absolute valuealgebra-precalculus
Question
The number of integral ordered pairs ($x,y$) satisfying the system of given equations
$|x+y-4| = 5$ ………(i)and
$|x-3|+|y-1| = 5$ …..(ii) is/are
(A)2
(B)4
(C)6
(D)12
My attempt
I first used $x+y-4=5$ as one equation and $x+y-=-5$ as other and similarly obtained four equations from equation (ii) but could not get to the answer please tell me if the method is wrong itself or maybe the last step miscalculations are the reasons of not getting the answer
If $a=3$ (orange line) then we have equality and $x=0$ which is a right boundary case. Hence we need $a<3$.
If $\displaystyle a=-\frac{13}{4}$ (violet line) then we have the left boundary where equality holds for $x<0$. So, we have $\displaystyle a>-\frac{13}{4}$ for inequality.
Observe that
$$ y^ 8 - x^8 = (y - x) + (y^ 7 - x^7) + xy (x^6 - y^6)$$
$\qquad$ If $x < y < 0 $, then the LHS is negative, but each term on the RHS is positive. Hence contradiction. Likewise if $ y < x < 0 $.
Note: A very common misconception when solving symmetric system of equations is thinking that all solutions to $x = f(y), y = f(x)$ are of the form $ y = x$.
In fact, what we truly require is $ f(f(x)) = x, y = f(x)$.
This is most obvious when looking at self-inverse functions like $f(x) = x, f(x) = -x, f(x) = \frac{1}{x}$. However, it's not about the function, but about the "self-inverse" point, namely $f(x) = f^{-1} (x)$.
Best Answer
Rewrite the first equation $|x+y-4| = 5$ as
$x = - y - 1$
or
$x = 9 - y$
and rewrite the second equation $|x-3|+|y-1| = 5$ as
$x = |y-1| - 2,\:\:\:\:$ $|y-1|<5$
or
$x = 8 - |y-1|,\:\:\:\:$ $|y-1|<5$
or
$x = 3\:$ and $\:|y-1|=5$
and work out the solution from there.