Adding and Subtracting Polynomials | MathHelp.com - By MathHelp.com
00:0-1 | in this problem we are asked to add the given | |
00:02 | polynomial . Then we're asked to subtract the second polynomial | |
00:07 | from the first , in part A . To add | |
00:11 | the given polynomial , we simply add parentheses t squared | |
00:17 | plus 60 minus nine plus parentheses , t squared plus | |
00:23 | 70 minus three . Notice that I use parentheses around | |
00:28 | the polynomial . This is a good habit to get | |
00:32 | into , even though the parentheses will not affect the | |
00:36 | addition . Next we simply add the like terms T | |
00:42 | squared plus t squared is two , T squared 60 | |
00:47 | plus 70 is 13 T And -9 -3 is -12 | |
00:55 | . So we have to t squared plus 13 T | |
00:59 | minus 12 . In part B were asked to subtract | |
01:05 | the second polynomial from the first . So we have | |
01:09 | parentheses t squared plus 60 minus nine minus parentheses , | |
01:16 | T squared plus 70 minus three . Notice that the | |
01:22 | second polynomial is subtracted from the first and again notice | |
01:28 | that we use parentheses around each polynomial now , it's | |
01:34 | important to understand that the minus sign outside the second | |
01:38 | set of parentheses can be thought of as a negative | |
01:42 | one . So we need to distribute the negative one | |
01:47 | through each of the terms in the second set of | |
01:50 | parentheses . So after rewriting our first polynomial T squared | |
01:56 | plus 60 minus nine , we have negative one times | |
02:02 | t squared or negative T squared -1 times positive 70 | |
02:09 | , Which is negative 70 and negative one times negative | |
02:14 | three , Which is positive three . Now we combine | |
02:19 | like terms T squared minus t squared , cancels out | |
02:25 | positive 60 minus 70 is negative one , T . | |
02:29 | Or negative T and negative nine plus three is negative | |
02:34 | six . So we have negative T -6 . Make | |
02:40 | sure to distribute the negative one through the parentheses when | |
02:45 | subtracting the second polynomial from the first . |
DESCRIPTION:
In this example, notice that each of our variables, x, y, and z, appears in all three equations. To solve this system, we use the addition method. In other words, letâs start with our first two equations, x + y + z = 4, and x â y + z = 2. Notice that if we add these equations together, the +y and ây will cancel out, and we have 2x + 2z = 6. So, in our new equation, 2x + 2z = 6, weâve eliminated the variable y. Unfortunately, we still havenât solved for any of our variables. However, if we can create another equation with just x and z in it, then weâll have a system of equations in two variables, which we can use to solve for x and z. To create another equation with just x and z in it, we need to eliminate y. We canât add the first and second equations together, because weâve already done that. However, notice that if we add the first and third equations together, the first equation has a +y and the third equation has a ây, so weâll be able to eliminate the y. So we have our first equation, x + y + z = 4, and our third equation, x â y â z = 0, and adding them together, notice that the +y â y cancels out, and, as a bonus, the +z â z also cancels out, so we have 2x = 4, and dividing both sides by 2, x = 2. Now, since we know that x = 2, notice that if we plug a 2 in for x in the equation that we created earlier, we can solve for z. And we have 2(2) + 2z = 6, or 4 + 2z = 6, and subtracting 4 from both sides, we have 2z = 2, and dividing both sides by 2, z = 1. So x = 2 and z = 1, and to find the value of y, we simply plug our values of x and z into any of the equations in the original system. Letâs use the first equation, x + y + z = 4. Since x = 2 and z = 1, we plug a 2 in for x and a 1 in for z, and we have 2 + y + 1 = 4, or 3 + y = 4, and subtracting 3 from both sides, y = 1. So x = 2, y = 1, and z = 1, and finally, we write our answer as the ordered triple, x, y, z, or (2, 1, 1).
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