Algebra

Solving systems of equations by elimination | Algebra Basics | Khan Academy

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Let’s look at a few more methods for solving systems of equations. Let’s say I have the equation 3x plus 4y is equal to 2.5. And I have another equation 5x minus 4y is equal to 25.5. We want to find the values \u200b\u200bof x and y that satisfy both of these equations. If you think about it graphically, it will be the intersection of the lines representing the solutions of both of the equations. So how can we do that? We saw that in the replacement, we want to remove one of the variables. We did it by substituting last time. But is there anything we can add or subtract – let’s focus of this yellow, of this above equation here – is there something we can add or subtract from both sides of this equation? Remember that every time you work with an equation, you have to add or subtract the same thing for both sides. But is there anything we could add or subtract from both sides of this equation, which can remove one of the variables? And then we will have an equation with one variable and we will be able to solve it. And this is probably not obvious, although it is true before your eyes. What if we just add this equation to this equation? What I mean is: what happens if we add 5x minus 4y to the left and add 25.5 to the right? If I had to literally add this to the left, to add this to the right side. And you’ll probably say, Sal, wait, how can you just to add two equations this way? Remember that when you do some equation, if I have some equation with the form – well, really any equation – Ah plus Wu is equal to C if I want to do something with it equation, you just have to add the same thing to both sides of the equation. So I could for example, I can add D to both sides of the equation. Since D is equal to D, I will not changes the equation. You will get Ax plus Wu, plus D is equal to C plus D. We have seen this many, many times. Everything you do with one side of the equation, you have to do it and to the other side. But you say, hey, Sal, wait, on the left side, you add 5x minus 4y to the equation. On the right side you add 25.5 to the equation. Don’t you add two different things to both sides of the equation? And my answer will be no. We know that 5x minus 4y is 25.5. This quantity and this quantity are the same. They are both 25.5. This second equation tells me explicitly. So, I can add this to the left. I’m essentially adding 25.5 to it. And I can add 25.5 to the right side. So, let’s do this. If we add the left side, 3x plus 5x is 8x. And then, what is 4y minus 4y? And that’s the whole point. When I looked at these two equations, I said, oh, I have 4y, I also have minus 4 y. If you just add these two together, they will are canceled. They will be plus 0y. Or this whole article will just disappear. And that’s going to be equal to 2.5 plus 25.5 is 28. So, you separate both sides. You get 8x is equal to 28. Divide the two sides by 8 and we get x is equal to 28 on 8 or divide the numerator and denominator of 4. This is equal to 7 over 2. This is our value of x. Now we want to find our value of y. We can replace this back in each one of these two equations. Let’s use the above. You can do the same with the bottom. So, we know that 3 over x, 3 over 7 over 2 – just I replace the value of x that we found above equation – 3 over 7 over 2, plus 4y is equal to 2.5. Let me just write this as 5/2. We will remain in the world of fractions. This will be 21 over 2 plus 4y equals 5/2. We subtract 21 over 2 on both sides. So, minus 21 over 2, minus 21 over 2. Left side – just stay with 4y because these two boys are canceled – is equal to – that’s 5 minus 21 over 2. Minus 16 over 2. That’s minus 16 over 2, which is the same as – ok i will write it as minus 16 over 2. Or we can write this as – let’s continue here above – 4y – just keep going in that line of thinking here – 4y is equal at minus 8. Divide the two sides by 4, and you get y is equal to minus 2. So the solution of this equation is x is equal to 7/2, y is equal to minus 2. These will be the coordinates of their intersection. And you can try them in both of these equations here. Let’s check that out also satisfies the following equation. 5 over 7/2 is 35 over 2, minus 4 over minus 2, so minus – 8. This equates to – let’s see, this is 17.5 plus 8. And that’s really equal to 25.5. So, this satisfies both equations. Now let’s see if we can use our newly discovered skills for coping with a verbal task, our newly discovered skills in elimination. It is said here that Nadia and Peter visit a pastry shop. Nadia buys 3 chocolate desserts and 4 fruit rolls for $ 2.84. Peter also buys 3 chocolate desserts, but can only afford 1 additional fruit roll. It costs $ 1.79. What is the price of each chocolate dessert and each fruit roll? Let’s define some variables. Let’s just use x and y. To take x is equal to the price of the chocolate dessert – I would put c and f for the fruit roll, but I will stay with x and y – price for chocolate dessert. And let y be equal to the price of a fruit roll. Okay. So what does this first statement tell us? Nadia buys 3 chocolate desserts, so the price of 3 desserts will be 3x. Three times the price of chocolate dessert. And 4 fruit rolls. Plus 4 per y, the price of a fruit roll. That’s how much Nadia spent. 3 chocolate desserts, 4 fruit rolls. And it will cost $ 2.84. This is what the first statement tells us. It becomes this equation. The second statement. Peter also buys 3 chocolate desserts, but can only afford 1 additional fruit roll. Plus 1 extra fruit roll. The value of his purchase is $ 1.79. What is the price of each chocolate dessert and each fruit roll? We will solve this with the help of elimination. You can solve this using any of the techniques we have seen so far – replacement, elimination, even graphics, although it’s a little hard to see things with drawing. How can we do this? Remember, with elimination you will add – let let’s focus on this above equation here. Is there anything we can add to both sides of this an equation that will help us to remove one of the variables? Or let me put it this way: is there anything we can add or subtract from both sides of this equation that will help us to eliminate one of the variables? As with the task we did a little earlier in the video, what will happen if we subtract this equation or what if we subtract 3x plus y from 3x plus 4y from left side and subtract $ 1.79 from the right side? And remember that in doing so, I will pull out the same thing something on both sides of the equation. That’s $ 1.79. How do I know that? Because it’s said to be $ 1.79. So if we do that, we’re going to pull out the same thing on both sides of the equation. So, let’s subtract 3x plus y from the left side of the equation. And let me do it up on the right. If I subtract 3x plus y, it’s the same thing as minus 3x minus y if you just distribute the negative sign. So let’s get it out. You get minus 3x minus y – maybe you should it is good to make it clear that this is not a plus sign; you can imagine that multiply the second equation by minus 1 – e equal to minus $ 1.79. I just take the second equation. You can imagine multiplying it by minus 1 and will now add to the left side, the left side of this equation and the right side to the right side of this equation. And what will we get? When you add 3x plus 4y, minus 3x, minus y, 3x are canceled. 3x minus 3x is 0x. I won’t even write it. You get 4x minus – sorry, 4y minus y. This is 3y. And that’s going to be $ 2.84 minus $ 1.79. How much is that? That’s $ 1.05. So, 3y equals $ 1.05. We divide the two sides into 3. y is equal to – how much is $ 1.05 divided by 3? 3 is contained in $ 1.05 … Contained in 1 zero times. 0 over 3 is 0. 1 minus 0 is 1. Download down 0. 3 is contained in 10 three times. 3 by 3 is 9. We take it out. 10 minus 9 is 1. We download 5 below. 3 is contained in 15 five times. 5 by 3 is 15. We take it out. We have no leftovers. So, y is equal to $ 0.35. The price of the fruit roll is $ 0.35. We can now substitute back into any of these equations, to find the price of the chocolate dessert. So, let’s use the equation below here. Which was originally, if you remember before I multiplied it minus 1, it was 3x plus y equals $ 1.79. This means that 3x plus the price of a fruit roll, 0.35 is equal to $ 1.79. If we subtract 0.35 from both sides, what will we get? On the left side – just stay 3x; these are canceled – is equal to – let’s see this is $ 1.79 minus $ 0.35. That’s $ 1.44. And 3 is contained in $ 1.44, I think it is contained – well, 3 is contained in $ 1.44, goes into 1 zero times. 1 by 3 is 0. Download below 1. We take it out. Download below 4. 3 enters 14 four times. 4 by 3 is 12. I make it messy. 14 minus 12 is 2. Download below 4. 3 is contained in 24 eight times. 8 by 3 is 24. There is no residue. So, x is equal to 0.48. So, we already have it. We found using elimination that the cost of chocolate dessert is equal to $ 0.48 and that the price of fruit roll is equal to $ 0.35.

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