# Is there an other way for it?

Consider two equations:

$$x+y=2$$
$$y+z=4$$

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 $$y=1 \Longrightarrow$$
$$x=1$$ ; $$z=3$$.
$$\therefore x+z = 1+3 = 4.$$

$$x+z=6$$
or
$$x+z=4$$

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

• Why does $x+y=2$ imply $y=1$ or $y=0$? – user239203 Jan 18 at 8:16
• i assumed it.....that's why i asked for a better way – Some 1 Jan 18 at 8:17
• $x+z$ is not determined. You have just two linear equations on three variables... – Henno Brandsma Jan 18 at 8:19
• i got it...thanks – Some 1 Jan 18 at 8:20
• 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) – 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