Find $x$ if $\frac {1} {x} + \frac {1} {y+z} = \frac {1} {2}$

Question

I found this question from past year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.

Find $x$ if \begin{align}\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac{1}{3}\\ \frac {1} {z} + \frac {1} {x+y} &= \frac {1} {4}\end{align}

$(A)\;\frac 32$

$(B)\;\frac {17}{10}$

$(C)\;\frac {19}{10}$

$(D)\;\frac {21}{10}$

$(E)\;\frac {23}{10}$

EDIT: I'm very sorry guys, it should be $\frac {1} {x+y}$ not $xy$, I'm sorry for the typos (idk what is wrong with me)

Answer 1

\begin{align} \frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac{1}{3}\\ \frac {1} {z} + \frac {1} {x+y} &= \frac {1} {4} \end{align}

Is equivalent to the system \begin{align} xy+xz&=2(x+y+z) \\ xy+yz&=3(x+y+z) \\ xz+yz&=4(x+y+z), \end{align} Solving this system as a linear system of equations in terms of $xy,xz$ and $yz$ and $x+y+z$, we arrive at

\begin{align} xy&=\tfrac12 (x+y+z), \quad(1) \\ yz&=\tfrac52 (x+y+z), \\ xz&=\tfrac32 (x+y+z). \end{align}

Dividing these equations pairwise, we get \begin{align} z&=5x \\ y&=\tfrac53 x, \end{align}

combining with (1) we have

\begin{align} \tfrac{5}{3}x^2-\tfrac{23}{6}x&=0 \end{align} and since $x\ne0$, the answer is $x=\tfrac{23}{10}$.

Answer 2

NOte: edit #1, equations changed when typo was noticed NOTE!!! This is not a complete solution, and there is likely something wrong with it. I'm posting it for 2 reasons: 1: To show the type of logic one would use, and 2: To see if anyone can spot the mistake so it can be fixed. (and 3, I spent so much time on it I don't want to delete it!)

Now:

Multiply through each equation by $xyz$ to get rid of the factions, and turn it into 3 equations in 3 unknowns: $$yz+x=\frac 1 2 xyz $$ $$xz +y=\frac 1 3 xyz $$ $$xy+z=\frac 1 4 xyz $$

Let's go backwards and use the third equation to solve for z: Subtract z from both sides and factor out the z, and you get $$xy=z(\frac 1 4 xy -1)$$, to clear the fractions I'm going to multiply through by 4 for convenious, giving $$4xy=z(xy-4)$$, thus $$z=\frac {4xy} {xy-4}$$ Now we plug that into our first two equations for $z$, giving us $$y(\frac {4xy} {xy-4})+x=\frac 1 2 xy(\frac {4xy} {xy-4}) $$ and $$x(\frac {4xy} {xy-4})+y=\frac 1 3 xy(\frac {4xy} {xy-4}) $$

Now, we're going to solve the second equation for $y$. Again, it's much easier to get rid of the fractions first, so I'm going to multiply the whole equation by $3(xy-4)$ to get $$12x^2y+3xy^2-12y^2=4x^2y^2$$ Every term has a $y$ and $y\ne 0$, so cancelling, we get $$12x^2+3xy-12y=4x^2y$$ Move the two terms with $y$ to the right hand side and factor out the $y$, and we get $$12x^2=y(4x^2+12-3x)$$, hence $$y=\frac {12x^2} {4x^2 +12 -3x}$$

Now I need to plug that into our first equation, but before I do that, I'm going to clear out the denominators in the first equations, so I multiply through by $2(xy-4)$ to give us $$8x^2y^2+2x(xy-4)=xy$$ Again at this step I can divide through by an $x$, giving us $$8xy^2+2xy-8=y$$

Now, we plug in what we have $y$ for in terms of x, giving us $$8x(\frac {12x^2} {4x^2 +12 -3x})^2+2x(\frac {12x^2} {4x^2 +12 -3x})-8=\frac {12x^2} {4x^2 +12 -3x}$$

Continuing our theme of not dealing with fractions, we're going to multipy through by $(4x^2-12x-3)^2$ to get

$$8x(12x^2)^2+24x^3(4x^2-12x-3)-8(4x^2-12x-3)^2=12x^2(4x^2-12x-3) $$

Multiply this out and we have $$1152x^5+96x^5-288x^4-72x^3-128x^4+768x^3-960x^2-576x-72=48x^4-144x^3-36x^2$$ Moving everything to the left hand side and combining powers of $x$, we have $$1248x^5-464x^4+840x^3-924x^2-582x-72=0$$ The only simplification that can be done here is dividing by 2, which still gives us a quintic, which has no easy way to solve. Wolframalpha shows is now showing 3 real solutions, none of which match

Answer 3

multiplying (2) by $x$ $(x \ne 0)$ we obtain $$\frac{1}{z}=\frac{1}{4}-\frac{1}{xy}$$ with (3) we obtain $$y=\frac{12(x^2-1)}{4x^2-3x}$$ (I) from (1) and (3) we get $$\frac{x^2y}{3}-x^2+1=\frac{xy}{4}$$ plugging (I) in this equation and simplifying we get $$-60 x^5+119 x^4+156 x^3-288 x^2-72 x+144=0$$ with five real solutions, no of them from (A) to (E) the system after the correction has the solution $$x=\frac{23}{10}$$

Answer 4

Choosing
$p = \frac{1}{x} $ $q = \frac{1}{y} $ $r = \frac{1}{z} $

will certainly help writing the equations more neatly, I guess. Then your equations can be written as:

$$2p + 2qr = 1 \\ 3q + 3pr = 1 \\ 4r + 4pq = 1$$

using first two equations you get:

$$ 2p + 2qr = 3q + 3pr \\ \implies r = \frac{3q - 2p}{3p - 2q}$$

Substitute in equations and solve as solved by Alan above.

Answer 5

Let $x=1/p$, $y=1/q$, and $z=1/r$. The three equations become

$$\begin{align} {1\over2}&=p+{qr\over q+r}={s\over q+r}\\ {1\over3}&=q+{rp\over r+p}={s\over r+p}\\ {1\over4}&=r+{pq\over p+q}={s\over p+q}\\ \end{align}$$

where

$$s=pq+qr+rp$$

Thus

$$\begin{align} 2s&=q+r\\ 3s&=r+p\\ 4s&=p+q\\ \end{align}$$

From this it follows that

$$(q+r)+(p+q)=2(r+p)\implies2q=r+p=3s\implies q={3\over2}s$$ and the others follow:

$$r=2s-q={1\over2}s\quad\text{and}\quad p=4s-q={5\over2}s$$

But now we have

$$s=pq+qr+rp={15\over4}s^2+{3\over4}s^2+{5\over4}s^2={23\over4}s^2$$

hence $s=4/23$, so $p=10/23$, and thus $x=23/10$.