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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

Originally Posted by AshleyRae09

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

Originally Posted by AshleyRae09

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. :)