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find the indicated diffences?
(this is what confuses me I have a 40 problem worksheet and im typing out the hard ones so mabey I can get some help)

(2a-3b)-(3a+5b)-a-8b


(12z-6)-(-3z+7)


(and then I'll have some with exponeents that are erally aggravating like this one)

(6z(exponent is 2)+3)-(5z(exponent of two)-2z-6)

(and the ones I really hate is)
(x-c)-(2x-4c)

2006-09-25 16:40:03 · 6 answers · asked by Z ten 3 in Science & Mathematics Mathematics

6 answers

(2a-3b) - (3a+5b) - a - 8b
=2a-3b-3a-5b-a-8b
=-a-8b-a-8b
=-2a - 16b

(12z-6)-(-3z+7)
=12z - 6 +3z-7
=15z - 13

(6z^2+3) - (5z^2-2z-6)
=6z^2 + 3 - 5z^2 + 2z +6
=z^2 + 2z + 9

(x-c)-(2x-4c)
=x-c - 2x + 4c
=-x + 3c

2006-09-25 17:13:07 · answer #1 · answered by Anonymous · 0 1

(2a - 3b) - (3a + 5b) - a - 8b
2a - 3b - 3a - 5b - a - 8b
(2 - 3 - 1)a + (-3 - 5 - 8)b
(-1 - 1)a + (-8 - 8)b
(-1 + (-1))a + (-8 + (-8))b
-2a - 16b

(12z - 6) - (-3z + 7)
12z - 6 + 3z - 7
(12 + 3)z + (-6 - 7)
15z + (-6 + (-7))
15z - 13

(6z^2 + 3) - (5z^2 - 2z - 6)
6z^2 + 3 - 5z^2 + 2z + 6
(6 - 5)z^2 + 2z + (3 + 6)
z^2 + 2z + 9

(x - c) - (2x - 4c)
x - c - 2x + 4c
(1 - 2)x + (-1 + 4)c
-x + 3c

2006-09-26 00:46:25 · answer #2 · answered by Sherman81 6 · 0 0

It's just a matter of grouping the like terms and remembering to watch the plus and minus signs:

(2a-3b)-(3a + 5b) -a -8b = (2a MINUS 3a - a) + NEGATIVE 3b MINUS 5b - 8b

(2a - 3a -a) + (-3b - 5b - 8b) = NEGATIVE 2a + NEGATIVE 16b or -2a - 16b

It's the same for all the rest. A term with an exponent is just another type of term to group:

(6z^2 + 3) - (5z^2 -2z - 6) = (6z^2 MINUS 5z^2) - (NEGATIVE 2z) + (3 MINUS NEGATIVE 6)

(6z^2 - 5z^2) - (-2z) + (3 + 6) = z^2 + 2z + 9

The only tricky part to these is the plus and minus signs.

2006-09-26 00:01:58 · answer #3 · answered by toby t 2 · 0 1

(2a-3b) - (3a+5b) - a - 8b =
2a-3b-3a-5b-a-8b =
-a-8b-a-8b =
-2*a - 16*b

(12z-6)-(-3z+7) =
12z - 6 +3z-7 =
15*z - 13

(6z(exponent is 2)+3)-(5z(exponent of two)-2z-6) =
(6z^2+3) - (5z^2-2z-6) =
6z^2 + 3 - 5z^2 + 2z +6 =
z^2 + 2*z + 9

(x-c)-(2x-4c) =
x-c - 2x + 4c =
-x + 3*c

2006-09-26 00:03:29 · answer #4 · answered by Joe C 3 · 0 1

you only need to group the like terms properly
like in this one
(2a-3b)-(3a+5b)-a-8b
remove the brackets

=2a-3b-3a-5b-a-8b

grouping like terms
=(2a-3a-a)-3b-5b-8b
= -2a-16b


it applies to the same in case of problems with exponents.
exponent is given by symbol-^. if exponent for the whole no then
((6z)^2+3)-((5z)^2-2z-6)
=36(z^2)+3-25z^2+2z+6
=36(z^2)-25(z^2)+2z+6
=11z^2+2z+9

if exponent is for z term onlythen it is-
6(z^2)+3-(5(z^2)-2z-6)
=6(z^2)-5(z^2)+2z+3+6
=z^2+2z+9


x-c-(2x-4c)
=x-c-2x+4c)
= -x+3c

2006-09-26 01:25:25 · answer #5 · answered by KSA 3 · 0 0

no help to satanists.

2006-09-26 01:16:38 · answer #6 · answered by kevin 2 · 0 1

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