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How To Solve Algebraic Equations - 'x' In One Place
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In the second equation times the X by -1 to make it
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How To Solve Algebraic Equations - 'x' In One Place
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Hi, I am Dr. Shah, I was the National Lector Competition winner in 1989, and I am the Math master at Mathscool. Now, ready for a new art in Math? I am going to show you how to solve algebraic equations. Now, there are two cases we need to consider. So Case 1: 'x' in one place. So if an equation has x in one place, we are going to use a method of a flow diagram, to help us rearranging it. So we will start with an example, 2(x-3)=12. 'x' is only in one place in this equation. So to use a flow diagram, flow diagram means just reading the order, the mathematical order of the equation. So looking at the 'x', the thing we see next to it is 'a subtract 3' [-3], then here are brackets, outside the brackets is a times 2 [X 2] and that, when the whole of that side of the equation is read, we get to the equal sign, equals 12[=12]. Now we are going to solve this equation, so basically, we want to unravel the 'x'. So what we are going to do is reverse the steps. So the opposites of times 2 is divided by 2, so we divide by 2, both sides, which gives x-3=6, and the times 2 and the divide by 2 cancel each other out, and now the opposite of minus 3 is add 3, so add 3 to both the sides, again the minus 3 and the plus 3 cancel each other out, and so this side we are left with only x and this side we get 9, and that is the answer, x=9. Okay. So, let's have another example, so using this Case 1, 'x' in one place. 3(2-x) =18. Slightly more tricky flow diagram this time, starting with the 'x'. You see here, in front of the x here is a negative sign, so to get that negative sign, we have to times it with minus 1. When you times it with minus 1 [-1], that means you are making the 'x' negative, and this is a plus 2 at the end here, so +2, this isn't x-2, it is 2-x, the flow diagram is slightly more complicated, times minus 1 makes the 'x' negative and add 2 it, to get +2, that is everything inside the brackets done, now, here is the times 3 out side the brackets, then way up to the go sign, and now you have to reverse the steps, times 3 is divided by 3 , and on this side times 3 and divided by 3 cancel each other out, just leaving 2-x, and 18 divided by 3 is 6, opposite of +2 is -2, and so -2 from both sides, again the +2 and -2 cancel each other out, so we are left with x = 4, and the opposite of times -1, is the same thing as times -1, times -1 changes the sign, and changes the sign again and you have reversed it. And so we have to times -1 both the sides, which leaves as, x=-4, so Case 1 dealt with 'x' in one place. Now, if you want to try more difficult equations, so Case 2: x in more than one place. So for these examples, you have to use several rules, the first rule is we want to get rid, if there is any fractions, so multiply by denominators to get rid of any fractions, then we want to multiply any brackets getting rid of the brackets in the equation, then we want to collect all the x terms to one side, and once we have done that, we should be able to solve it by rearranging it the the same way as we did before. So I am going to give you a different example here for Case 2. .
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