Obviously I don't know what I'm doing here. The book gives an answer of-8, so any hints on where my method is flawed would be appreciated.

z + 9 = 1

z + 9 = + (-1) = 1 + (-1)

z + 8 = 0

z = -1

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- Feb 19th 2006, 02:40 PMEuclid AlexandriaSolving Equations - using addition property
Obviously I don't know what I'm doing here. The book gives an answer of

**-8**, so any hints on where my method is flawed would be appreciated.

**z + 9 = 1**

z + 9 = + (-1) = 1 + (-1)

z + 8 = 0

z = -1 - Feb 19th 2006, 02:51 PMThePerfectHackerQuote:

Originally Posted by**Euclid Alexandria**

$\displaystyle z+9=1$

Thus,

$\displaystyle z+9+(-9)=1+(-9)$

Thus,

$\displaystyle z+(9-9)=-8$

Thus,

$\displaystyle z+0=-8$

Thus,

$\displaystyle z=-8$ - Feb 19th 2006, 03:05 PMEuclid Alexandria
My next question is, why do we use -9 to solve the equation? What is the rule we use to get that number?

The addition property of equality confuses me. From the book: "Let*a*,*b*and*c*represent numbers.

If*a*=*b*, then*a + c = b + c*and*a - c = b - c*."

How is this applied to**z + 9 = 1**? Is this to say that**z + 9**is*a*, and**1**is*b*? Or something else?

Is it to say that**z**=*a*,**9**=*c*, and**1**=*b*? By my understanding of that reasoning,**z + 9 = 1 + 9**, which does not result in**-8**. But it seems, at first glance, to be what the rule implies. - Feb 19th 2006, 03:25 PMThePerfectHackerQuote:

Originally Posted by**Euclid Alexandria**

- Feb 19th 2006, 08:54 PMearbothQuote:

Originally Posted by**Euclid Alexandria**

the first aim when solving an equation: get the unknown alone (and maybe homesick) on one side of the equation. That means on the other hand: Take off all what is not unknown: The only way to do so (as you've written above) is to add a number, a term**on both sides of the equation**:

x+9=1 Take off the 9 by adding the (additive) inverse number to get zero (that means the 9 has vanished on the LHS) and you'll get the answer, which ThePerfectHacker has already given to you.

Greetings

EB - Feb 20th 2006, 07:49 AMEuclid AlexandriaEquations containing a variable on both sides
Ok, I see part of how we're using the addition property of equality to solve for z. We're using

*a*-*c*=*b*-*c*, where*c*is the**9**on both sides.

How about a problem where both sides contain the variable? What is the rule for deciding on which side to isolate the variable? Without clearly understanding the rule, there are two ways I can solve for x in this equation:

10x = -2 + 9x

**1)**10x + (-9)x = -2 + 9x + (-9)x

1x (or x) = -2

**2)**10x + (-10)x = -2 + 9x + (-10)x

x = -2 + (-1)x

I can read option**2)**in two ways: If 1x (or -1x) is always to be regarded as just x, then both methods are correct. But if -2 + (-1)x is -3x, then I can see by checking that option**1)**is the only correct method.

Basically, from looking at this problem, I am wondering if the rule goes something like:*isolate the variable by subtracting it from the side containing the most numbers.*Somehow I doubt this is the rule, however. - Feb 20th 2006, 08:02 AMearbothQuote:

Originally Posted by**Euclid Alexandria**

what you proposed is OK.**You**can decide on which side of the equation you'll have the variable. For instance take your example:10x = -2 + 9x.

You now would like to have the variable**on the RHS**of your equation. So you have to move the 10x from the LHS by adding -10x on both sides: 10x-10x=-2+9x-10x. You'll get: 0=-2-x. Now move the -2 by adding +2 on both sides and you'll get: 0+2=-2-x+2 or 2=-x

I suppose that you don't want to know the value of -x, so you have to change the sign by multiplying by (-1) (on both sides!)

2*(-1)=-x*(-1) will give:-2=x and that's exactly your result.

I hope this will help a little bit further on.

Greetings

EB - Feb 20th 2006, 12:48 PMtopsquarkQuote:

Originally Posted by**Euclid Alexandria**

$\displaystyle 10x =-2+9x$

$\displaystyle 10x+(-10)x=-2+9x+(-10)x$

$\displaystyle 0=-2+(-1)x$

$\displaystyle 0+x=-2+(-1)x+x$

$\displaystyle x=-2+0$

$\displaystyle x=-2$.

Obviously, which method you choose should not matter, but some methods work alot faster than others!

-Dan - Feb 21st 2006, 04:50 AMEuclid Alexandria
Thank you so much PerfectHacker, Earboth and Topsquark. This concept has been my biggest obstacle so far, but it is finally making some sense. I can see the couple places where I was off track, and it's relieving to know I was closer to being correct than I thought.

Topsquark, I never would have thought about adding an x back to both sides, but I can see how it balances the equation just like a number would. As far as multiplication, Earboth, that property of equality is covered in the next section of my book, so your example was very helpful too!