Example 13. Solve the system of equations

$$
\left\{\begin{array}{l}
|x+y-4|=5 \\
|x-3|+|y-1|=5
\end{array}\right.
$$
Solution. The given system is equivalent to the system

$$
\left\{\begin{array}{l}
|x+y-4|=5 \\
|x-3|+|y-1|=5 \\
|x+y-4|=|x-3|+|y-1|
\end{array}\right.
$$

which (since $x+y-4=(x-3)+(y-1)$) by statement 2 is equivalent to the mixed system

$$
\left\{\begin{array}{l}
|x+y-4|=5 \\
|x-3|+|y-1|=5 \\
(x-3)(y-1) \geqslant 0
\end{array}\right.
$$

The last system is equivalent to the following system of systems:

$$
\left\{\begin{array} { l } 
{ | x + y - 4 | = 5 , } \\
{ | 3 - x | + | y - 1 | = 5 , } \\
{ x - 3 \geqslant 0 , } \\
{ y - 1 \geqslant 0 , }
\end{array} \quad \left\{\begin{array}{l}
|x+y-4|=5 \\
|x-3|+|y-1|=5 \\
x-3 \leqslant 0 \\
y-1 \leqslant 0
\end{array}\right.\right.
$$

The first system of this conjunction is equivalent to the system

$$
\left\{\begin{array} { l } 
{ x + y - 4 = 5 } \\
{ ( x - 3 ) + ( y - 1 ) = 5 , } \\
{ x - 3 \geqslant 0 } \\
{ y - 1 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+y=9 \\
x \geqslant 3 \\
y \geqslant 1
\end{array}\right.\right.
$$

all solutions of which can be written in the form $(t ; 9-t)$, where $t$ is any number from the interval $[3 ; 8]$.

The second system of the conjunction is equivalent to the system

$$
\left\{\begin{array}{l}
x+y=-1 \\
x \leqslant 3 \\
y \leqslant 1
\end{array}\right.
$$

the set of solutions of which consists of numerical pairs ( $a ;-1-a$ ), where $a$ is any number from the interval $[-2 ; 3]$.

Thus, the solutions of the system (19) are all possible numerical pairs

$$
\begin{aligned}
& (t ; 9-t), \quad \text { where } t \in[3 ; 8] ; \\
& (a ;-1-a), \text { where } a \in[-2 ; 3] .
\end{aligned}
$$

Geometric interpretation of the system (19). If a rectangular coordinate system $X O Y$ is given on the plane, then (Fig. 2.5):

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-046.jpg?height=731&width=965&top_left_y=1925&top_left_x=548)

Fig. 2.5
a) the set of all points on the line $y=-x+9$ (i.e., points $(t ; 9-t)$, where $t \in(-\infty ;+\infty)$) and the set of all points on the line $y=-x-1$ (i.e., points $(a ;-1-a)$, where $a \in(-\infty ;+\infty)$) is the set of all solutions of the first equation of the system (19);

b) the set of all points of the square $A B C D$, including its vertices $A(3 ; 6), B(8 ; 1), C(3 ;-4)$ and $D(-2 ; 1)$ (i.e., points $(t ; 9-t)$, where $t \in[3 ; 8]$ - points on side $A B$; points ( $\alpha ; \alpha-7$ ), where $\alpha \in[3 ; 8]$ - points on side $B C$; points $(a ;-1-a)$, where $a \in[-2 ; 3]$ - points on side $C D$; points $(\beta ; \beta+3)$, where $\beta \in[-2 ; 3]$ - points on side $D A$ ), is the set of all solutions of the second equation of the system (19);

c) the set of all points on side $A B$ of the square $A B C D$, lying on the line $y=-x+9$, and the set of all points on side $C D$, lying on the line $y=-x-1$ (i.e., points $(t ; 9-t)$, where $t \in[3 ; 8]$, and points ( $a ;-1-a$ ), where $a \in[-2 ; 3]$ ), is the set of all solutions of the system (19).

## EXERCISE 1

Solve the equation:

1) $|x-1|=3 ; 2)|x|=-3 x-5$
2) $x^{2}+3 x+|x+3|=0$; 4) $|x-1|+|x-3|=2$;
3) $|x-1|+|x-2|+|x-3|=2$.

## EXERCISE 2

Solve the equation:

1) $|x-7|=2$
2) $\left(x^{2}-5 x+6\right)^{2}-5\left|x^{2}-5 x+6\right|+6=0$;
3) $\left|x^{2}-4 x+3\right|=-(4+2 \sqrt{\overline{3}}) x$;
4) $|x-1|+|x-3|=3$
5) $\left|x^{2}-4 x+3\right|+\left|x^{2}-5 x+6\right|=1$.

## EXERCISE 3

1. Solve the equation:
1) $|3 x-5|=|5-2 x|$
2) $2 x^{2}-5|x|+3=0$
3) $\frac{\left|x^{2}-4 x\right|+3}{x^{2}+|x-5|}=1$;
4) $x^{2}-6 x+|x-4|+8=0$.
2. Find all roots of the equation

$$
\left|x^{2}+x-1\right|=2 x-1
$$

satisfying the inequality $x0$.

4. Find all solutions of the system of equations

$$
\left\{\begin{array}{l}
|x+1 / y|+|10 / 3-x+y|=10 / 3+y+1 / y \\
x^{2}+y^{2}=82 / 9
\end{array}\right.
$$

satisfying the condition $x>0, y>0$.

5. For each value of $a$ find all $x$ satisfying the condition:
1) $|i+3|-a|x-1|=4 ; \quad 2)|x-2|+a|x+3|=5$;
2) $a|x+3|+2|x+4|=2$;
3) $3|x-2|-a|2 x+3|=21 / 2$.
6. Find all solutions of the equation

$$
1+x+\left|x^{2}-x-3\right|=0
$$

satisfying the inequality $x+\sqrt{14} / 3>0$.

7. Find all values of $a$ for which the equation has a unique root:
1) $|1-a x|=1+(1-2 a) x+a x^{2}$;
2) $|(a+1) x-2|=(1+a) x^{2}-2 a x+2$.
8. Find all values of $a$ for which there exists only one value of $x$ satisfying the system of equations:
1) $\left\{\left|x^{2}-5 x+4\right|-9 x^{2}-5 x+4+10 x|x|=0\right.$, $\left\{x^{2}-2(a-1) x+a(a-2)=0\right.$ 2) $\left\{\begin{array}{l}\left|x^{2}-7 x+6\right|+x^{2}+5 x+6-12|x|=0, \\ x^{2}-2(a-2) x+a(a-4)=0 .\end{array}\right.$
9. Find all values of $a$ for which the equation:
1) $\left.x-a=2|2| x\left|-a^{2}\right| ; \quad 2\right) x-a / 2=4|4| x\left|-a^{2}\right|$;
2) $x-a / 3=9|9| x\left|-a^{2}\right| ; \quad$ 4) $x-a / 2=2|2| x\left|-a^{2}\right|$ has three distinct roots. Find these roots.
10. Find all values of $a$ for which the equation has exactly two distinct roots:
1) $x|x+2 a|+1-a=0$
2) $x^{2}+4 x-2|x-a|+2-a=0$.
11. Find all values of $a$ for which the system of equations

$$
\left\{\begin{array}{l}
y(a x+1)+13 x-a(1+y)=0 \\
x-x y+|2+y|=0
\end{array}\right.
$$

has a solution.

## § 2. Irrational Equations

In this section, we consider equations containing the unknown under the radical sign. Note that

1. All roots of even degree, entering the equation, are arithmetic. In other words, if the radicand is negative, the root is meaningless; if the radicand is zero, the root is also zero; if the radicand is positive, the value of the root is positive.
2. All roots of odd degree, entering the equation, are defined for any real value of the radicand. In this case, the root