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Let's start with #1:
x + y = 6
y = 2x.



There are a few ways to solve these, but the substitution method is often easiest.



1. Substitute 2x in for y in the 1st equation: x + (2x) = 6.
(This can be done because the 2nd equation says y and 2x are interchangeable).



2. Simplify: 3x = 6.



3. Solve for x: x = 2.



4. Plug 2 in for x in the 1st equation: 2 + y = 6.



5. Solve for y: y = 4.



6. Final Answer: (2, 4). (x = 2, y = 4).



That's the substitution method - the next example has a few more steps:



2x-3y=-13
y=2x+7



1. Substitute 2x + 7 in for y in the 1st equation: 2x - 3(2x + 7) = -13.
(This can be done because the 2nd equation says y and 2x + 7 are interchangeable).



2. Simplify: -4x - 21 = -13.



3. Solve for x: x = -2.



4. Plug -2 in for x in the 1st equation: 2(-2) - 3y = -13.



5. Solve for y: y = 3.



6. Final Answer: (-2, 3). (x = -2, y = 3).

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Good question. The difference between expressions and equations is key to the answer.

A simple example of an expression is: .

And a simple example of an equation is: .

If the equation were a scale, the left side and right side would balance each other perfectly. Now if the same weight (say 3) is added to both sides of the equation:
,

you get an equation that is equivalent to (basically the same as) the original equation .

is basically the same equation as (they have the same answer).

In an equation, the left and right side are balanced. The main idea in solving equations is: if you start with a balanced scale and then do the same thing to both sides of the scale, the scale will still end up balanced.


With an expression, however there's no scale. An expression is like a weight just sitting there on it's own. So if 3 is added to an expression, it's no longer the same expression.
is not the same expression as .

Moving to your question, the reason you can multiply both sides of by 6 is that you're preserving the balance by doing the same thing (multiplying by 6) to both sides.

Multiplying the expression by 6, however, results in a different expression that's not equivalent. So multiplying an expression by a number typically changes the expression and so isn't allowed.



Sometimes, though you do multiply an expression by 1. Multiplying by any other number usually changes the expression, but multiplying by 1 doesn't change it. That's why multiplying by 1 is allowed. Usually you multiply by 1 in a different form, such as or . For example, starting with

,

you can multiply the by and the by :



to get the common denominator:

.


I hope this helps - let me know if you have questions about any part of it.

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An absolute value equation such as

|y| = 10

is really two equations:

y = 10 and y = -10.

Similarly, the equation

|2x+6|= 10

is equivalent to the two equations

2x + 6 = 10 and 2x + 6 = -10.

Solving these both for x gives the two answers.

The first one, for example is: 2x + 6 = 10. You can solve it at mathick.com, or as follows:

Subtract 6 from both sides: 2x = 4

Divide both sides by 2: x = 2.

Solving the second equation gives x = -8, so the two answers are 2 and -8.

You can plug these back into the original equation to verify that they are correct.

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I need helping solving this for x:
(ln x)^3=ln x^4
Can I rewrite it as (ln x)^3=4lnx ?

Yes.

Then can I divide both sides by lnx leaving (ln x)^2=4 ?

Yes, but this assumes that you're not dividing both sides by 0, i.e. that ln(x) is not 0. This step wouldn't be valid in the case that ln(x) = 0, so this case (ln(x) = 0) needs to be treated separately.

Can I now square (root) both sides leaving me with ln x=2 ?

Right, ln (x) = 2, and also ln(x) = -2. (Taking the square root of both sides gives ln x = +2 and ln x = -2.)

Now I'm not sure what to do next

To solve ln(x) = 2 for x, exponentiate both sides:
.

The left side simplifies, giving one of the final answers:

.

The equation ln (x) = -2 can be solved similarly.

Finally, there is the case when ln(x) = 0. This happens when x = 1. To verify that this is a solution, you can plug it into the original equation and see if it checks out (gives a true equation).