7
$\begingroup$

I want to solve this system of equations, I have been out of Maths for a long!!

$$\sqrt{x}+y=7$$

$$\sqrt{y}+x=11$$

Just wondering easiest step to find values for $x$ and $y$ from the above equations?

$\endgroup$
11
  • 3
    $\begingroup$ This is one of those `awkward' questions. Equations of this form (with $a$ and $b$ replacing 7 and 11) reduce to quartics, which do have analytic solutions, but they're pretty horrible. On the other hand, these particular equations have an integral solution. $\endgroup$ Feb 13, 2012 at 18:28
  • 2
    $\begingroup$ @ChrisTaylor (9,4) is the only one. $\endgroup$
    – Pedro
    Feb 13, 2012 at 18:31
  • 2
    $\begingroup$ Why is "slope of a line" in the title? $\endgroup$ Feb 13, 2012 at 19:15
  • 2
    $\begingroup$ @AméricoTavares , I should have explained Derivation for x and y !! My initial hunch on deriving the above equation was based on y = mx + b !! Its my fault, $\endgroup$
    – goofyui
    Feb 13, 2012 at 23:17
  • 1
    $\begingroup$ @Medex: I changed the title and retagged your question. $\endgroup$ Feb 14, 2012 at 0:03

7 Answers 7

7
$\begingroup$

First rearrange the equations like this: $$y-4=3-\sqrt{x} \\x-9=2-\sqrt{y}$$ then multiply two sides of these equations to each other; $$(y-4)(x-9)=(3-\sqrt{x})(2-\sqrt{y})$$ Since $x,y\ge 0$ we can factor LHS : $$(\sqrt{y}-2)(\sqrt{y}+2)(\sqrt{x}-3)(\sqrt{x}+3)=(3-\sqrt{x})(2-\sqrt{y})$$ Implies that$$ (\sqrt{y}-2)(\sqrt{x}-3)((\sqrt{y}+2)(\sqrt{x}+3)-1)=0 $$ at least one of the factors must be zero.

1. $$(\sqrt{y}-2)=0 \Rightarrow y=4$$

by substitution $y=4$ in the first eq. we have $x=9$.

and by checking in the second eq. implies that $$x=9\qquad and \qquad y=4$$ is a solution.

or

2. $$(\sqrt{x}-3)=0 \Rightarrow x=9$$

by substitution in the first eq. we have $y=4$ and by checking in the second eq. implies that this is not a new solution.

or

3. $$(\sqrt{y}+2)(\sqrt{x}+3)-1=0 \Rightarrow (\sqrt{y}+2)(\sqrt{x}+3)=1$$ but it has no solution for $x$ and $y$ because $$ (\sqrt{y}+2)\ge 2 ; \quad (\sqrt{x}+3) \ge 3 \Rightarrow (\sqrt{y}+2)(\sqrt{x}+3)\ge 6 $$ as a result $(\sqrt{y}+2)(\sqrt{x}+3)=1$ has no solution for $x$ and $y$.

$\endgroup$
5
$\begingroup$

Note that : $x,y \geq 0$

$(7-y)^2=11-\sqrt y$

substitute $~\sqrt y = t~$ , so :

$(7-t^2)^2=11-t \Rightarrow t^4-14 \cdot t^2+49=11-t \Rightarrow t^4-14 \cdot t^2+t+38=0$

Note that possible nonnegative integer solutions of equation $~\sqrt x+y=7~$ are :

$(x,y) \in \{(1,6),(4,5),(9,4),(16,3),(25,2),(36,1),(49,0) \}$

Since pair $(x,y)$ has to be solution of equation $\sqrt y +x =11$ also we have that :

$(x,y)=(9,4) \Rightarrow \sqrt y=2 \Rightarrow t=2 ~$

Therefore polynomial $~t^4-14 \cdot t^2+t+38~$ has to be divisible by $(t-2)$

If you divide this polynomial by $(~t-2)~$ using Polynomial long division method you will get polynomial :

$t^3+2t^2-10t-19$

So you should solve equation $~t^3+2t^2-10t-19=0~$ to obtain other solutions .

This can be done using general formula of roots .

$\endgroup$
4
  • $\begingroup$ I am able to derive upto this equation t ^ 4 −14 ^ t ^ 2 +t+38=0 , from then i am lost !! Can you please explain the rest $\endgroup$
    – goofyui
    Feb 13, 2012 at 22:08
  • $\begingroup$ I said upto this step : t 4 −14⋅t 2 +t+38=0 $\endgroup$
    – goofyui
    Feb 13, 2012 at 22:10
  • $\begingroup$ @Medex,see edit... $\endgroup$
    – Pedja
    Feb 14, 2012 at 5:03
  • $\begingroup$ @TPedja Great Thank you !! Your Derivation helped me a lot !! $\endgroup$
    – goofyui
    Feb 14, 2012 at 17:14
5
$\begingroup$

There are no easy steps to solve the given system of two equations $$ \begin{eqnarray} \sqrt{x}+y &=&7 \tag{1} \\ \sqrt{y}+x &=&11, \end{eqnarray}$$ which is equivalent to $$ \begin{eqnarray*} \sqrt{x} &=&7-y \\ \sqrt{y} &=&11-x. \end{eqnarray*}$$ If you square both equations you get $$ \begin{eqnarray*} x &=&\left( 7-y\right) ^{2} \\ y &=&\left( 11-x\right) ^{2}, \end{eqnarray*}$$ whose solutions include the solutions of the given system. Now substitute the second in the first and vice-versa $$ \begin{eqnarray*} x &=&\left( 7-\left( 11-x\right) ^{2}\right) ^{2}\Leftrightarrow \left( 7-\left( 11-x\right) ^{2}\right) ^{2}-x=0 \\ y &=&\left( 11-\left( 7-y\right) ^{2}\right) ^{2}\Leftrightarrow \left( 11-\left( 7-y\right) ^{2}\right) ^{2}-y=0 \end{eqnarray*}$$ and expand the left hand sides to get two quartics $$ \begin{eqnarray} x^{4}-44x^{3}+712x^{2}-5017x+12\,996 &=&0 \tag{2} \\ y^{4}-28y^{3}+272y^{2}-1065y+1444 &=&0. \end{eqnarray} $$

Instead of solving these two equations by the standard method, we can first apply the rational root theorem, from which we know that the possible integer roots are of the form $x=\pm p,y=\pm q$, where $p$ and $q$ are integer factors of the constant terms, respectively, $ 12996=2^{2}3^{2}19^{2}$ and $1444=2^{2}19^{2}$. By direct computation we conclude that only $x=9$ and $y=4$ are integer solutions $$ \begin{eqnarray*} \sqrt{9} &=&7-4 \\ \sqrt{4}+9 &=&11. \end{eqnarray*} $$ Since $$\begin{eqnarray*}x^{4}-44x^{3}+712x^{2}-5017x+12996=\left( x-9\right) \left( x^{3}-35x^{2}+397x-1444\right) \end{eqnarray*} $$ and $$ \begin{eqnarray*}y^{4}-28y^{3}+272y^{2}-1065y+1444 =\left( y-4\right) \left( y^{3}-24y^{2}+176y-361\right) \end{eqnarray*} $$ we still need to find the roots of two cubic equations, one in $x$ and the other in $y$ $$ \begin{eqnarray} x^{3}-35x^{2}+397x-1444 &=&0 \tag{3} \\ y^{3}-24y^{2}+176y-361 &=&0. \end{eqnarray}$$

It would be possible to find the roots algebraically, but the computation is so long that I computed the roots numerically in SWP $$ \begin{eqnarray*} x &\approx &7.8687,\quad x \approx 12.848,\quad x \approx 14.283 \\ y &\approx &3.4156,\quad y \approx 9.8051,\quad y \approx 10.779. \end{eqnarray*} $$

Edited. However these solutions are not solutions of the original system. For instance, for $x\approx 7.8687$, from $\sqrt{7.868\,7}+y=7$, we find $ y\approx 4.1949$, which is not a solution of $\sqrt{y}+x=11$ $$ \sqrt{4.1949}+7.8687\approx 9.9168\neq 11, $$ and similarly for the other solutions. So the only solution is $\left( x,y\right) =\left( 9,4\right) $.

Added. Here is a plot of the two curves, $\sqrt{x}+y=7$ (blue) and $\sqrt{y}+x=11$ (black), which cross each other at $(x,y)=(9,4)$.

enter image description here

$\endgroup$
4
  • $\begingroup$ Thanks for your comment, I have noticed that you have included a more detailed answer with a graph, so I will remove my previous answer. $\endgroup$
    – NoChance
    Feb 14, 2012 at 12:46
  • $\begingroup$ @EmmadKareem You are welcome. $\endgroup$ Feb 14, 2012 at 13:40
  • 1
    $\begingroup$ @AméricoTavares, Great help ! You also have changed the title and made it easier !! Thank you so much !! $\endgroup$
    – goofyui
    Feb 14, 2012 at 17:15
  • $\begingroup$ @Medex You are welcome. $\endgroup$ Feb 14, 2012 at 17:20
2
$\begingroup$

I'll solve a related system of equations: $$A + B^2 = 7$$ and $$B +A^2=11$$

Solving for $A$ in the first gives $$A=7-B^2$$ Substituting this into the second equation gives $$B+(7-B^2)^2 = 11$$ from which we arrive at $$B^4 -14B^2+B+38=0$$


Added Here, my goal is to find some solution(s) to this quartic (i.e. fourth degree) equation. There is a theoretical result that tells us what the rational (this includes integers) solutions will be, if there are any.

Basically, the rational root theorem says that in this case, any rational root will divide $38$.

We (the royal we ... I) then employ a technique known as synthetic (polynomial) division. I will not go into much detail, except to mention that I usually test smaller potential roots first... so testing $\pm 1, \pm 2 $ gives us that $B = 2$ is a solution. When we have exhausted our potential rational roots, we see that this is the only rational solution.

It then follows that $A=3$, since if $B = 2$, the first equation can be rewritten as $$A + 2^2 = 7$$

$\endgroup$
4
  • $\begingroup$ Rationality of solutions is not an issue here. The solution $(9,4)$ has already been identified. It remains to prove that there are no other solutions. $\endgroup$ Feb 13, 2012 at 20:21
  • $\begingroup$ @Christian: the OP asked for steps in determining solutions. While you might not accept my solution of a related problem, neither do I accept an identification of a solution when a student has asked for a method. $\endgroup$ Feb 13, 2012 at 21:16
  • $\begingroup$ @TheChaz , I am able to understand your last part of the derivation .. B4−14B2+B+38=0 , From then can you please explain me how you are able to generate B=2 and A=3 $\endgroup$
    – goofyui
    Feb 13, 2012 at 23:09
  • 1
    $\begingroup$ @TheChaz Great Thank you !! Your Derivation was simple and easy to follow !! Thank you for your help !! $\endgroup$
    – goofyui
    Feb 14, 2012 at 17:13
2
$\begingroup$

The OP asked for the "easiest steps" to solve this system of equations. Here is what I would do:

Given that the occurring expressions are so simple I would try to get a global overview by drawing a figure. This means that we have to intersect the two graphs $$y=7-\sqrt{x}\quad(x\geq0)\ ,\qquad y=(11-x)^2\quad(x\leq11)$$ (note restriction of $x$ for the second graph!). Looking at the figure one immediately sees that there is exactly one intersection point; one guesses that this is the point $(9,4)$. Plugging this point into the original system of equations one verifies that it is indeed a solution.

In order to prove that we were not cheated by our eyes we have to prove that the auxiliary function $$f(x)\ :=\ (11-x)^2-(7-\sqrt{x})=114-22x+x^2+\sqrt{x}$$ has exactly one zero in the interval $[0,11]$. Unfortunately the derivative $$f'(x)=-22+2x+{1\over 2\sqrt{x}}$$ is not strictly negative in this interval; so we have to take some extra measures: When $0\leq x\leq5$ then $f(x)\geq 114-22x\geq4$, and when $10\leq x\leq11$ then $f(x)\leq 1-7+\sqrt{x} < -2$. For the intermediate interval $[5,10]$ we now can show that $f$ is strictly decreasing: $$f'(x)\leq -22+20+{1\over 2\sqrt{5}} < -1\ .$$

$\endgroup$
0
$\begingroup$

we can rewrite equations as $(x-4)+\sqrt{y}-3=0$,$(y-9)+\sqrt{x}-2=0$ so we have $(\sqrt{x}-2)(\sqrt{x}+2)+\sqrt{y}-3=0$ ,$(\sqrt{y}-3)(\sqrt{y}+3)+\sqrt{x}-2=0$ substitute $\sqrt y$ from first equation and factor we have $(\sqrt{x}-2)((\sqrt{x}+2)(\sqrt{y}+3)-1)=0$ so we have $x=4,y=9$

$\endgroup$
0
$\begingroup$

Let $x=\tan^2\theta$ and $y=\sec^2\theta$

  1. $\tan\theta+\sec^2\theta=7$
  2. $\tan^2\theta+\sec\theta=11$

Take 1) $\tan\theta+1/\cos^2\theta=7$ Solve it and you will get

$$7\sin^2\theta+\sin\theta-6=0$\tag{A}$$

Now similarly take equation (2) solve it you will get

$$12\cos^2\theta-\cos\theta-1=0\tag{B}$$

Take (A). Let $\sin\theta=t$, $7t^2+t-6=0$.

Solve it you will get $t=-1$ or $t=6/7$

$\sin\theta=-1$ or $\sin\theta=6/7$.

Now

Take eq. (B)

Do similarly As equation (A)

$12\cos^2\theta-\cos\theta-1=0$

Solve it as above $\cos\theta=1/3$ or $\cos\theta=-1/4$

$\cos\theta$ value exists in 3 quadrant not -ve

So $-1/4$ not possible.

$\cos\theta=1/3$

$\sin\theta$ value exist in 3 quadrant not +ve so $\sin\theta=-1$.

Now $\tan\theta=\sin\theta/\cos\theta=-1/(1/3)=-3$.

$x=\tan^2\theta=(-3)^2=9$

Put in equation 1 $y=4$

Therefore $x=9$ and $y=4$ is answer.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .