3(x+3y)+4(2x-4y)
distribute : 3x+9y+8x-16y
combine like terms: 11x-7y
answer: 11x-7y
2006-12-26 04:25:55
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answer #1
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answered by Anonymous
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Q : 3(x+3y) +4(2x-4y)
A :step 1; multiply
3x+9y+8x-16y
step2; add the coeffecients of the same variables
i.e., 3x+8x+9y-16y
11x -7y
3(x+3y)+4(2x-4y) = 11x-7y is the required answer.
2006-12-26 05:27:57
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answer #2
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answered by Anonymous
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First: use distribution:
3(x) + 3(3y) + 4(2x) - 4(4y)
3x + 9y + 8x - 16y
Second: combine "like" terms:
3x + 8x + 9y - 16y
11x - 7y
2006-12-26 12:53:00
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answer #3
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answered by ♪♥Annie♥♪ 6
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Your first step would be to distribute the 3 and the 4 over their respective set of brackets.
3x + 9y + 8x - 16y
Now, group like terms.
11x - 7y
2006-12-26 04:29:51
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answer #4
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answered by Puggy 7
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Distribute the first terms:
3x+9y+8x-16y
Simplify:
11x-7y
2006-12-26 04:27:25
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answer #5
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answered by John C 1
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3(x + 3y) + 4(2x - 4y) = 0
3x + 9y + 8x - 16y = 0
11x - 7y = 0
The answer is 11x - 7y = 0
- - - - - - - - -s-
2006-12-26 05:46:58
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answer #6
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answered by SAMUEL D 7
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=3x +9y + 8x -16 y = 11x-7y
2006-12-26 04:26:21
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answer #7
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answered by Frederic R 3
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a million.. 3x^2 + 4x + 3x^2 - 4y = 6x^2 + 4( x-y) = 2 [ 3x^2 + 2 (x-y) 2.. (5x-2y) - (4x+8y) = 5x -2y - 4x - 8y = x -10y 3. 3x^3(2x^2) = 6x^6 4. 5x^3y^4z/10xy^2z^3 = x^2y^2/ 2z^2 5.(-6a^3b^4)^2= 36a^6b^8 6.(x+9)(x-9)= (x^2-80 one) 7.(3x-7)^2= 9x^2 - 42x + 40 9= (3x-7)(3x-7) 8.(3x+6)(4x-7)=12x^2 -21x +24x -40 two= 12x^2 +3x -40 two =3(4x^2+x -14) 9. 5x^3y + xy^4 +2xy/xy =xy(5x^2 +y^3 +2)/ xy =5x^2 +y^3 +2 10..15ab +6ab^4 +24b/3b =3b(5a + 2ab^3 +8)/3b =5a+2ab^3 +8= a(5+2b^3)+8 whats up there!! it quite is my answer yet you should use the single that extra wholesome you and stay away from something
2016-11-23 17:57:02
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answer #8
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answered by Anonymous
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= 3x + 9y + 8x - 16 y = 3x + 8x + 9y - 16y = 11x -7y
2006-12-26 04:27:46
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answer #9
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answered by saudipta c 5
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3x+9y+8x-16y
11x-7y My final answer♥
2006-12-26 04:26:38
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answer #10
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answered by ♥USMCwife♥ 5
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