a. 5y + 20 =
b. 7x - 28 =
c. 8a + 10b - 18c =
d. 9a + 27b + 81 =
I'm not sure how to set it up
2007-09-25 03:42:20 · 6 answers · asked by Tiffany S 1 in Science & Mathematics ➔ Mathematics
a. 5y + 20 = 5(y + 4)
b. 7x - 28 = 7(x - 4)
c. 8a + 10b - 18c = 2(4a + 5b - 9c)
d. 9a + 27b + 81 = 9(a + 3b + 9)
2007-09-25 03:49:33 · answer #1 · answered by N E 7 · 0⤊ 1⤋
Use the distributive property backward to factor. Find something that you can take out of every term.
a. 5 goes evenly into both 5 and 20, so you can take a 5 out. When you take 5 out of 5y, you only have y left. When you take 5 out of 20, you have 4 left.
5y + 20 = 5(y + 4)
b. 7 goes into both 7 and 28: 7(x-4)
c. 2 will go into all three terms here.
2(4a + 10b - 18c)
d. 9 will go into all three terms.
9(a + 3b + 9)
2007-09-25 11:00:25 · answer #2 · answered by Anonymous · 0⤊ 0⤋
a. 5y + 20 =
Common factor is 5.
5 (y + 4)
b. 7x - 28 =
Common factor is 7
7 (x - 4)
c. 8a + 10b - 18c =
Common factor is 2
2 (4a + 5b - 9c)
d. 9a + 27b + 81 =
Common factor is 9
9 (a + 3b + 9)
2007-09-25 10:53:11 · answer #3 · answered by edith p 3 · 0⤊ 1⤋
5y+20=5(y+4)
7x-28=7(x-4)
8a+10b-18c=2(4a+5b-9c)
9a+27b+81=9(a+3b+9)
2007-09-25 10:48:41 · answer #4 · answered by Grampedo 7 · 0⤊ 1⤋
Sure. Remove the common factor in each expression. For example
5y + 20 = 5(y + 4)
Doug
2007-09-25 10:51:59 · answer #5 · answered by doug_donaghue 7 · 0⤊ 1⤋
5y+20
therefore 5(y+4)=0;
y+4=0
therefore y= -4.
2007-09-25 10:52:15 · answer #6 · answered by Anonymous · 1⤊ 2⤋