Can someone explain how to do this problem?

Question

I have always hated math, and now I am stuck with Algebra..

Here is a sample problem that I have to do and I am LOST!!

12+3(x-4)-21=5[5-4(4-x)]

If you can expalin this so that a stupid person can understand then I would be grateful!

Answer 1

12+3(x-4)-21=5[5-4(4-x)]

first multiply the 5 out side the bracket by the stuff in the bracket so

5*5 and 5*-4(4-x)

12+3(x-4)-21=25-20(4-x)

now take the #'s out side the perenticies and multiply the stuff inside so

3*x  and  3*-4   and   -20*4  and -20*-x

12+3x-12-21=25-80+20x

    12-12-21      25-80

3x-21=-55+20x    now add 55 to both sides and subtract 3x from both

-3x     -3x

+55    +55

34=17x

34/17=x

2=x

Answer 2

Everything in front of a parenthesis get distributed to both terms

so we have

12 + 3*x - 3*4 - 21 = 5[ 5 - 4*4 + 4*x]

12 - 12 - 21 + 3x = 5[ 5 - 16 + 4x]

-21 + 3x = 5 (-11 + 4x)

-21 + 3x = -55 + 20x

-21 +55 = 20x - 3x

34 = 17x

x = 2

Answer 3

i hope this will be of some...help...i'll do the sum first and then i'll try to explain...if you do not get any point you can mail me...so here it goes...

12+3(x-4)-21=5{5-4(4-x)}

Break the bracket and mulitply the whole thing with (x-4); on the other side of the equation multiply break the other bracket and multiply 4 by (4-x) to get:

12+3x-12-21=5{5-(16-4x)}

Remove the first bracket on the right hand side of the equation and change the sign and u acquire this:

12+3x-12-21=5(5-16+4x)

Break the last bracket and multiply the whole thing with 5(no sign changes this time as there are no negative's in front of the bracket

12+3x-12-21=25-80+20x

Rearranging both sides(taking the x's on one side and the integers on the right side) of the equation:

-3x+20x = 12-12-21-25+80

Subtracting and adding on both sides you get:

17x=34

Therefore by dividing 34 by 17:

x=2

Answer 4

Lets start by combing terms on each side of equation

on left we have 21  + 3(x-4)

or -9 +3(x-4)

diistribute and we have -9 +3x+12

or 3 + 3x

on right side we need to distribute twice

starting inside 5[5-(16-4x)]

5[5-16+4x]

or 5[-11+4x] = -55+20x

so 3+3x=-55+20x

add 55 to both sides

58+3x=20x

subtract 3x

58=17x

x=58/17

Answer 5

12 + 3(x - 4) - 21 = 5[5 - 4(4 - x) ],

=> 12 + 3x - 12 - 21 = 5[5 - 16 + 4x) ],

=> 12 + 3x - 12 - 21 = 25 - 80 + 20x,

=> 12 + 3x - 33 = 25 - 80 + 20x,

=> 0 = 25 - 80 + 20x - 3x + 33 - 12

=> 0 = -55 + 17x + 21

=> 0 = 25 - 80 + 20x - 3x + 33 - 12

=> 0 = - 34 + 17x

=> 34 = 17x

=> x = 2...

Just simply the ones in the parenthesis and then multiply each of them with the signs and you'll get the answer....

Answer 6

12+3x-12-21 = 5[5-16+4x]

Distribute the parenthesis [3(x-4) = 3 times x minus 3 times four]

Distribute parenthesis before the brackets.

12+3x-33 = 25-80+20x

Simplify? (dont forget 12-33)

-21+3x = -55+20x

Subtract 3x from BOTH sides

-21 = -55+17x

Add 21 to both sides since its a negative

0=-34+17x

Just to keep it easier, add 24 to both sides

34=17x

Divide 17 from both sides

34/17 = 2

X=2

Check your work.

Answer 7

The first thing to do is to expand the innermost expressions within the parentheses:

12+3x-12-21=5[5-16+4x]

12+3x-12-21=25-80+20x

Now combine like terms and simplify

3x-21=-55+20x

3x-20x=-55+21

-17x=-34

x = 2

Answer 8

12+3(x-4)-21=5[5-4(4-x)]

12 + 3x - 12 - 21= 5 ( 5-16+4x)

12 + 3x -33   = 5 (-11 + 4x)

-21 + 3x = -55 +20x

3x - 20x = - 55 + 21

- 17x = - 34

x = 34 /17

x = 2

i hope you understand it now

Answer 9

12+3(x-4)-21=5[5-4(4-x)]

12+3x-12-21=5[5-16+4x]

12+3x-33=25-80+20x

-21+3x=-55+20x

3x-20x=21-55

-17x=-34

17x=34

x=34:17

x=2