Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $u + 2v = 2, \;\; \left|2u - 3v\right| = 1$

Given system of equations: $\;$ $u + 2v = 2$ $\;\;\; \cdots \; (1)$

$\left|2u - 3v\right| = 1$ $\;\;\; \cdots \; (2)$

When $\;$ $2u - 3v > 0$, $\;$ i.e. $\;$ when $\;$ $2u > 3v$, $\;$ $\left|2u - 3v\right| = 2u - 3v$

Then equation $(2)$ becomes

$2u - 3v = 1$ $\;\;\; \cdots \; (3)$

Solving equations $(1)$ and $(3)$ simultaneously gives

$-7v = -3$ $\implies$ $v = \dfrac{3}{7}$

Substituting $v = \dfrac{3}{7}$ in equation $(1)$ gives

$u + \dfrac{6}{7} = 2$ $\;$ i.e. $\;$ $u = 2 - \dfrac{6}{7} = \dfrac{8}{7}$

Now, when $u = \dfrac{8}{7}$ and $v = \dfrac{3}{7}$,

$2u = \dfrac{16}{7}$, $3v = \dfrac{9}{7}$ and the condition $2u > 3v$ is satisfied.

When $\;$ $2u - 3v < 0$, $\;$ i.e. $\;$ when $\;$ $2u < 3v$, $\;$ $\left|2u - 3v\right| = -2u + 3v$

Then equation $(2)$ becomes

$-2u + 3v = 1$ $\;\;\; \cdots \; (4)$

Solving equations $(1)$ and $(4)$ simultaneously gives

$7v = 5$ $\implies$ $v = \dfrac{5}{7}$

Substituting $v = \dfrac{5}{7}$ in equation $(1)$ gives

$u + \dfrac{10}{7} = 2$ $\;$ i.e. $\;$ $u = 2 - \dfrac{10}{7} = \dfrac{4}{7}$

Now, when $u = \dfrac{4}{7}$ and $v = \dfrac{5}{7}$,

$2u = \dfrac{8}{7}$, $3v = \dfrac{15}{7}$ and the condition $2u < 3v$ is satisfied.

$\therefore \;$ The solution to the given pair of equations is $\;$ $\left(u, v\right) = \left\{\left(\dfrac{8}{7}, \dfrac{3}{7}\right), \; \left(\dfrac{4}{7}, \dfrac{5}{7}\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x + 2y - 6 = 0, \;\; \left|x - 3\right| - y = 0$

Given system of equations: $\;$ $x + 2y - 6 = 0$ $\;\;\; \cdots \; (1)$

$\left|x - 3\right| - y = 0$ $\;\;\; \cdots \; (2)$

When $\;$ $x - 3 > 0$, $\;$ i.e. $\;$ when $\;$ $x > 3$, $\;$ $\left|x - 3\right| = x - 3$

Then equation $(2)$ becomes

$x - 3 - y = 0$ $\;\;\; \cdots \; (3)$

Solving equations $(1)$ and $(3)$ simultaneously gives

$3x - 12 = 0$ $\implies$ $x = 4$

The condition for this solution is $\;$ $x > 3$

and $\;$ $x = 4 > 3$

$\implies$ $x = 4$ $\;$ is a valid solution.

Substituting the value of $x$ in equation $(1)$ gives

$4 + 2y - 6 = 0$ $\;$ i.e. $\;$ $2y = 2$ $\;$ i.e. $\;$ $y = 1$

When $\;$ $x - 3 < 0$, $\;$ i.e. $\;$ when $\;$ $x < 3$, $\;$ $\left|x - 1\right| = - \left(x - 3\right) = 3 - x$

Then $(2)$ becomes

$3 - x - y = 0$ $\;\;\; \cdots \; (4)$

Solving equations $(1)$ and $(4)$ simultaneously gives

$x = 0$

The condition for this solution is $\;$ $x < 3$

and $\;$ $x = 0 < 3$

$\implies$ $x = 0$ $\;$ is a valid solution.

Substituting the value of $x$ in equation $(1)$ gives

$2y - 6 = 0$ $\implies$ $y = 3$

$\therefore \;$ The solution to the given pair of equations is $\;$ $\left(x, y\right) = \left\{\left(4, 1\right), \; \left(0, 3\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\left|x - 1\right| + y = 0, \;\; 2x - y = 1$

Given system of equations: $\;$ $\left|x - 1\right| + y = 0$ $\;\;\; \cdots \; (1)$

$2x - y = 1$ $\;\;\; \cdots \; (2)$

When $\;$ $x - 1 > 0$, $\;$ i.e. $\;$ when $\;$ $x > 1$, $\;$ $\left|x - 1\right| = x - 1$

Then $(1)$ becomes

$x - 1 + y = 0$ $\;\;$ i.e. $\;$ $x + y = 1$ $\;\;\; \cdots \; (3)$

Solving equations $(2)$ and $(3)$ simultaneously gives

$3x = 2$ $\implies$ $x = \dfrac{2}{3}$

The condition for this solution is $\;$ $x > 1$

But $\;$ $x = \dfrac{2}{3} < 1$

$\implies$ $x = \dfrac{2}{3}$ $\;$ is not a valid solution.

When $\;$ $x - 1 < 0$, $\;$ i.e. $\;$ when $\;$ $x < 1$, $\;$ $\left|x - 1\right| = - \left(x - 1\right) = 1 - x$

Then $(1)$ becomes

$1 - x + y = 0$ $\;$ i.e. $\;$ $x - y = 1$ $\;\;\; \cdots \; (4)$

Solving equations $(2)$ and $(4)$ simultaneously gives

$x = 0$

The condition for this solution is $\;$ $x < 1$

and $\;$ $x = 0 < 1$

$\implies$ $x = 0$ $\;$ is a valid solution.

Substituting the value of $x$ in equation $(2)$ gives

$-y = 1$ $\implies$ $y = -1$

$\therefore \;$ The solution to the given pair of equations is $\;$ $\left(x, y\right) = \left\{\left(0, -1\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x + 3 \left|y\right| - 1 = 0, \;\; x + y + 3 = 0$

Given system of equations: $\;$ $x + 3 \left|y\right| - 1 = 0$ $\;\;\; \cdots \; (1)$

$x + y + 3 = 0$ $\;\;\; \cdots \; (2)$

When $\;$ $y > 0$, $\;$ $\left|y\right| = + y$

Then $(1)$ becomes

$x + 3y - 1 = 0$ $\;\;\; \cdots \; (3)$

Solving equations $(2)$ and $(3)$ simultaneously gives

$2y - 4 = 0$ $\implies$ $y = 2$

Substituting $\;$ $y = 2$ $\;$ in equation $(2)$ gives

$x + 2 + 3 = 0$ $\implies$ $x = -5$

When $\;$ $y < 0$, $\;$ $\left|y\right| = -y$

Then $(1)$ becomes

$x - 3y - 1 = 0$ $\;\;\; \cdots \; (4)$

Solving equations $(2)$ and $(4)$ simultaneously gives

$-4y - 4 = 0$ $\implies$ $y = -1$

Substituting the value of $y$ in equation $(2)$ gives

$x - 1 + 3 = 0$ $\implies$ $x = -2$

$\therefore \;$ The required solution is $\;$ $\left(x, y\right) = \left\{\left(-5, 2\right), \; \left(-2, -1\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $y + x - 1 = 0, \; \left|y\right| - x - 1 = 0$

Given system of equations: $\;$ $y + x - 1 = 0$ $\;\;\; \cdots \; (1)$

$\left|y\right| - x - 1 = 0$ $\;\;\; \cdots \; (2)$

When $\;$ $y > 0$, $\;$ $\left|y\right| = + y$

Then $(2)$ becomes

$y - x - 1 = 0$ $\;\;\; \cdots \; (3)$

Adding equations $(1)$ and $(3)$ gives

$2y - 2 = 0$ $\implies$ $y = 1$

Substituting $\;$ $y = 1$ $\;$ in equation $(1)$ gives

$x = 1 - 1 = 0$

When $\;$ $y < 0$, $\;$ $\left|y\right| = -y$

Then $(2)$ becomes

$-y - x - 1 = 0$

i.e. $\;$ $x + y = -1$ $\;\;\; \cdots \; (4)$

From equation $(1)$, $\;$ $x + y = 1$ $\;\;\; \cdots \; (1a)$

$\therefore \;$ When $\;$ $\left|y\right| = - y$, $\;$ we see from equations $(1a)$ and $(4)$ that equations $(1)$ and $(4)$ cannot be solved simultaneously.

$\therefore \;$ The required solution is $\;$ $\left(x, y\right) = \left\{\left(0, 1\right) \right\}$

Algebra - Equations and Inequations

Solve the equation: $\;$ $\left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right) = 15$

Given equation: $\;$ $\left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right) = 15$

i.e. $\;$ $\left(x^2 - 5x + 4\right) \left(x^2 - 5x + 6\right) = 15$ $\;\;\; \cdots \; (1)$

Let $\;$ $x^2 - 5x = p$ $\;\;\; \cdots \; (2)$

Then $(1)$ becomes

$\left(p +4\right) \left(p + 6\right) = 15$

i.e. $\;$ $p^2 + 10p + 24 = 15$

i.e. $\;$ $p^2 + 10p + 9 = 0$

i.e. $\;$ $\left(p + 9\right) \left(p + 1\right) = 0$

i.e. $\;$ $p = -9$ $\;$ or $\;$ $p = -1$

When $\;$ $p = -9$, $(2)$ becomes

$x^2 - 5x = -9$

i.e. $\;$ $x^2 - 5x + 9 = 0$ $\;\;\; \cdots \; (3)$

Discriminant of $(3)$ is

$\Delta = \left(-5\right)^2 - 4 \times 1 \times 9 = = 25 - 36 = - 9 < 0$

$\implies$ $(3)$ does not have any real roots.

When $\;$ $p = -1$, $\;$ $(2)$ becomes

$x^2 - 5x = -1$

i.e. $\;$ $x^2 - 5x + 1 = 0$ $\;\;\; \cdots \; (4)$

Solving $(4)$ for $x$ gives

$x = \dfrac{5 + \sqrt{21}}{2}$ $\;$ or $\;$ $x = \dfrac{5 - \sqrt{21}}{2}$

$\therefore \;$ Solution of the given equation is

$x = \left\{\dfrac{5 + \sqrt{21}}{2}, \; \dfrac{5 - \sqrt{21}}{2} \right\}$

Algebra - Equations and Inequations

Solve the equation: $\;$ $\left(x - \sqrt{3}\right)^4 - 5 \left(x - \sqrt{3}\right)^2 + 4 = 0$

Given equation: $\;$ $\left(x - \sqrt{3}\right)^4 - 5 \left(x - \sqrt{3}\right)^2 + 4 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $\left(x - \sqrt{3}\right)^2 = p$ $\;\;\; \cdots \; (2)$

Then $(1)$ becomes

$p^2 - 5p + 4 = 0$

i.e. $\;$ $\left(p - 4\right) \left(p - 1\right) = 0$

i.e. $\;$ $p = 4$ $\;$ or $\;$ $p = 1$

When $\;$ $p = 4$, $\;$ $(2)$ becomes

$\left(x - \sqrt{3}\right)^2 = 4$

i.e. $\;$ $x - \sqrt{3} = \pm 2$

i.e. $\;$ $x = \sqrt{3} + 2$ $\;$ or $\;$ $x = \sqrt{3} - 2$

When $\;$ $p = 1$, $\;$ $(2)$ becomes

$\left(x - \sqrt{3}\right)^2 = 1$

i.e. $\;$ $x - \sqrt{3} = \pm 1$

i.e. $\;$ $x = \sqrt{3} + 1$ $\;$ or $\;$ $x = \sqrt{3} - 1$

$\therefore \;$ The solutions of $(1)$ are

$x = \left\{\sqrt{3} + 2, \sqrt{3} - 2, \sqrt{3} + 1, \sqrt{3} - 1 \right\}$