Solution to Math problem #245:
Employing Cramer's rule, we note first that since
$$\Delta_x = \left|\begin{array}{cc} 12 & - 8 \\ 15 & - 6 \end{array}\right|\not= 0$$
the system does not have solutions only if
$$\Delta = \left|\begin{array}{cc} a & - 8 \\ 2 & - 6 \end{array}\right| = - 6a + 16 = 0,$$
that is when $a = 8/3$.
Consider the following system of linear equations
$$\begin{cases}a_1x + b_1y = c_1, \\ a_2x + b_2y = c_2,\end{cases}\quad\quad (1)$$
where $a_1^2 + b_1^2 \not=0$, $a_2^2 + b_2^2 \not=0$. Next, let us define the following three determinants
$$\Delta = \left|\begin{array}{cc} a_1 & b_1 \\ a_2 & b_2 \end{array}\right| = a_1b_2 - a_2b_1;$$
$$\Delta_x = \left|\begin{array}{cc} c_1 & b_1 \\ c_2 & b_2 \end{array}\right| = c_1b_2 - c_2b_1;$$
$$\Delta_y = \left|\begin{array}{cc} a_1 & c_1 \\ a_2 & c_2 \end{array}\right| = a_1c_2 - a_2c_1.$$
Then the system (1) has a unique solution if and only if the determinant $\Delta \not=0$. In this case the solution is given by the formulas
$$x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta},$$
which are called Cramer's formulas, or Cramer's rule. If all of the coefficients $a_1$, $b_1$, $a_2$ and $b_2$ are not equal to zero, then the condition $\Delta \not=0$ is equivalent to
$$\frac{a_1}{a_2} \not= \frac{b_1}{b_2}.$$
Conversely, the system (1) has no solutions if and only if $\Delta = 0$ and at least one of the determinants $\Delta_x$ and $\Delta_y$ is not equal to zero.
If all of the coefficients $a_1$, $b_1$, $a_2$ and $b_2$ are not equal to zero, then the condition $\Delta = 0$, $\Delta_x \not=0$ (or, $\Delta = 0$, $\Delta_y \not=0$) is equivalent to the following condition
$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \not=\frac{c_1}{c_2}.$$
The system (1) has infinitely many solutions if and only if $\Delta = \Delta_x = \Delta_y = 0.$ If all of the coefficients $a_1$, $b_1$, $a_2$ and $b_2$ are not equal to zero, then the condition $\Delta = \Delta_x = \Delta_y = 0$ is equivalent to the following condition
$$\frac{a_1}{a_2} = \frac{b_1}{b_2} =\frac{c_1}{c_2}.$$
If the conditions $a_1^2 + b_1^2 \not= 0$, $a_2^2 + b_2^2 \not=0$ are not satisfied, then $\Delta = \Delta_x = \Delta_y = 0$ may not imply that the system (1) has infinitely many solutions. For example, all of the three determinants of the linear system
$$\begin{cases} 0\cdot x + 0\cdot y = 4, \\ 0\cdot x + 0\cdot y = 15\end{cases}$$
vanish, but the system has no solutions.
Math problem: Find all values of the parameter $a$, for which the linear system
$$\begin{cases} ax - 8x = 12, \\ 2x - 6y = 15 \end{cases}$$
has no solutions.
Discussion and hints: Investigate the coefficient matrix of the system, employ Cramer's rule.
Solution
Solution to Math problem #244:
Since $(x^2 - x) - (2-x) = x^2 - 2$, the inequality in question can be rewritten as
$$|x^2 - x| > |x^2 - x| - |2 - x|.$$
In view of the basic algebraic inequality BI2, we conclude that the set of solution to the given inequality coincides with the set of solutions to the following inequality
$$(2-x)(x^2 - x) < 0,$$
or,
$$(x-2)(x - \sqrt{2})(x + \sqrt{2}) > 0.$$
The last inequality holds true for $ - \sqrt{2} < x < \sqrt{2}$ and $x > 2$, which is also the solution to the original inequality.
Math problem: Solve the following inequality
$$|x^2 - x| < |2-x| + |x^2 - 2|.$$
Discussion and hints: Employ the Basic algebraic inequalities as appropriate.
Solution
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