Systems of Three Equations | MathHelp.com - By MathHelp.com
00:0-1 | in this example notice that each of our variables X | |
00:04 | , Y and Z appears in all three equations to | |
00:09 | solve this system . We use the addition method . | |
00:14 | In other words , let's start with our first two | |
00:16 | equations , X plus Y plus Z equals four , | |
00:21 | an x minus Y plus Z equals two . Notice | |
00:26 | that if we add these equations together , the plus | |
00:29 | Y and minus Y will cancel out . And we | |
00:33 | have two , X plus two , Z equals six | |
00:37 | . So in our new equation two X plus two | |
00:41 | Z equals six . We've eliminated the variable . Why | |
00:46 | ? Unfortunately we still haven't solved for any of our | |
00:50 | variables . However , if we can create another equation | |
00:55 | with just X and Z in it , then we'll | |
00:58 | have a system of equations in two variables , which | |
01:02 | we can use to solve for X and Z to | |
01:06 | create another equation with just X and Z in it | |
01:10 | , we need to eliminate why we can't add the | |
01:14 | 1st and 2nd equations together because we've already done that | |
01:19 | . However , notice that if we add the 1st | |
01:22 | and 3rd equations together , the first equation has a | |
01:26 | plus Y And the 3rd equation has a -Y . | |
01:30 | So we'll be able to eliminate the why . So | |
01:34 | we have our first equation , X plus Y plus | |
01:37 | Z equals four . And our 3rd equation x minus | |
01:42 | y minus Z equals zero . And adding them together | |
01:47 | . Notice that the plus y minus Y cancels out | |
01:51 | and as a bonus the plus z minus Z also | |
01:55 | cancels out . So we have two . x equals | |
01:59 | four And dividing both sides by two , x equals | |
02:04 | two . Now , since we know that X equals | |
02:08 | two , notice that if we plug a two in | |
02:11 | for X in the equation that we created earlier , | |
02:14 | we can solve for Z and we have two times | |
02:19 | two plus two , Z equals six or four plus | |
02:24 | two , Z equals six . And subtracting four from | |
02:27 | both sides , we have two . Z equals two | |
02:32 | and dividing both sides by two , Z equals one | |
02:36 | , So x equals two , z equals one . | |
02:40 | And to find the value of why we simply plug | |
02:43 | our values of X and Z into any of the | |
02:47 | equations in the original system . Let's use the first | |
02:51 | equation , X plus Y plus Z equals four , | |
02:57 | since X equals two and Z equals one . We | |
03:00 | plug a two in for X and a one in | |
03:03 | for Z . And we have two plus Y plus | |
03:08 | one equals four or three plus Y equals four . | |
03:13 | And subtracting three from both sides , Y equals one | |
03:18 | , So x equals two , y equals one and | |
03:22 | z equals one . And finally we write our answer | |
03:26 | as the ordered triple , X , Y , Z | |
03:30 | Or 2 . 1 , 1 . |
DESCRIPTION:
Here weâre asked to graph the following function and use the horizontal line test to determine if it has an inverse. And if so, find the inverse function and graph it. So letâs start by graphing the given function, f(x) = 2x â 4, and remember that f(x) is the same as y, so we can rewrite the function as y = 2x â 4. Now, we simply graph the line y = 2x â 4, which has a y-intercept of -4, and a slope of 2, or 2/1, so we go up 2 and over 1, plot a second point and graph our line, which weâll call f(x). Next, weâre asked to use the horizontal line test to determine if the function has an inverse. Since thereâs no way to draw a horizontal line that intersects more than one point on the function, the function does have an inverse. So we need to find the inverse and graph it. To find the inverse, we switch the x and the y in original function, y = 2x â 4, to get x = 2y â 4. Next, we solve for y, so we add 4 to both sides to get x + 4 = 2y, and divide both sides by 2 to get 1/2x + 2 = y. Next, letâs flip our equation so that y is on the left side, and we have y = 1/2x + 2. Finally, we replace y with the notation that we use for the inverse function of f, as shown here. And remember that weâre asked to graph the inverse as well, so we graph y = ½ x + 2. Our y-intercept is positive 2, and our slope is ½, so we go up one and over 2, plot a second point, graph the line, and label it as the inverse function of f. Notice that the graph of the inverse function is a reflection of the original function in the line y = x.
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