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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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
Since,Quote:
Originally Posted by Euclid Alexandria
Thus,
Thus,
Thus,
Thus,
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.
Because we are trying to isolateQuote:
Originally Posted by Euclid Alexandria
on one side. By using
you are not isolating the
Hello,Quote:
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
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.
Hello,Quote:
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
Your method in 2) is correct (if a bit longer) but you made a mistake, which is why it didn't work. 10x + (-10)x = 0, not x. It should read:Quote:
Originally Posted by Euclid Alexandria
.
Obviously, which method you choose should not matter, but some methods work alot faster than others!
-Dan
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!