Solve the following system of equations, both by calculation and graphically.

$$
|x-1|+|x+y|=6
$$

$$
|y-1|+|x+y+1|=4
$$
I. In the graphical solution, we separately produce the images of equations (1) and (2) in the $x, y$ Cartesian coordinate system, then read off the coordinates of the common points of the two images. The absolute value is non-negative, so one of the terms on the left side of (1) cannot be greater than 6, and the terms on the left side of (2) cannot be greater than 4.

Let $x-1 \geqq 0$, that is, $0 \leqq x-1 \leqq 6, 1 \leqq x \leqq 7$. Depending on whether

$$
x+y \geqq 0, \quad \text { or } \quad x+y \leqq 0
$$

(1) becomes:

$$
2 x+y-7=0, \quad \text { or } \quad(x-1)-(x+y)-6=-y-7=0
$$

From the first, substituting the lower and upper limits of $x$, when $x=1$, $y=5$, and when $x=7$, $y=-7$, thus under the condition $x-1 \geqq 0$, (1) is satisfied by the points on the segments $P_{1} P_{2}$ and $P_{2} P_{3}$, where

$$
P_{1}(1,5), \quad P_{2}(7,-7), \quad P_{3}(1,-7)
$$

Similarly, for $x-1 \leqq 0$, more precisely $-5 \leqq x \leqq 1$, under the further assumption that

$$
x+y \geqq 0, \quad \text { or } \quad x+y \leqq 0
$$

from the equations $y=5$ and $y=-5-2 x$, we get the segments $P_{4} P_{1}$ and $P_{3} P_{4}$, where $P_{4}(-5, -5)$. Therefore, the image of (1) is the perimeter of the quadrilateral $P_{1} P_{2} P_{3} P_{4}$ (which is clearly a parallelogram).

![](https://cdn.mathpix.com/cropped/2024_05_02_8db03fb91b052bae9ca9g-1.jpg?height=772&width=767&top_left_y=980&top_left_x=668)

Similarly, the image of (2) is the perimeter of the parallelogram $Q_{1} Q_{2} Q_{3} Q_{4}$, where $Q_{1}(-6,5), Q_{2}(2,1), Q_{3}(2,-3), Q_{4}(-6,1)$, and the sides are parts of the lines with equations $x+2 y-4=0, x=2, x+2 y+4=0, x=-6$. Therefore, the solutions to the system of equations are given by the coordinates of the two intersection points of the perimeters of the two parallelograms:

$$
\begin{gathered}
M_{1} \text { gives: } x=-2, y=-1 \\
M_{2} \text { gives: } x \approx-4.7, y \approx 4.3 .
\end{gathered}
$$

II. The solution can also be found by calculation, by considering every possible sign of the expressions within the absolute values and checking whether the roots obtained satisfy these conditions. This would involve checking 12 cases, taking into account the inequality $x+y < x+y+1$. However, we can reduce the number of cases by first considering only the signs of $x-1$ and $y-1$ and finding an equation for $x+y=a$ in each case.
a) $x \geqq 1, y \geqq 1$. In this case, $x+y \geqq 2$, so all the terms within the absolute value signs are positive:

$$
\begin{aligned}
& 2 x+y=7 \\
& x+2 y=7
\end{aligned}
$$

The solution to the system of equations is the point $R_{1}\left(\frac{10}{1}, \frac{1}{3}\right)$, which does not satisfy the condition $y \geqq 1$.
b) $x \geqq 1, y<1$. Taking the absolute value of the known terms, we get the system of equations

$$
\begin{gathered}
|x+y|=7-x \\
|x+y+1|=3+y
\end{gathered}
$$

Subtracting the first from the second,

$$
|x+y+1|-\mid x+y=(x+y)-4
$$

Introducing the notation $a=x+y$, we get the equation

$$
|a+1|-|a|=a-4
$$

The value of the left side is between -1 and +1, so $a$ can only be in the interval $(3,5)$, and $a$ and $a+1$ are positive:

$$
(a+1)-a=a-4
$$

Thus, $a=x+y=5$, substituting this into (3) gives the point $R_{2}(2,3)$, which does not satisfy the condition $y<1$.
c) $x<1, y \geqq 1$. Taking the absolute value of the known terms,

$$
\begin{gathered}
|x+y|=5+x \\
|x+y+1|=5-y
\end{gathered}
$$

The difference of these two equations, using the notation $a=x+y$, leads to the equation

$$
|a|-|a+1|=a
$$

The value of the left side, and thus $a$, is between -1 and +1, so the value of $a+1$ is in the interval $(0,2)$, and its sign is known:

$$
|a|=2 a+1
$$

$a$ cannot be positive, because then $a=2 a+1$, so $a<0$; therefore, $3 a+1=0, a=-1 / 3$. Substituting this into (4) gives the point $M_{2}\left(-\frac{14}{3}, \frac{13}{3}\right)$, which satisfies our assumptions and is a solution.
d) $x<1, y<1$, similarly to the above,

$$
\begin{gathered}
|x+y|=5+x \\
|x+y+1|=3+y
\end{gathered}
$$

The sum of these two equations leads to the equation

$$
|a|+|a+1|=a+8
$$

Since $x+y<2, a<2$. If $-1 \leqq a \leqq 2$, then the left side is at most 5, and the right side is at least 7, so only $a<-1$ is possible: $-2 a-1=a+8$, thus $a=-3$, and from (5) we get the point $M_{1}(-2,-1)$, which satisfies the conditions.