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here are the problems
(2x-3y)+(5x+4y)
a-(2a+b)+2(a-2b)
3(5x+4)-(3x-5)
(8x-3y)-(4x+3y-z)+2z
i already did them but im not sure i did it right!
step by step instructions please!!

2007-11-27 16:55:15 · 6 answers · asked by Aims 3 in Education & Reference Homework Help

also with (8a-7b)-(4a-5b)

2007-11-27 17:17:11 · update #1

6 answers

The first problem as I read it is: (2x - 3y) + (5x + 4y)

Step 1. Add the variables in X: 2X + 5X = 7 X

Step 2. Add the variables in Y: -3Y + 4Y= Y

Answer: 7X + Y


The second problem as I read it is: a -(2a +b) + 2(a -2b)

Step 1. perform multiplication on the variables inside the parentheses. The result is: a -2a -b + 2a -4b.

Step 2 add the terms: the result is: a - 5b


The third problem as I read it is::

3(5X+4) -(3X -5)

Step 1 multiply: 15X + 12 -3X +5

Step 2 combine terms: 15X -3X = 12X and 12 +5 =17

The result is: 12X +17

The fourth problem as I read it is: (8X - 3Y) -(4X +3Y -Z) +2Z

Step 1 multiply: 8X -3Y -4X -3Y + Z +2Z

Step 2 combine terms: 8X -4X = 4X, -3Y -3Y = -6Y, Z+2Z=3Z

The result is: 4X -6Y +3Z


(edit) the last one as I read it is (8A -7B) - (4A -5B)

Step 1 multiply: Result: 8A -7B -4A +5B

Step 2 combine terms: A -4A = 4A, -7B +5B = -2B

The result is: 4A -2B

2007-11-27 18:36:15 · answer #1 · answered by Anonymous · 0 1

You haven't actually defined the question so I'll just assume its to simplify them ;)

(2x - 3y) + (5x + 4y)
Ok here we have 2 "types" (the x's and the y's) of terms, group them seperately and add.
2x + 5x = 7x
-3y + 4y = y
7x + y

a-(2a+b)+2(a-2b)
a little trickier. remember bedmas.
brackets come first, with multiplication, so expand the brackets
2(a-2b) = 2a - 4b

so a-(2a+b)+2(a-2b) = a - 2a - b + 2a - 4b (we can remove all brackets now)

-a + b + 2a - 4b = a + b - 4b = a - 5b

3(5x+4)-(3x-5), same sort of problem , deal with the brackets first.

3(5x+4)-(3x-5) = 15x + 12 - (3x - 5), one slight catch here, when we remove the brackets on "(3x + 5)" we are subtracting, so gotta make those negatives positive.

15x + 12 - (3x - 5) = 15x + 12 - 3x - 5 = 12x + 17

(8x-3y)-(4x+3y-z)+2z = 8x - 3y - 4x - 3y + z + 2z

remember just like making postives negative when subtracting we also make negatives positive (bet your head is positively spinning right now) so - (-z) = + z.

8x - 3y - 4x - 3y + z + 2z = 4x - 6y + 3z

hope that helps a little.

2007-11-27 17:18:53 · answer #2 · answered by pelirojo60 1 · 0 0

1. (2x-3y)+ (5x+4y) = 2x-3y+ 5x+4y =
7x+1y

2. a-(2a+b)+2(a-2b) = a-2a+b +2*(a-2b)=
-a+b+2a-4b= a-3b

3.3*(5x+4)-(3x-5) = 15x +12 - 3x+5=
12x + 17

4. (8x-3y) -*(4x+3y-z)+2z = 8x-3y-4x-3y+z+2z=
4x-6y+3z

(8a-7b) -*(4a-5b)= 8a-7b-4a+5b= 4a-2b

well, its hard to show steps but i hope u understand

2007-11-27 17:15:17 · answer #3 · answered by moon 3 · 0 0

1. (2x-3y)+(5x+4y)
= 2x-3y+5x+4y (not much to explain on this one, but since you're adding, the signs remain the same)
= 7x+y

2. a-(2a+b)+2(a-2b)
= a+(2a*[-1])+(b*[-1])+(2*a)-(2*2b) (distribute the -[which represents -1] and the 2)
= a-2a-b+2a+4b
= a-2a+2a-b-4b (reorganize them so it's easier to see)
= a-5b

3. 3(5x+4)-(3x-5)
= (3*5x)+(3*4)+(3x*[-1])+([-5]*[-3]) (distribute like in #2)
= 15x+12-3x+5
= 15x-3x+12+5 (group similar terms so it's easier to see)
= 12x+17

4. (8x-3y)-(4x+3y-z)+2z
= 8x-3y+(4x*[-1])+(3y*[-1])+([-z]*[-1])+2z (distribute)
= 8x-3y-4x-3y+z+2z
= 8x-4x-3y-3y+z+2z (group similar terms)
= 4x-6y+3z

2007-11-27 17:08:28 · answer #4 · answered by PhuKi 6 · 0 0

(2x - 3y) + (5x +4y)
combine like signs yields 7x + y
and that's as far as we can take that one
******
now let's mutliply thru both parentheses
whooops is that -a(2a.....
or a - (2........ makes a big difference

if it's the 2nd then we only have to multiply the second PP
a-2a -b +2a -4b...leaves us
a-5b

**********

2007-11-27 17:08:19 · answer #5 · answered by tom4bucs 7 · 0 0

1. =2x-3y+5x+4y
=(2x+5x)+(-3y+4y)
=7x+y

2. =a-2a-b+2a-4b
=(a-2a+2a) + (-b-4b)
=a-5b

3. =15x+12-3x+5
=(15x-3x)+(12+5)
=12x+17

4. =8x-3y-4x-3y+z+2z
=(8x-4x)+(-3y-3y)+(z+2z)
=4x-6y+3z

2007-11-27 17:07:59 · answer #6 · answered by Anonymous · 0 0

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