Solve the following equation:

$$
|2 x-3|=\frac{6 x+17}{11}
$$
I. solution. On the left side, we can omit the absolute value sign if $2 x-3 \geq 0$, that is, $x \geq 3 / 2$, and if $x<3 / 2$, then $|2 x-3|=-(2 x-3)=3-2 x$.

Thus, the equation

$$
2 x-3=\frac{6 x+17}{11}
$$

has a root that satisfies if it is not less than 3/2, so the number $x_{1}=25 / 8$ is a solution to the equation; in addition, the

$$
3-2 x=\frac{6 x+17}{11}
$$

equation has a root if it is less than 3/2. This provides the root $x_{1}=4 / 7$.

Since we have considered every sign possibility, there are no more solutions.

Csaba Zsuzsanna (Celldömölk, Berzsenyi D. g. II. o. t.)

II. solution. We can solve our problem by the method of graphing the equations. It is sufficient to graph the expressions on both sides for the values where the right side is not negative, that is, if $x \geq -17 / 6$, because the left side cannot be negative.

![](https://cdn.mathpix.com/cropped/2024_05_02_0e7aa441db5f3027f05fg-1.jpg?height=534&width=765&top_left_y=823&top_left_x=669)

Unnamed submission

We note that it would be difficult to read the value of the smaller root precisely from the graph.

III. solution: The squares of equal numbers are also equal. The square of the left side is, regardless of the sign of $2 x-3$, $(2 x-3)^{2}$. Thus, from the equation

$(2 x-3)^{2}=\frac{(6 x+17)^{2}}{121}$, or $56 x^{2}-207 x+100=0$, $x$ cannot be anything other than $4 / 7$ and $25 / 8$. Verification shows that both values are solutions to the given equation.

Héray Tibor (Bp. IX., Fáy A. g. I. o. t.)