[zak07]

I am stuck on the equations with division
like this one 4x+4/2=3(x+2)-2
my brain locks up and I have no idea how to get rid of the fraction and I know that is what I should do first right??

some one please help me

Zak:(

[worthbeads]

Zak, first of all remember that 4/2 is 4 halves, or 2 (order of operations;parenthesis, then powers, then multiply and divide left to right, the add and subtract left to right), so there is one problem solved. So, now substitute...

4x+2=3(x+2)-2
Now use the distributive property. 3(x=2)= 3*x+3*2, which equals 3x+6

Now substitute again... 4x+2=3x+4 (6-2)

Now you subtract a number from each side

4x+2=3x+4
__ -2 _ -2 (pay no attention to the underscores)

4x=3x+2

and get rid of the variable from the right side by subtracting 3x from each side


4x=3x+2
-3x -3x

x=2

Now check

if x=2, then 4*2+2=10
AND 3*2+3*2-2=10 (keep in mind order of operations)

and 10=10

If you have any other troubles, ask your teacher.

[dmatos]

Normally, with a fraction in an algebra equation, you can multiply both sides of the equation by the denominator in the fraction. If 2x=1, then 4x=2. Similarly, if

(x+5)/3 = 2x

then

x+5 = 3 * 2x

If there is more than one denominator in the equation, just keep multiplying both sides by the denominators until there are none left.

The example that you gave is fairly trivial, as the denominator divides easily into the numerator of your fraction, leaving you with a whole number. There is no reason to multiply both sides, if you can just perform the division.

[cherrymerd]

I'm in algebra 2 and we just learned about the factor of z I need help because I don't understand it
x+y-2z=5
x+2y+z=8
2x+3y-z=1
what do I do

[dmatos]

A linear equation with three variables describes a plane in 3-D space, similar to how a linear equation with two variables describes a line in 2-D space. What is the question asking you to do with these equations?

Since you have three equations, it may be asking you to solve for x, y, and z. You have three equations, and three unknowns. For the most part, you can always solve for n unknowns if you have n equations. There are a couple of ways to do it:

1. Substitution - manipulate one of the equations so that z is alone on the left. Then, take that value for z and substitute it into the other two equations. Repeat for y to solve for x. Sub the value you get for x back into the equation for y, then x & y back into the equation for z.

2. Equation Subtraction - subtract the left side of equation 1 from the left side of equation 2, and the right side of equation 1 from the right side of equation 2. The resulting equation will no longer have an x in it. Do the same thing with eq'ns 1 and 3 or 2 and 3 (note that you are allowed to subtract scalar multiples of equations as well) for a second equation with no x in it. Then, do some multiplying and subtracting of these two new equations to get one that has only a y or z in it. Solve, and back-substitute.

These are the easiest two to perform.