kyotsu-test 2012 QCourse1-I-Q2

kyotsu-test · Japan · eju-math__session2 Modulus function Solving equations involving modulus
Consider the following equation in $x$
$$|ax - 11| = 4x - 10, \tag{1}$$
where $a$ is a constant.
(1) Equation (1) can be rewritten without using the absolute value symbol as
$$\begin{aligned} & \text{when } ax \geqq 11, \quad \text{then } (a - \mathbf{N})x = \mathbf{O}; \\ & \text{when } ax < 11, \quad \text{then } (a + \mathbf{P})x = \mathbf{QR}. \end{aligned}$$
(2) When $a = \sqrt{7}$, the solution of equation (1) is
$$x = \frac{\mathrm{S}}{\mathrm{T}} - \sqrt{\mathrm{U}}.$$
(3) Let $a$ be a positive integer. When equation (1) has a positive integral solution, we have $a = \mathbf{W}$, and that solution $x = \mathbf{X}$.
Consider the following equation in $x$

$$|ax - 11| = 4x - 10, \tag{1}$$

where $a$ is a constant.

(1) Equation (1) can be rewritten without using the absolute value symbol as

$$\begin{aligned}
& \text{when } ax \geqq 11, \quad \text{then } (a - \mathbf{N})x = \mathbf{O}; \\
& \text{when } ax < 11, \quad \text{then } (a + \mathbf{P})x = \mathbf{QR}.
\end{aligned}$$

(2) When $a = \sqrt{7}$, the solution of equation (1) is

$$x = \frac{\mathrm{S}}{\mathrm{T}} - \sqrt{\mathrm{U}}.$$

(3) Let $a$ be a positive integer. When equation (1) has a positive integral solution, we have $a = \mathbf{W}$, and that solution $x = \mathbf{X}$.