View Full Version : Solve the solution
AshleyRae09
Sep 25, 2009, 07:31 PM
how do you solve for the solution
x+9y=2(1)
x=9-y(2)
ohsohappy
Sep 25, 2009, 07:34 PM
What level of math is this?
AshleyRae09
Sep 25, 2009, 07:35 PM
Its algebra in college
ohsohappy
Sep 25, 2009, 07:38 PM
Haa crap once sec, I'll ask my boyfriend. Tutors Calculus ane linear Algebra. So he's a smarty pants.
AshleyRae09
Sep 25, 2009, 07:39 PM
Thank you because I'm stuck on that one and another one I posted if you could look at that one for me? Thanks
ohsohappy
Sep 25, 2009, 07:40 PM
thank you because im stuck on that one and another one i posted if you could look at that one for me? thanks
Maybe copy and paste it?
AshleyRae09
Sep 25, 2009, 07:42 PM
solve with add and multiply 5+6x<53
ohsohappy
Sep 25, 2009, 07:45 PM
When's it due? Monday? He might be taking his time. I had to text it to him.
AshleyRae09
Sep 25, 2009, 07:54 PM
Yeah its due Monday
Thankss
Unknown008
Sep 26, 2009, 01:29 AM
Here, the easiest way to solve it is by substitution.
x+9y=2
x=9-y
The second equation shows you the 'value' of x. So, use that in the first equation.
x+9y=2
Since 'x' is 9-y, replace the value of 'x' like this:
(9-y)+9y=2
Did you see it? I replaced the x by its equal value, that is 9-y.
Now, it's easy to find y.
(9-y)+9y=2
9-y+9y=2
- y + 9y gives + 8y
9+8y=2
Now remove 9 from both sides:
9+8y - 9=2 - 9
The 9 cancels out on the left hand side:
\cancel{9}+8y \cancel{- 9}=2 - 9
8y=-7
Divide both sides by 8:
\frac{8y}{8}=\frac{-7}{8}
The 8 cancel out on the left hand side:
\frac{\cancel{8}y}{\cancel{8}}=\frac{-7}{8}
y=-\frac{7}{8}
Here is the value of y.
Now replace that value in the previous equation, x=9-y
x=9-(-\frac78)
x=9+\frac78
x=\large{9}\frac78
Unknown008
Sep 26, 2009, 01:36 AM
5+6x<53
This one now.
5+6x<53
Remove 5 from both sides:
5+6x-5<53-5
6x<48
Divide both sides by 6:
\frac{6x}{6}<\frac{48}{6}
x<8
Here it is! :)
If you were to divide or multiply by a negative number at any time, you will have to change the inequality sign. '>' becomes '<' and vice versa.
I'll give you an example:
Let's say 5 < 6
That's true right?
Ok, let's multiply that by -2:
5\times -2 < 6 \times -2
-10 < -12
Is -10 really smaller than -12? No. If you plot these on a number line, -10 is larger than -12. So, when you multiply or divide by a negative number, always invert the inequality sign.
I hope it helped! :)
If you have more problems, don't hesitate to ask. Next time though, post what you did and got wrong, so that we know where the problem is. :)
ohsohappy
Sep 26, 2009, 08:15 AM
HAHAHA OH GOSH! WHY didn't I think of that. I must have been tired last night, I could have doen that easily. :)