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
$3m^2-16m-64$.

 

[chwala]

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.

The third solution will be given using $m=-2.67$

We then have,

$x+y =-2.0025$
$x^3+y^3 = -24.03$

$x^3+(-x^3-6x^2-12x-8)+24.03=0$

$-6x^2-12x+16.03=0$

$x=-2.91$ and $y=0.91$ correct to two decimal points.

 

[pbuk]

$x=-2.91$ and $y=0.91$ correct to two decimal points.

The words "correct to two decimal [places]" should never appear anywhere in the answer to an algebra problem. PUT YOUR CALCULATOR AWAY.

$m= \dfrac{144±288}{54}$

Think about what you are writing down: what can you immediately see about all the terms in this equation?

$m_3=-1.37 ×10^{-10}$

Do you really think that is true? Can you not see that a calculator is no use to you - in fact it is almost always a hindrance - in algebra?

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
$3m^2-16m-64$.

You need to replace your calculator skills with skills in spotting common factors:
- Even numbers should be obvious
- If the digits sum to a multiple of 3 the whole is a multiple of 3
- If the digits sum to a multiple of 9 the whole is a multiple of 9
- If the last digit is 0 or 5 the whole is a multiple of 5
- If the smallest number is a multiple of 7 (you should be able to spot these immediately for 2 digit numbers) it can be worth checking all the numbers for divisibility by 7 (this is the only time your calculator can be helpful)

 

[chwala]

The words "correct to two decimal [places]" should never appear anywhere in the answer to an algebra problem. PUT YOUR CALCULATOR AWAY.


Think about what you are writing down: what can you immediately see about all the terms in this equation?


Do you really think that is true? Can you not see that a calculator is no use to you - in fact it is almost always a hindrance - in algebra?


You need to replace your calculator skills with skills in spotting common factors:
- Even numbers should be obvious
- If the digits sum to a multiple of 3 the whole is a multiple of 3
- If the digits sum to a multiple of 9 the whole is a multiple of 9
- If the last digit is 0 or 5 the whole is a multiple of 5
- If the smallest number is a multiple of 7 (you should be able to spot these immediately for 2 digit numbers) it can be worth checking all the numbers for divisibility by 7 (this is the only time your calculator can be helpful)

Didn't think much into it... I was quickly looking for solutions to the problem,...with that said I'll most definitely work out the solutions as you've rightly put it... will get back on this later. Thanks.

 

[pbuk]

I was quickly looking for solutions to the problem

But you were going about it the wrong way: the way to quickly find solutions to algebra problems is to practice quickly simplifying expressions, including extracting common factors, not to reach for a calculator.

 

[bob012345]

FYI, here is a graph showing the intersection of the two curves.

 

[chwala]

But you were going about it the wrong way: the way to quickly find solutions to algebra problems is to practice quickly simplifying expressions, including extracting common factors, not to reach for a calculator.

It was not much of a big deal... i thought i had missed out on something real important...like i said i just was seeking solutions but i do get your point all the same...

using, $m=-\dfrac{8}{3}$

$x+y=\dfrac{3}{4} xy$
$x+y=-2$

We then have, from my previous working...

$x+y =-2$
$x^3+y^3 = -\dfrac{72}{3}$

$x^3+(-x^3-6x^2-12x-8)+\dfrac{72}{3}=0$

$6x^2+12x-16=0$

$3x^2+6x-8=0$

...

$x=-2.91$ and $y=0.91$ correct to two decimal points... or solutions can also be given in exact form using the quadratic formula. Cheers man.

 

[chwala]

View attachment 356433FYI, here is a graph showing the intersection of the two curves.

It could be more useful if you could demonstrate/find alternative ways to solve the problem algebraically...the graph illustration is quite clear with me...

 

[bob012345]

It could be more useful if you could demonstrate/find alternative ways to solve the problem algebraically...the graph illustration is quite clear with me...

You can use the lines from the origin to the intercept points which have the form $y=ax$. Substitute into each equation and get two solutions for $x=f(a)$. Set those equal to each other and you get a polynomial in $a$. Solving that gives four values for the slopes ${1/2,2,1/4(sqrt(33)-7),-1/4(sqrt(33)+7)}$. Use those to get the nonzero $x,y$ values.

 

[D H]

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

Just piling on: Please refrain from doing this. Instead keep the 0.75 as $3/4$. Doing so will benefit you in the long run. Plus, this is a sign that you are relying on your calculator far too much.

Let $m=xy$

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

Keeping this to rational factors, this is better expressed as $\frac{27}{64}m^3 - \frac94m^2-9m=0$, or
(m)(\frac{27}{64}m^2 - \frac94m - 9) = 0\tag{1}
You should not have fed the equation you found to a calculator. It will blindly see that it was given a cubic and use its cubic solver. This is a cubic, but it also is $m$ times a quadratic in $m$. The only solutions are $m=0$ and the two solutions to the quadratic, $3m^2-16m-64 = 0$.

The trivial solution, $m=0$, can and must be ruled out due to the problem statement, which involves divisions by $x$ and by $y$. This means that both x and y must be non-zero for the problem statement to make sense, and hence so must their product $xy\equiv m$ be non-zero. Sometimes making substitutions or equating two expressions can introduce false solutions. That definitely is the case here, where the trivial solution is a false solution. This is something to be aware of and be on the watch for.

The quadratic equation $3m^2-16m-64 = 0$ can be factored fairly easily. If a root is an integer, it must be a factor of 64. 1, 2, and 4 are easy to rule out. $m=8$ works perfectly, yielding $(m-8)(3m+8) = 0$ as the factorization. Alternatively, the generic quadratic equation solver works nicely here. Both approaches yield $m=8$ as one solution and $m=-\frac83$ (rather than $m\approx-2.6$ or $m\approx-2.7$) as the second solution.

 

[D H]

View attachment 356433 FYI, here is a graph showing the intersection of the two curves.

The one solution that can easily be ascertained from the graph is $y=y=0$. This is a false solution. There are four points that represent true solutions where neither $x$ nor $y$ is zero.

 

[bob012345]

The one solution that can easily be ascertained from the graph is $y=y=0$. This is a false solution. There are four points that represent true solutions where neither $x$ nor $y$ is zero.

I think the upper right quadrant solutions are pretty clear visually from the plot.

 

[MatinSAR]

@chwala I did it this way, similar to your original post somehow, except I didn't use a calculator at all. This led to an error in your calculations, which is why you didn't consider $xy=0$ as a possibility. From your right equation you have:$$ \dfrac 1 x + \dfrac 1 y = \dfrac 3 4$$$$\dfrac {x+y}{xy}=\dfrac 3 4$$This means you can consider: $$x+y=3k$$$$xy=4k$$ All that remains is to express the left equation in terms of $x+y$ and $xy$. $$ x^3+y^3 = 9xy \rightarrow (x+y)(-3xy+(x+y)^2)=9xy $$ $$ (3k)(-12k+9k^2)=36k$$ $$ k(3k^2-4k-4)=0 $$ $$k=0,2,-\dfrac 2 3$$Now you need to solve : $$x+y=0,6,-2$$$$xy= 0,8,-8/3$$ The first scenario is not acceptable. The second situation is the one you derived: $$x=4 \rightarrow y=2$$ or $$ x=2 \rightarrow y=4$$ Third situation is $x+y=-2$ and $xy =-8/3$ so your $(x,y)$ pair is going to be: $$ \left(-1 + \frac{\sqrt{33}}{3},\ -1 - \frac{\sqrt{33}}{3}\right) \quad \text{or} \quad \left(-1 - \frac{\sqrt{33}}{3},\ -1 + \frac{\sqrt{33}}{3}\right) $$

 

[chwala]

@chwala I did it this way, similar to your original post somehow, except I didn't use a calculator at all. This led to an error in your calculations, which is why you didn't consider $xy=0$ as a possibility. From your right equation you have:$$ \dfrac 1 x + \dfrac 1 y = \dfrac 3 4$$$$\dfrac {x+y}{xy}=\dfrac 3 4$$This means you can consider: $$x+y=3k$$$$xy=4k$$ All that remains is to express the left equation in terms of $x+y$ and $xy$. $$ x^3+y^3 = 9xy \rightarrow (x+y)(-3xy+(x+y)^2)=9xy $$ $$ (3k)(-12k+9k^2)=36k$$ $$ k(3k^2-4k-4)=0 $$ $$k=0,2,-\dfrac 2 3$$Now you need to solve : $$x+y=0,6,-2$$$$xy= 0,8,-8/3$$ The first scenario is not acceptable. The second situation is the one you derived: $$x=4 \rightarrow y=2$$ or $$ x=2 \rightarrow y=4$$ Third situation is $x+y=-2$ and $xy =-8/3$ so your $(x,y)$ pair is going to be: $$ \left(-1 + \frac{\sqrt{33}}{3},\ -1 - \frac{\sqrt{33}}{3}\right) \quad \text{or} \quad \left(-1 - \frac{\sqrt{33}}{3},\ -1 + \frac{\sqrt{33}}{3}\right) $$

Smart move!