Algebra: Linear equations 4 | Linear equations | Algebra I | Khan Academy

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I also present “Linear Equations”, stage 4. Let’s start with a few tasks. Yes. Let’s say I have the following situation: 3 / x = 5. This task is a little different of those we have seen so far. Because here, instead of x being in the numerator, he is in the denominator. I personally prefer x not to be in my denominator, so I want to move it from the denominator to the numerator or at least not leave it in the denominator as quickly as possible. One way to do this is by multiplying both sides of the equation by x and so on the left side the two x will be shortened. And 5x will remain on the right side. This is equal to … hicks are shortened. And we get 3 = 5x. We can also write it as 5x = 3. We can think about this in two ways. Or we just multiply both sides by 1/5, or divide by 5. If we multiply both sides by 1/5, the left side becomes x. Right side: 3 by 1/5 is 3/5. What did we do here? This has become a stage 2 task, or even from stage 1, very quickly. We just had to multiply both sides of this equation according to x. And we got rid of x in the denominator. Let’s solve another problem. Let us have (x + 2) on (x + 1) be equal to 7. Instead of having x in the denominator, we have (x + 1). But we will do the same. To remove this (x + 1) from the denominator, we multiply the sides of the equation by (x + 1) / 1. Because we did it on the left side, we have to do it on the right side as well. 7 by (x + 1). On the left side (x + 1) it is shortened. Only (x + 2) remains. It is on a unit, but we can ignore it! And that’s equal to 7 (x + 1). Thus, (x + 2), remember that it is 7 over the whole (x + 1). Ie we must use the distribution property. It is equal to 7x + 7. I think this has become a linear equation from stage 3. Now we just have to transfer the hicks to one side. And transfer the constants, such as 2 and 7, on the other side of the equation. I will put x on the left. We put 7x on the left. We subtract 7x from both sides. Minus 7x. They are destroyed on the right. On the left we have (-7x + x). Becomes (-6x + 2), and on the right remains 7. Now we just have to get rid of the couple, by subtracting 2 from both sides. Remains -6x = 5 This is a stage 1 task. We only multiply both sides by the reciprocal of the coefficient on the left. The coefficient is -6. Multiply both sides by -1/6. -1/6. Left side: -1/6 to -6. This is equal to 1. It remains for us that x is equal to 5 over -1/6. This is – 5/6. We’re ready. If we want to check it, we just take this x = -5/6 and replace it in the initial equation, to make sure it’s correct. Let’s do another one. I’m making them up right now, excuse me. Let me think about it. 3 / (x + 5) = 8 (x + 2). We do exactly the same thing although we now have two expressions that we want to remove from the denominators. We want these (x + 5) and (x + 2) to disappear. Let’s start with (x + 5). As before, we will multiply both sides of the equation for (x + 5). You can say ‘by (x + 5) / 1’. By (x + 5) / 1. On the left side they are shortened. We are left with 3 = 8 (x + 5) And this whole on (x + 2). On top, let me say again, we just multiply by 8 the whole expression. This is (8x + 40) / (x + 2). We want to remove this (x + 2). Maybe in the same way. Multiply both sides of the equation by (x + 2) / 1. x + 2 We can say that we multiply both sides by (x + 2). This 1 is a bit unnecessary. On the left it becomes 3x + 6. We always multiply the whole expression by 3, because otherwise the equation changes. 3 by (x + 2). And on the right. These (x + 2) will be shortened. Only 8x + 40 remains. This is a stage 3 task. If we subtract 8x from both sides, minus 8x, I’m running out of space. Minus 8x. They are shortened on the right. On the left we have -5x + 6, only 40 remained on the right. Now we just subtract 6 from both sides of the equation. I will write it here. Minus 6. I hope I don’t confuse you, by moving here. If we subtract 6 from both sides, on the left we get -5x, and on the right 34. This is a stage 1 task. Multiply both sides by -1/5. By -1/5. On the left we have x. On the right we have -34/5. If I didn’t make a stupid mistake, I think it’s true. And if he understands what we did here, you can now deal with linear equations from stage 4! Have fun!

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