2007-01-18 11:35:18 · 5 answers · asked by Anonymous in Education & Reference ➔ Homework Help
1. -7a-9b+4a+3b. ok first you would have to common the numbers with the same coefficients. So, -7a and positive 4a have the same coefficients. So, combine those which is -7a+4a=3a. So far, the equation looks like 3a-9b+3b. Do the same by combine the same coefficients with the letter b. -9b+3b=-2b. What is left over is 3a-2b, which is your answer
2. -4(2y^2-3y+7). First you must distribute the -4 to each of the numbers. So, -4 times 2y^2 is -8y^2. -4 times -3y is 12y. -4 times 7 is -28. So, you put those together which makes -8y^2+12y-28 that is your answer.
3. 21x-14y+7/-7. First, you can see that the numbers 21, -14, and 7 can be easily divisible by -7. So, divide each of the numbers one by one. 21x divided by -7 is -3x. -14y divided by -7 equals 2. 7 divided by -7 is -1. Put all the left over numbers to get -3x+2y-1 which is your answer.
2007-01-18 11:46:28 · answer #1 · answered by Amanda 4 · 0⤊ 0⤋
1.-3(a+2b)
3. 7(3x-2y)-1
2.Sorry, I don't understand the ^ sign
More explanation on #3- I wasn't sure it if the 7/-7 or if the entire problem was divided by 7
2007-01-18 19:47:39 · answer #2 · answered by lou53053 5 · 0⤊ 0⤋
-7a-9b+4a+3b
-3a-6b
-4(2y^2-3y+7)
-8y^2-12y+28
21x-14y+7/-7
3x-2y+1
2007-01-18 19:38:42 · answer #3 · answered by Christine 2 · 0⤊ 0⤋
-3a-6b,-8y^2-6y+-28,21x-14y+1
im guesing this is rite im in honers algebra 1(in highschool) and even though i did this a month ago i dont think i did it right
2007-01-18 19:41:48 · answer #4 · answered by Britanie 3 · 0⤊ 1⤋
first one is -3a-6b
second is -8y^2+12y-28
third one is -3x+2y-1
2007-01-18 19:46:56 · answer #5 · answered by hina s 2 · 0⤊ 0⤋