Since,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.
Hello,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,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: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!