Solving Problems Using Systems Of Equations

Solving Problems Using Systems Of Equations-3
Let's explore a few more methods for solving systems of equations. Let's say I have the equation, 3x plus 4y is equal to 2.5.

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We'll go back to our same example to illustrate this.

Find the ordered pair for which From the graph, we can see that the solution to the system is .

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?

And I have another equation, 5x minus 4y is equal to 25.5.

Solving Problems Using Systems Of Equations

And we want to find an x and y value that satisfies both of these equations. We'll show how to solve the same problem from the elimination section using substitution. Plugging this into either of the equations and solving for yields .The third method for solving a system of linear equations is to graph them in the plane and observe where they intersect.There is a whole field of mathematics devoted to the study of linear equations called linear algebra.Convenient systems usually seem very tough to solve at first.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. 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. 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?

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