Solve the simultaneous equations

[chwala]

Homework Statement

See attached

Relevant Equations

understanding of simultaneous equations

In my approach i have,

$x+y = \dfrac{3}{4}xy$

and
$(x+y)^3=x^3+y^3+3xy(x+y)$

$(0.75xy)^3=9xy+ 3xy(0.75xy)$

Let $m=xy$

$0.421875m^3-2.25m^2-9m=0$

$m_1=8, m_2=-2.6$ and $m_3=-1.37 ×10^{-10}$

using

$m_1=8 = xy$

we shall have,

$x^3+y^3=9xy$

$x^3+y^3=72$

since,

$x+y=6$

then it follows that

$x^3+(6-x)^3-72=0$

$x^3+216-36x-72x+12x^2+6x^2-x^3-72=0$

$⇒18x^2-108x+144=0$

$x=4, ⇒y=2$

$x=2, ⇒y=4$

... other values can be found in a similar manner...There may be a better approach than mine hence my post. Cheers.

 

[PeroK]

I think $m_3 = 0$.

You can check for yourself that $2,4$ is a solution.

The third solution is the tricky one. I think you need to find and check that one.

 

[SammyS]

In my approach i have,

$x+y = \dfrac{3}{4}xy$

and
$(x+y)^3=x^3+y^3+3xy(x+y)$

$(0.75xy)^3=9xy+ 3xy(0.75xy)$

If you write that using rational fractions, rather than decimal fractions, you have:

$\displaystyle \left( \frac 3 4 \right)^3 (xy)^3 = 9xy + 3xy \left( \frac 3 4 \right) xy$

Rearranging and simplifying gives:

$\displaystyle \left( \frac {27} {64} \right) (xy)^3 - \left( \frac 9 4 \right) (xy)^2 - 9(xy) =0$

Multiplying through by $\displaystyle \frac {64} {9}$ gives:

$\displaystyle 3 (xy)^3 - 16 (xy)^2 - 64(xy) =0$

Factoring the left hand side gives:

$\displaystyle (xy-8)(3 xy + 8) (xy)=0$

 

[topsquark]

Homework Statement: See attached
Relevant Equations: understanding of simultaneous equations

$0.421875m^3-2.25m^2-9m=0$

$m_1=8, m_2=-2.6$ and $m_3=-1.37 ×10^{-10}$

That you wrote that second line is a sign that you are depending on your calculator too much.

And, as I mentioned in another thread, it would be a good idea to not use decimals.

-Dan

 

[chwala]

That you wrote that second line is a sign that you are depending on your calculator too much.

And, as I mentioned in another thread, it would be a good idea to not use decimals.

-Dan

@topsquark thanks let me check on how to minimise calculator in my working. Your other points are well noted..i do skip some steps as I assume that my audience (you guys) are quite highly intelligent people to see in between the lines...my bad. I'll work on showing the required steps man!

Cheers mate.

 

[Mark44]

let me check on how to minimise calculator in my working.

Not much to check. When you're working on a problem in algebra, don't use a calculator, or at least not until the final step.

i do skip some steps as I assume that my audience (you guys) are quite highly intelligent people to see in between the lines

Sure, we can probably fill in missing details, but we're working for free, and might not want to take the time to do so.

 

[chwala]

That you wrote that second line is a sign that you are depending on your calculator too much.

And, as I mentioned in another thread, it would be a good idea to not use decimals.

-Dan

Dan,
Ok we shall have (without calc),

$(0.75xy)^3=9xy+ 3xy(0.75xy)$

letting $xy=m$

$\dfrac{27}{64} m^3 = 9m + \dfrac{9}{4} m^2$

$27m^3=576m+144m^2$

$27m^3-576m-144m^2=0$

$m(27m^2-144m-576)=0$

$m_1=0$

$27m^2-144m-576=0$

$m= \dfrac{144±\sqrt{20736+62208}}{54}$


$m= \dfrac{144±\sqrt{82944}}{54}$

$m= \dfrac{144±288}{54}$

$m_2=8$ and $m_3=-2.67$ to two decimal places.

From here the steps to solution will follow. Cheers man.

 

[WWGD]

Dan,
Ok we shall have (without calc),

$(0.75xy)^3=9xy+ 3xy(0.75xy)$

letting $xy=m$

$\dfrac{27}{64} m^3 = 9m + \dfrac{9}{4} m^2$

$27m^3=576m+144m^2$

$27m^3-576m-144m^2=0$

$m(27m^2-144m-576)=0$

$m_1=0$

$27m^2-144m-576=0$

$m= \dfrac{144±\sqrt{20736+62208}}{54}$


$m= \dfrac{144±\sqrt{82944}}{54}$

$m= \dfrac{144±288}{54}$

$m_2=8$ and $m_3=-2.67$ to two decimal places.

From here the steps to solution will follow. Cheers man.

Don't know if this is the input you're looking for, but you could have divided your equation through by 9, to end with
$3m^2-16m-64$.