# Solve two algebra problems

• Feb 2nd 2009, 09:53 PM
Shandy
Solve two algebra problems

2 8
X = -
15

and:
2
X x 3x = -X + 4

Thank you !
• Feb 2nd 2009, 10:00 PM
Hello Shandy
Quote:

Originally Posted by Shandy

2 8
X = -
15

and:
2
X x 3x = -X + 4

Thank you !

I don't understand what you mean. Can you write these more clearly? Use x^2 to mean $\displaystyle x^2$ if you like, and write the word 'times' to mean multiply, so that we don't think it's a letter x.

• Feb 2nd 2009, 10:01 PM
Jhevon
Quote:

Originally Posted by Shandy

2 8
X = -
15

and:
2
X x 3x = -X + 4

Thank you !

please fix the format of your question. just use / to indicate fractions. example, type 1/x to mean $\displaystyle \frac 1x$ and 1/(x + 1) to mean $\displaystyle \frac 1{x + 1}$ etc, also, use ^ to indicate powers. type x^2 to mean $\displaystyle x^2$ for instance. you can use * to mean multiply. what you posted is confusing
• Feb 3rd 2009, 12:29 AM
Algebra
Hello Shandy

I'm just guessing that your first equation might be

$\displaystyle x^2 = \frac{8}{15}$

If it is, the answers are

$\displaystyle x = \pm \sqrt{\frac{8}{15}}$

So get your calculator working! Divide 8 by 15. Then find the square root. Then put a $\displaystyle \pm$ sign in front, because there will be two answers.

The second one, I think, might be:

$\displaystyle x \times 3x = -x^2 + 4$

If so, then you solve it like this:

$\displaystyle x \times 3x = -x^2 + 4$

$\displaystyle \Rightarrow 3x^2 = -x^2 + 4$

$\displaystyle \Rightarrow 4x^2 = 4$

$\displaystyle \Rightarrow x^2 = 1$

$\displaystyle \Rightarrow x = \pm 1$

How am I doing? Did I guess right?

• Feb 3rd 2009, 04:30 AM
stapel
Quote:

Originally Posted by Shandy

2 8
X = -
15

and:
2
X x 3x = -X + 4

Looking at your post inside "Quote" tags, some of the original formatting is displayed while in message-entry mode. Assuming you are using "X" and "x" to actually be the same variable, your equations appear to be:

. . . . .x^2 = 8/15

. . . . .(x)(3x) = -x^2 + 4

If so, then you can solve the first equation by taking the square root of each side of the equation, remembering the "plus-minus" on the right-hand side. You'll probably need to "rationalize the denominator" to get the answer in the "right" format.

For the second one, simplify the left-hand side to get 3x^2. Get the variable term together, and then divide 4x^2 = 4 on both sides by 4. Then you can solve by taking the square root of either side.

Have fun! :D