# finding x

• December 14th 2010, 07:09 AM
finding x
2(3x + 4) = 4x + 16 find x

2y = 1/2(3y + 5) find y

d + 2 = 1/2(4d - 6) find d

4q - 2 = 2(3q - 6) find q

Can someone offer help?
• December 14th 2010, 07:10 AM
e^(i*pi)
The Distributive law is used here. Can you expand the brackets?
• December 14th 2010, 07:14 AM
Expand the brackets?
• December 14th 2010, 05:54 PM
Prove It
You should know that $\displaystyle a(b + c) = ab + ac$. See how what's on the outside has been multiplied by everything on the inside?

If you do that then that simplifies the problems greatly.

This is called "expanding the brackets".
• December 14th 2010, 06:01 PM
bigwave
Quote:

2(3x + 4) = 4x + 16 find x

as mentioned your first step should look like this

$6x + 8 = 4x +16$

it easy from here

expanding the brackets means distribution
• December 14th 2010, 06:07 PM
pickslides
If the distributive law is not desired then as a first step for each

Q1. divide both sides by 2

Q2. times both sides by 2

Q3. times both sides by 2

Q1. divide both sides by 2
• December 14th 2010, 06:08 PM
Prove It
Quote:

Originally Posted by pickslides
If the distributive law is not desired then as a first step for each

Q1. divide both sides by 2

Q2. times both sides by 2

Q3. times both sides by 2

Q1. divide both sides by 2

Even after doing this, you will still need to expand some brackets...
• December 14th 2010, 06:53 PM
arccos
Quote:

2(3x + 4) = 4x + 16 find x

The idea of these problems is to get all of the unknown(in this case x) on one side and the numbers on the other.
Firstly , expand all the brackets.
This can be done by multiplying 2(the factor) into the bracket.
$= 6x + 8$ (nothing changes on the right side)
$6x + 8 = 4x + 16$
Now, 'move' the x to the left side and the 8 to the right side.
Subtract 8 and 4x from both sides.
$6x +8 - 8 - 4x = 4x - 4x -8 + 16$
$2x = 8$
$x = \dfrac{8}{2} = 4.$

Solve the other problems in the similar fashion. Hope this helps.