# Help solve, I got quiz tomorow on this stuff need quick help!

• Jul 8th 2009, 10:10 PM
elitescouter
Help solve, I got quiz tomorow on this stuff need quick help!
Allright so tomorow I got a quiz on this stuff. I just need someone to show me step by step how to do this!

$\displaystyle \frac{bx}{a}+\frac{1}{5}=2x$ SOLVE FOR X

$\displaystyle \frac{3}{M}=\frac{M}{5}$ SOLVE FOR M

$\displaystyle \frac{1}{U}=\frac{y}{k}+\frac{1}{h}$ SOLVE FOR K

$\displaystyle \frac{V-6}{5}+\frac{V-8}{15}+\frac{V}{10}=0$ SOLVE FOR V

$\displaystyle \frac{5}{w+3}+4=15$ SOLVE FOR w

and some similar problems that I need to isolate

$\displaystyle \frac{K}{M}=P+\frac{a}{M^2}$ ISOLATE a

Thanks!!!
• Jul 8th 2009, 10:29 PM
Stroodle
$\displaystyle \frac{bx}{a}+\frac{1}{5}=2x$

First establish a common denominator:

$\displaystyle \frac{5bx}{5a}+\frac{a}{5a}=\frac{10ax}{5a}$

Then solve for $\displaystyle x$

$\displaystyle 10ax-5bx=a$

$\displaystyle x(10a-5b)=a$

$\displaystyle x=\frac{a}{10a-5b}$

$\displaystyle x=\frac{a}{5(2a-b)}$

Use the same process for this one:

$\displaystyle \frac{1}{U}=\frac{y}{k}+\frac{1}{h}$

$\displaystyle \frac{hk}{hkU}=\frac{hUy}{hkU}+\frac{kU}{hkU}$

$\displaystyle hk-kU=hUy$

$\displaystyle k(h-U)=hUy$

$\displaystyle \therefore k=\frac{hUy}{h-U}$

Can you now use this process to do the other ones? Each term just needs to have the same denominator.

For the last question:

$\displaystyle \frac{K}{M}=P+\frac{a}{M^2}$

$\displaystyle \frac{a}{M^2}=\frac{K}{M}-P$

$\displaystyle a=M^2(\frac{K}{M}-P)$

$\displaystyle a=\frac{KM^2}{M}-PM^2$

$\displaystyle a=KM-PM^2$

$\displaystyle a=M(K-PM)$
• Jul 8th 2009, 11:00 PM
Stroodle
$\displaystyle \frac{5}{w+3}+4=15$

For this question you can multiply each term by $\displaystyle (w+3)$ to get rid of the fraction.

$\displaystyle 5+4(w+3)=15(w+3)$

$\displaystyle 4w+17=15w+45$

$\displaystyle 11w=-28$

$\displaystyle \therefore w=-\frac{28}{11}$
• Jul 8th 2009, 11:35 PM
elitescouter
Sorry to bother u but u think u can help me with

$\displaystyle$\displaystyle
\frac{3}{M}=\frac{M}{5}$SOLVE FOR M$

I get stuck half way through it.
• Jul 8th 2009, 11:45 PM
yeongil
Quote:

Originally Posted by elitescouter
Sorry to bother u but u think u can help me with

$\displaystyle \frac{3}{M}=\frac{M}{5}$ SOLVE FOR M

I get stuck half way through it.

Just cross multiply:
\displaystyle \begin{aligned} \frac{3}{M} &= \frac{M}{5} \\ M^2 &= 15 \end{aligned}

Now find M.

01
• Jul 8th 2009, 11:46 PM
arze
Quote:

Originally Posted by elitescouter
Sorry to bother u but u think u can help me with

$\displaystyle$\displaystyle
\frac{3}{M}=\frac{M}{5}$SOLVE FOR M$

I get stuck half way through it.

make every term have the same denominator.
M^2=3*5
M=sqrt (15)
• Jul 8th 2009, 11:50 PM
Stroodle
Don't forget the $\displaystyle \pm$

$\displaystyle m^2=15$

$\displaystyle \therefore m=\pm \sqrt{15}$
• Jul 8th 2009, 11:51 PM
arze
oops forgot that!
• Jul 9th 2009, 06:45 AM
allyourbass2212
Quote:

Originally Posted by yeongil
Just cross multiply:
\displaystyle \begin{aligned} \frac{3}{M} &= \frac{M}{5} \\ M^2 &= 15 \end{aligned}

Now find M.

01

What other alternatives other than cross multiplying are possible here? Because if it were me I would multiply by the reciprocal $\displaystyle \frac{m}{3}*\frac{3}{m}=\frac{m}{5}*\frac{m}{3}$. Which would equal $\displaystyle \frac{m^2}{15}$
• Jul 9th 2009, 07:16 AM
Stroodle
You can use a common denominator:

$\displaystyle \frac{3}{M}=\frac{M}{5}$

$\displaystyle \frac{15}{5M}=\frac{M^2}{5M}$

$\displaystyle M^2=15$

$\displaystyle \therefore M=\pm\sqrt{15}$
• Jul 9th 2009, 07:21 AM
Prove It
Quote:

Originally Posted by allyourbass2212
What other alternatives other than cross multiplying are possible here? Because if it were me I would multiply by the reciprocal $\displaystyle \frac{m}{3}*\frac{3}{m}=\frac{m}{5}*\frac{m}{3}$. Which would equal $\displaystyle \frac{m^2}{15}$

Your method would be fine as long as you keep your eye on BOTH sides of the equation.

$\displaystyle \frac{3}{M} = \frac{M}{5}$

$\displaystyle \frac{3}{M}\cdot\frac{M}{3} = \frac{M}{5}\cdot\frac{M}{3}$

$\displaystyle 1 = \frac{M^2}{15}$

$\displaystyle 15 = M^2$

$\displaystyle \pm\sqrt{15} = M$.
• Jul 9th 2009, 07:23 AM
Prove It
Quote:

Originally Posted by Stroodle
$\displaystyle \frac{5}{w+3}+4=15$

For this question you can multiply each term by $\displaystyle (w+3)$ to get rid of the fraction.

$\displaystyle 5+4(w+3)=15(w+3)$

$\displaystyle 4w+17=15w+45$

$\displaystyle 11w=-28$

$\displaystyle \therefore w=-\frac{28}{11}$

It's easier if you subtract 4 from both sides first...

$\displaystyle \frac{5}{w + 3} + 4 = 15$

$\displaystyle \frac{5}{w + 3} = 11$

$\displaystyle \frac{w + 3}{5} = \frac{1}{11}$

$\displaystyle w + 3 = \frac{5}{11}$

$\displaystyle w = \frac{5}{11} - 3$

$\displaystyle w = -\frac{28}{11}$.
• Jul 9th 2009, 07:46 AM
allyourbass2212
Embarrassingly enough the basic algebra book I am reading through does not cover how to solve when the equation looks as such.

$\displaystyle 1 = \frac{M^2}{15}$

$\displaystyle 15 = M^2$

Would you please explain the steps involved in how $\displaystyle 1 = \frac{M^2}{15}$ equals $\displaystyle 15 = M^2$
?

Many thanks
• Jul 9th 2009, 07:50 AM
Prove It
Quote:

Originally Posted by allyourbass2212
Embarrassingly enough the basic algebra book I am reading through does not cover how to solve when the equation looks as such.

$\displaystyle 1 = \frac{M^2}{15}$

$\displaystyle 15 = M^2$

Would you please explain the steps involved in how $\displaystyle 1 = \frac{M^2}{15}$ equals $\displaystyle 15 = M^2$
?

Many thanks

Can you see that $\displaystyle M^2$ is divided by 15?

To undo this, we have to multiply both sides by 15.

So $\displaystyle 1 = \frac{M^2}{15}$

$\displaystyle 1\cdot 15 = \frac{M^2}{15}\cdot 15$

$\displaystyle 15 = M^2$

Then to undo the squaring, we have to take the square root of both sides.

So $\displaystyle \pm \sqrt{15} = M$.
• Jul 9th 2009, 07:51 AM
allyourbass2212
Ah yes very good thank you.