Consider two equations:


Find the value of $(x+z)$.
($x,y,z$ all are positive real numbers)

My Approach:
$\because x+y=2 \Longrightarrow y=0$ or $y=1$.

Case 1:
taking $y=0$$ \Longrightarrow$
$x=2$ ; $z=4$.
$\therefore x+z = 2+4 = 6.$

Case 2:
taking $y=1 \Longrightarrow$
$x=1$ ; $z=3$.
$\therefore x+z = 1+3 = 4.$


I want to know if my approach is correct or is there more better way to evaluate this ?

  • 2
    $\begingroup$ Why does $x+y=2$ imply $y=1$ or $y=0$? $\endgroup$ – user239203 Jan 18 at 8:16
  • $\begingroup$ i assumed it.....that's why i asked for a better way $\endgroup$ – Some 1 Jan 18 at 8:17
  • 1
    $\begingroup$ $x+z$ is not determined. You have just two linear equations on three variables... $\endgroup$ – Henno Brandsma Jan 18 at 8:19
  • $\begingroup$ i got it...thanks $\endgroup$ – Some 1 Jan 18 at 8:20
  • 2
    $\begingroup$ You also seem to be assuming that x, y, and z are natural numbers, since you only consider cases 0 and 1, but the problem says they can be real numbers, which include fractions and irrational numbers as well. There are in fact infinitely many possible solutions for (x + z), with the answer stated in terms of y, such that whatever y you choose will determine the value for (x+z) $\endgroup$ – epte Jan 18 at 8:24

Adding the equations gives $x+2y+z= 6$ or $x+z=6-2y$. So for every $y$ you choose you'll get a different sum and all values $\le 6$ can be assumed. So taking these two specific ones proves or shows nothing. You can just say that $x+z \le 6$ as $y\ge 0$.


Given $\qquad x+y=2\space\land\space y+z=4\qquad$ we can start by eliminating $y$. $$ y+z=4\implies z=4-y\\ (x+y=2)\implies x=2-y\\ z-x=(4-y)-(2-y)=2\qquad\longrightarrow (z-x=2)\\ (x+z=4)-(z-x=2)\implies 2x=2\implies x=1\\ (x+z=4)+(z-x=2)\implies 2z=6\implies z=3\\ \therefore\quad x+z=1+3=4 $$ At this point we can also see that $y=1$ but who cares? Ha ha


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