Removing brackets How do i remove them from this inequality?

Question

Like i said i dont know how to remove brackets. I took a fare shot and tryed on this problem: 2[3z-5(2z-3)] = 3z -4 How do i got about solving this. Like how do i handle those brackets i thought they were like parenthesis but thy are not i guess i keep on getting -10z = -21. so i guess im not handling the brackets right. Help please....

Answer 1

2 [ 3 z - 5 (2 z - 3) ] = 3 z - 4
2 [ 3 z - 10 z + 15 ] = 3 z - 4
2 [15 - 7 z ] = 3 z - 4
30 - 14 z = 3 z - 4
30 + 4 = 3z + 14 z
34 = 17 z
z = 2

Answer 2

To remove brackets, you must first solve for all numbers in parenthesis, starting with the inner most one. So:

2[3z-5(2z-3)]=3z-4

becomes

2[3z-10z+15]=3z-4

because you use the distributive property for "-5 multiplied by 2z" and then "-5 multiplied by -3". Remember, when there is a minus sign (or plus) before a number, you cannot separate them! you must use them together. Now you use the distributive property again and get:

6z-20z+30=3z-4

which simplifies to

-14z+30=3z-4

Because you distribute 2 throughout the parenthesis. Now just solve like an algebra equation by combining like terms. So lets add 14 to each side:

30=17z-4

Now isolate the variable by first adding 4 to each side:

34=17z

Now just solve for z by dividing both sides by 17

2=z

Just remember, brackets are just like parenthesis. As long as you solve for any parenthesis in them, starting with the the inner most, you should be fine.

example: [5-(4(3-1))] Solve 3-1, then multiply by 4, then subtract from 5.

I hope I helped you understand this. If not, someone else maybe able to clear this up.

Answer 3

The brakets are like parenthesis, but you just handle the brackets after you handle the parenthesis.

So first, handle the inner part of the equation:

-5(2z-3) = -10z - 15

Now put ^^^ into your equasion.

2[3z -10z -15] = 3z -4

2[-7z -15] = 3z - 4

-14z - 30 = 3z - 4

17z = 34

z = 2

You just forgot to distribute the 2 in your equasion

Answer 4

The rule is you work on the innermost parentheses/ brackets/ braces first, and then work your way outwards.
2[3z-5(2z-3)] = 3z -4

We remove the parentheses from 2z - 3 first by distributing -5 into the terms inside the parentheses. Note that -5(2z-3) is equal to -10z+15:
2[3z-10z+15] = 3z -4

Remove the [ ] this time:
6z - 20z + 30 = 3z - 4

Simplify:
-14z +30 = 3z - 4
-14z - 3z = -4 -30
-17z = -34
z = -34 / -17
z = 2

Answer 5

You always do the farthest-in parenthesis first. Brackets are just a farther-out parenthesis than the others.

2[3z-5(2z-3)]=3z-4

First, use distributive property for the inner-most paenthesis:

-5(2z-3)
-10z+15
2[3z-10z+15]=3z-4

Then use distributive property for the brackets.

2[3z-10z+15]=3z-4
6z-20z+30=3z-4

Then combine like terms

6z-20z+30=3z-4
minus 3z from the right
3z-20z+30=-4
minus 30 from left side
3z-20z=-34
combine z's
-17z=-34
divide by -17 on both sides
z=2

Re-work it by plugging in 2 for all the z's in the oringal problem.

*most likely, you would want to combine your variables last...in some cases. like this one.

Answer 6

it's not an inequality unless you wrote it wrong... what you have is
2[3z-5(2z-3)] = 3z -4
first, distribute the 5 in the ()
2[3z - 10z + 15] = 3z - 4
2[-7z + 15] = 3z - 4
now, distribute the 2 in the ()
-14z + 30 = 3z - 4
34 = 17z
2 = z

Answer 7

z=1.9
the brackets are parenthesis. the problem could be rewritten:
2(3z-5(2z-3))=3z-4
2(3z-10z+15)=3z-4
-14z+30=3z-4
34=17z
z=2

Answer 8

2[3z-5(2z-3)]=3z-4 -Multiply -5(2z-3)
2[3z-10z+15]=3z-4 - Add like terms
2[-7z+15]=3z-4 - Multiply -7z+15 by 2
-14z+30=3z-4 - Add 4 to both sides
-14z+34=3z - Add 14z to both sides
34=17z - Divide by 17
z=2