear equations.

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Comments Solving Problems Using Systems Of Equations

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