Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\sqrt{\dfrac{x}{y}} + \sqrt{\dfrac{y}{x}} = \dfrac{5}{2}$, $\;$ $x^2 + y^2 = 15$

Given system of equations:

$\sqrt{\dfrac{x}{y}} + \sqrt{\dfrac{y}{x}} = \dfrac{5}{2}$ $\;\;\; \cdots \; (1)$, $\;\;$ $x^2 + y^2 = 15$ $\;\;\; \cdots \; (2)$

Let $\;$ $\sqrt{\dfrac{x}{y}} = p$ $\;\;\; \cdots \; (3a)$

Then $\;$ $\sqrt{\dfrac{y}{x}} = \dfrac{1}{p}$ $\;\;\; \cdots \; (3b)$

In view of equations $(3a)$ and $(3b)$, equation $(1)$ becomes

$p + \dfrac{1}{p} = \dfrac{5}{2}$

i.e. $\;$ $p^2 + 1 = \dfrac{5}{2} p$

i.e. $\;$ $2p^2 - 5p + 2 = 0$

i.e. $\;$ $\left(p - 2\right) \left(2p - 1\right) = 0$

i.e. $\;$ $p = 2$ $\;\;$ or $\;\;$ $p = \dfrac{1}{2}$

Substituting the values of $p$ in equation $(3a)$ give

Case 1:

When $\;$ $p = 2$, $\;\;\;$ then $\;\;$ $\sqrt{\dfrac{x}{y}} = 2$

i.e. $\;$ $\dfrac{x}{y} = 4$

i.e. $\;$ $x = 4y$ $\;\;\; \cdots \; (4a)$

Substituting $\;$ $x = 4y$ $\;$ in equation $(2)$ gives

$\left(4y\right)^2 + y^2 = 15$

i.e. $\;$ $17 y^2 = 15$ $\implies$ $y = \pm \sqrt{\dfrac{15}{17}}$

Substituting the value of $y$ in equation $(4a)$ gives $\;\;\;$ $x = \pm 4 \sqrt{\dfrac{15}{17}}$

Case 2:

When $\;$ $p = \dfrac{1}{2}$, $\;\;\;$ then $\;\;$ $\sqrt{\dfrac{x}{y}} = \dfrac{1}{2}$

i.e. $\;$ $\dfrac{x}{y} = \dfrac{1}{4}$

i.e. $\;$ $y = 4x$ $\;\;\; \cdots \; (4b)$

Substituting $\;$ $y = 4x$ $\;$ in equation $(2)$ gives

$x^2 + \left(4x\right)^2 = 15$

i.e. $\;$ $17 x^2 = 15$ $\implies$ $x = \pm \sqrt{\dfrac{15}{17}}$

Substituting the value of $x$ in equation $(4b)$ gives $\;\;\;$ $y = \pm 4 \sqrt{\dfrac{15}{17}}$

$\therefore \;$ The solution to the given system of equations is

$\left(x, y\right) =$
$ \left\{\left(\dfrac{4 \sqrt{15}}{\sqrt{17}}, \dfrac{\sqrt{15}}{\sqrt{17}}\right), \left(\dfrac{-4 \sqrt{15}}{\sqrt{17}}, \dfrac{-\sqrt{15}}{\sqrt{17}}\right), \left(\dfrac{\sqrt{15}}{\sqrt{17}}, \dfrac{4 \sqrt{15}}{\sqrt{17}}\right), \left(\dfrac{- \sqrt{15}}{\sqrt{17}}, \dfrac{-4 \sqrt{15}}{\sqrt{17}}\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\sqrt{\dfrac{2x - 1}{y + 2}} + \sqrt{\dfrac{y + 2}{2x - 1}} = 2$, $\;$ $x + y = 2$

Given system of equations:

$\sqrt{\dfrac{2x - 1}{y + 2}} + \sqrt{\dfrac{y + 2}{2x - 1}} = 2$ $\;\;\; \cdots \; (1)$, $\;\;$ $x + y = 2$ $\;\;\; \cdots \; (2)$

We have from equation $(2)$, $\;\;\;$ $x = 2 - y$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(1)$ becomes

$\sqrt{\dfrac{2 \left(2 - y\right) - 1}{y + 2}} + \sqrt{\dfrac{y + 2}{2 \left(2 - y\right) - 1}} = 2$

i.e. $\;$ $\sqrt{\dfrac{3 - 2y}{y + 2}} + \sqrt{\dfrac{y + 2}{3 - 2y}} = 2$

i.e. $\;$ $\dfrac{3 - 2y + y + 2}{\sqrt{\left(y + 2\right) \left(3 - 2y\right)}} = 2$

i.e. $\;$ $5 - y = 2 \sqrt{6 - y - 2y^2}$ $\;\;\; \cdots \; (4)$

Squaring equation $(4)$ gives

$25 - 10y + y^2 = 24 - 4y -8y^2$

i.e. $\;$ $9y^2 -6y + 1 = 0$

i.e. $\;$ $\left(3y - 1\right)^2 = 0$

i.e. $\;$ $3y - 1 = 0$

i.e. $\;$ $y = \dfrac{1}{3}$

Substituting $\;$ $y = \dfrac{1}{3}$ $\;$ in equation $(3)$ gives $\;\;$ $x = 2 - \dfrac{1}{3} = \dfrac{5}{3}$

$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(\dfrac{5}{3}, \dfrac{1}{3}\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\sqrt{\dfrac{x}{y}} + \sqrt{\dfrac{y}{x}} = \dfrac{5}{2}$, $\;$ $x + y = 5$

Given system of equations:

$\sqrt{\dfrac{x}{y}} + \sqrt{\dfrac{y}{x}} = \dfrac{5}{2}$ $\;\;\; \cdots \; (1)$, $\;\;$ $x + y = 5$ $\;\;\; \cdots \; (2)$

We have from equation $(1)$,

$\dfrac{x + y}{\sqrt{xy}} = \dfrac{5}{2}$

i.e. $\;$ $\dfrac{5}{\sqrt{xy}} = \dfrac{5}{2}$ $\;\;\;$ [in view of equation $(2)$]

i.e. $\;$ $\dfrac{1}{\sqrt{xy}} = \dfrac{1}{2}$

i.e. $\;$ $\sqrt{xy} = 2$

i.e. $\;$ $xy = 4$ $\implies$ $x = \dfrac{4}{y}$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(2)$ becomes

$\dfrac{4}{y} + y = 5$

i.e. $\;$ $y^2 - 5y + 4 = 0$

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

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

When $\;$ $y = 1$, $\;$ we have from equation $(2)$, $\;$ $x = 5 - 1 = 4$

When $\;$ $y = 4$, $\;$ we have from equation $(2)$, $\;$ $x = 5 - 4 = 1$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations and investigate them with respect to $a$: $\;$ $2x + 3y = 5$, $\;$ $x - y = 2$, $\;$ $x + 4y = a$

Given system of equations:

$2x + 3y = 5$ $\;\;\; \cdots \; (1)$, $\;\;$ $x - y = 2$ $\;\;\; \cdots \; (2)$, $\; \;$ $x + 4y = a$ $\;\;\; \cdots \; (3)$

Multiplying equation $(2)$ with $3$ gives

$3x - 3y = 6$ $\;\;\; \cdots \; (4)$

Adding equations $(1)$ and $(4)$ gives

$5x = 11$ $\implies$ $x = \dfrac{11}{5}$

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

$y = \dfrac{11}{5} - 2 = \dfrac{1}{5}$

Substituting the values of $x$ and $y$ in equation $(3)$ gives

$\dfrac{11}{5} + \dfrac{4}{5} = a$ $\implies$ $a = \dfrac{15}{5} = 3$

$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(\dfrac{11}{5}, \dfrac{1}{5}\right) \right\}$ $\;$ when $\;$ $a = 3$

When $\;$ $a \neq 3$, $\;$ then equation $(3)$ is not satisfied and therefore the solution of the given system of equations is $\phi$.

Algebra - System of Equations and Inequalities

For what value of $a$ is the sum of the squares of the numbers, constituting the solution of the system of equations $\;$ $3x - y = 2 - a$ $\;$ and $\;$ $x + 2y = a + 1$ $\;$ the least?

Given system of equations:

$3x - y = 2 - a$ $\;\;\; \cdots \; (1)$, and $\;$ $\;$ $x + 2y = a + 1$ $\;\;\; \cdots \; (2)$

Multiplying equation $(1)$ with $2$ gives

$6x - 2y = 4 - 2a$ $\;\;\; \cdots \; (3)$

Adding equations $(2)$ and $(3)$ gives

$7x = 5 - a$ $\implies$ $x = \dfrac{5 - a}{7}$

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

$\dfrac{5 - a}{7} + 2y = a + 1$

i.e. $\;$ $2y = a + 1 - \dfrac{5 - a}{7}$

i.e. $\;$ $2y = \dfrac{7a + 7 - 5 + a}{7}$

i.e. $\;$ $2y = \dfrac{8a + 2}{7}$ $\implies$ $y = \dfrac{4a + 1}{7}$

Let $\;$ $s$ $\;$ be the sum of squares of the numbers constituting the solution of the given system of equations. Then,

$s = \left(\dfrac{5 - a}{7}\right)^2 + \left(\dfrac{4a + 1}{7}\right)^2$

i.e. $\;$ $s = \dfrac{25 + a^2 - 10 a + 16 a^2 + 1 + 8a}{49}$

i.e. $\;$ $s = \dfrac{17 a^2 - 2a + 26}{49}$ $\;\;\; \cdots \; (4)$

For $\;$ $s$ $\;$ to be minimum, $\;\;$ $\dfrac{ds}{da} = 0$

$\therefore \;$ We have from equation $(4)$,

$\dfrac{ds}{da} = \dfrac{17}{49} \times 2a - \dfrac{2}{49} + 0 = 0$

i.e. $\;$ $\dfrac{17 \times 2a}{49} = \dfrac{2}{49}$

i.e. $\;$ $a = \dfrac{1}{17}$

Algebra - System of Equations and Inequalities

For what values of $a$ does the following systems of equations $\;$ $x + ay = 1, \;\; ax + y = a^2$ $\;$ have solutions? Find those solutions.

Given system of equations:

$x + ay = 1$ $\;\;\; \cdots \; (1)$, and $\;$ $\;$ $ax + y = a^2$ $\;\;\; \cdots \; (2)$

We have from equation $(1)$, $\;\;$ $x = 1 - ay$ $\;\;\; \cdots \; (3)$

Then, in view of equation $(3)$, equation $(2)$ becomes

$a \left(1 - ay\right) + y = a^2$

i.e. $\;$ $a - a^2 y + y = a^2$

i.e. $\;$ $y \left(1 - a^2\right) = a^2 - a$ $\;\;\; \cdots \; (4)$

Case 1:

If $\; \;$ $1 - a^2 \neq 0$, $\;\;$ i.e $\;$ if $\;$ $a^2 \neq 1$ $\;\;$ i.e. $\;$ if $\;$ $a \neq \pm 1$, $\;$ then we have from equation $(4)$

$y = \dfrac{a \left(a - 1\right)}{1 - a^2}$

i.e. $\;$ $y = \dfrac{a \left(a - 1\right)}{\left(1 + a\right) \left(1 - a\right)}$

i.e. $\;$ $y = \dfrac{-a}{1 + a}$ $\;\;\;$ when $\;$ $a \neq \pm 1$

Substituting the value of $y$ in equation $(3)$ gives

$x = 1 - a \left(\dfrac{-a}{1 + a}\right)$

i.e. $\;$ $x = \dfrac{1 + a + a^2}{1 + a}$ $\;\;\;$ when $\;$ $a \neq \pm 1$

Case 2:

When $\;$ $a = + 1$, $\;$ we have from equations $(1)$ and $(2)$

$x + y = 1$ $\;\;\; \cdots \; (1a)$ $\;\;\;$ and $\;\;\;$ $x + y = 1$ $\;\;\; \cdots \; (2a)$

$\therefore \;$ If $x$ is any number $c, \; c \in R$, then from equation $(1a)$,

$y = 1 -x = 1- c$

Case 3:

When $\;$ $a = -1$, $\;$ we have from equations $(1)$ and $(2)$

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

Adding equations $(1b)$ and $(2b)$ gives $\;\;$ $0 = 2$, $\;\;$ which is not possible.

$\therefore \;$ The solution to the given system of equations is

$\left(x, y\right) = \phi$ $\;$ when $\;$ $a = -1$

$\left(x, y\right) = \left\{\left(c, 1 - c\right) \mid c \in R \right\}$ $\;$ when $\;$ $a = +1$

and $\;$ $\left(x, y\right) = \left\{\left(\dfrac{1 + a + a^2}{1 + a}, \dfrac{-a}{1 + a}\right) \right\}$ $\;$ when $\;$ $a \neq \pm 1$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^2 = \left(x - a\right)y, \;\; y^2 - ay = 9ax$

Given system of equations:

$x^2 = \left(x - a\right)y$ $\;\;\; \cdots \; (1)$, and $\;$ $\;$ $y^2 - xy = 9ax$ $\;\;\; \cdots \; (2)$

Case 1: When $a \neq 0$

We have from equation $(1)$,

$y = \dfrac{x^2}{x - a}$ $\;\;\; \cdots \; (3)$

Substituting the value of $y$ from equation $(3)$ in equation $(2)$ gives

$\left(\dfrac{x^2}{x - a}\right)^2 - x \left(\dfrac{x^2}{x - a}\right) = 9ax$

i.e. $\;$ $\dfrac{x^3}{\left(x - a\right)^2} - \dfrac{x^2}{x - a} = 9a$

i.e. $\;$ $x^3 - x^2 \left(x - a\right) = 9a \left(x - a\right)^2$

i.e. $\;$ $x^3 - x^3 + ax^2 = 9ax^2 - 18a^2 x + 9a^3$

i.e. $\;$ $8ax^2 - 18 a^2 x + 9a^3 = 0$

i.e. $\;$ $8ax^2 - 12 a^2 x - 6 a^2 x + 9a^3 = 0$

i.e. $\;$ $4ax \left(2x - 3a\right) - 3a^2 \left(2x - 3a\right) = 0$

i.e. $\;$ $\left(2x - 3a\right) \left(4ax - 3a^2\right) = 0$

i.e. $\;$ $x = \dfrac{3a}{2}$ $\;\;$ or $\;\;$ $x = \dfrac{3a}{4}$

When $\;$ $x = \dfrac{3a}{2}$, $\;$ we have from equation $(3)$

$y = \dfrac{\left(\dfrac{3a}{2}\right)^2}{\dfrac{3a}{2} - a}$ $\implies$ $y = \dfrac{9a^2}{4} \times \dfrac{2}{a} = \dfrac{9a}{2}$

When $\;$ $x = \dfrac{3a}{4}$, $\;$ we have from equation $(3)$

$y = \dfrac{\left(\dfrac{3a}{4}\right)^2}{\dfrac{3a}{4} - a}$ $\implies$ $y = \dfrac{9a^2}{16} \times \dfrac{2}{-a} = \dfrac{-9a}{4}$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(\dfrac{3a}{2}, \dfrac{9a}{2}\right), \left(\dfrac{3a}{4}, \dfrac{-9a}{4}\right) \right\}$

Case 2: When $a = 0$

Then equation $(1)$ becomes $\;\;\;$ $x^2 = xy$ $\;\;\; \cdots \; (4)$

and equation $(2)$ becomes $\;\;\;$ $y^2 - xy = 0$ $\;\;$ i.e. $\;$ $y^2 = xy$ $\;\;\; \cdots \; (5)$

$\therefore \;$ We have from equations $(4)$ and $(5)$,

$x^2 = y^2$ $\;\;\;$ i.e. $\;$ $x = y$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(k, k\right) \mid k \in R \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^2 + 2xy + y^2 - x - y = 6, \;\; x - 2y = 3$

Given system of equations:

$x^2 + 2xy + y^2 - x - y = 6$ $\;\;\; \cdots \; (1)$, and $\;$ $\;$ $x - 2y = 3$ $\;\;\; \cdots \; (2)$

We have from equation $(2)$,

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

Substituting the value of $x$ from equation $(2a)$ in equation $(1)$ gives

$\left(2y + 3\right)^2 + 2 \left(2y + 3\right) y + y^2 - \left(2y + 3\right) - y = 6$

i.e. $\;$ $4y^2 + 12y + 9 + 4y^2 + 6y + y^2 - 2y - 3 - y = 6$

i.e. $\;$ $9y^2 + 15y = 0$

i.e. $\;$ $3y \left(3y + 5\right) = 0$

i.e. $\;$ $y = 0$ $\;\;$ or $\;\;$ $y = \dfrac{-5}{3}$

Substituting $y = 0$ in equation $(2a)$ gives $\;\;$ $x = 3$

Substituting $y = \dfrac{-5}{3}$ in equation $(2a)$ gives $\;\;$ $x = \dfrac{-10}{3} + 3 = \dfrac{-1}{3}$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(3, 0\right), \left(\dfrac{-1}{3}, \dfrac{-5}{3}\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{xy}{x + y} = 1, \;\; \dfrac{xz}{x + z} = 2, \;\; \dfrac{yz}{y + z} = 3$

Given system of equations:

$\dfrac{xy}{x + y} = 1$ $\;\;\; \cdots \; (1)$, $\;$ $\dfrac{xz}{x +z} = 2$ $\;\;\; \cdots \; (2)$ $\;$ and $\;$ $\dfrac{yz}{y + z} = 3$ $\;\;\; \cdots \; (3)$

The given equations can be respectively simplified as

$\dfrac{x + y}{xy} = 1$ $\;\;\;$ i.e. $\;$ $\dfrac{1}{x} + \dfrac{1}{y} = 1$ $\;\;\; \cdots \; (1a)$

$\dfrac{x + z}{xz} = \dfrac{1}{2}$ $\;\;\;$ i.e. $\;$ $\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{1}{2}$ $\;\;\; \cdots \; (2a)$

$\dfrac{y + z}{yz} = \dfrac{1}{3}$ $\;\;\;$ i.e. $\;$ $\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{3}$ $\;\;\; \cdots \; (3a)$

Subtracting equations $(1a)$ and $(2a)$ gives

$\dfrac{1}{y} - \dfrac{1}{z} = \dfrac{1}{2}$ $\;\;\; \cdots \; (4)$

Adding equations $(3a)$ and $(4)$ gives

$\dfrac{2}{y} = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$ $\implies$ $y = \dfrac{12}{5}$

Substituting $\;$ $y = \dfrac{12}{5}$ $\;$ in equation $(4)$ gives

$\dfrac{1}{12/5} - \dfrac{1}{z} = \dfrac{1}{2}$

i.e. $\;$ $\dfrac{1}{z} = \dfrac{5}{12} - \dfrac{1}{2} = \dfrac{-1}{12}$ $\implies$ $z = -12$

Substituting $\;$ $z = -12$ $\;$ in equation $(2a)$ gives

$\dfrac{1}{x} - \dfrac{1}{12} = \dfrac{1}{2}$

i.e. $\;$ $\dfrac{1}{x} = \dfrac{1}{12} + \dfrac{1}{2} = \dfrac{7}{12}$ $\implies$ $x = \dfrac{12}{7}$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y, z\right) = \left\{\left(\dfrac{12}{7}, \dfrac{12}{5}, -12\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\left(x^2 + y^2\right)xy = 78, \;\; x^4 + y^4 = 97$

Given system of equations:

$\left(x^2 + y^2\right) xy = 78$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x^4 + y^4 = 97$ $\;\;\; \cdots \; (2)$

Equation $(2)$ can be written as

$\left(x^2 + y^2\right)^2 - 2 x^2 y^2 = 97$ $\;\;\; \cdots \; (2a)$

Let $\;\;$ $x^2 + y^2 = u$ $\;\;\; \cdots \; (3a)$

and $\;\;$ $xy = v$ $\;\;\; \cdots \; (3b)$ $\;\;\;$ i.e. $\;$ $x = \dfrac{v}{y}$ $\;\;\; \cdots \; (3c)$

In view of equations $(3a)$ and $(3b)$, equations $(1)$ and $(2a)$ become

$uv = 78$ $\;\;\;$ i.e. $\;$ $u = \dfrac{78}{v}$ $\;\;\; \cdots \; (4)$

and $\;\;$ $u^2 - 2v^2 = 97$ $\;\;\; \cdots \; (5)$

Substituting for $u$ from equation $(4)$ in equation $(5)$ gives

$\left(\dfrac{78}{v}\right)^2 - 2v^2 = 97$

i.e. $\;$ $6084 - 2v^4 = 97v^2$

i.e. $\;$ $2v^4 + 97v^2 - 6084 = 0$

i.e. $\;$ $\left(2v^2 + 169\right)$ $\left(v^2 - 36\right) = 0$

i.e. $\;$ $v^2 = \dfrac{-169}{2} = - 84.5$ $\;\;$ or $\;\;$ $v^2 = 36$

Since $v^2$ cannot be negative, the only possible solution is

$v^2 = 36$ $\;\;\;$ i.e. $\;$ $v = \pm 6$

Substituting the value of $v$ in equation $(4)$ gives

when $\;$ $v = +6$, $\;\;$ $u = \dfrac{78}{6} = 13$ $\;\;$ and when $\;$ $v = -6$, $\;\;$ $u = \dfrac{78}{-6} = - 13$

Substituting the values of $u$ and $v$ in equations $(3a)$ and $(3c)$ gives

1. when $\;\;$ $u = 13, \; v = 6$, $\;\;$ we have
   
   $x^2 + y^2 = 13$ $\;\;\; \cdots \; (6)$ $\;\;$ and $\;\;$ $x = \dfrac{6}{y}$ $\;\;\; \cdots \; (7)$
   
   In view of equation $(7)$, equation $(6)$ becomes
   
   $\left(\dfrac{6}{y}\right)^2 + y^2 = 13$
   
   i.e. $\;$ $y^4 - 13y^2 + 36 = 0$
   
   i.e. $\;$ $\left(y^2 - 9\right) \left(y^2 - 4\right) = 0$
   
   i.e. $\;$ $y^2 = 9$ $\;\;$ or $\;\;$ $y^2 = 4$
   
   i.e. $\;$ $y = \pm 3$ $\;\;$ or $\;\;$ $y = \pm 2$
   
   Substituting the value of $y$ in equation $(7)$ gives
   
   when $\;\;$ $y = \pm 3$, $\;\;$ $x = \dfrac{6}{\pm 3} = \pm 2$
   
   and when $\;\;$ $y = \pm 2$, $\;\;$ $x = \dfrac{6}{\pm 2} = \pm 3$



2. when $\;\;$ $u = -13, \; v = -6$, $\;\;$ we have
   
   $x^2 + y^2 = -13$ $\;\;\; \cdots \; (8)$ $\;\;$ and $\;\;$ $x = \dfrac{-6}{y}$ $\;\;\; \cdots \; (9)$
   
   Since the sum of squares of any two numbers cannot be negative, $\;$ $u = -13, \; v = -6$ $\;$ are not valid solutions.

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(2, 3\right), \; \left(-2, -3\right), \left(3, 2\right), \left(-3, -2\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^3 + y^3 = 7, \;\; x^3 y^3 = -8$

Given system of equations:

$x^3 + y^3 = 7$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x^3 y^3 = -8$ $\;\;\; \cdots \; (2)$

Equation $(2)$ can be written as

$\left(xy\right)^3 = \left(-2\right)^3$

i.e. $x \; y = -2$

i.e. $\;$ $x = \dfrac{-2}{y}$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(1)$ becomes

$\left(\dfrac{-2}{y}\right)^3 + y^3 = 7$

i.e. $\;$ $\dfrac{-8}{y^3} + y^3 = 7$

i.e. $\;$ $y^6 - 7y^3 - 8 = 0$ $\;\;\; \cdots \; (4)$

Let $\;\;$ $y^3 = p$ $\;\;\; \cdots \; (5)$

The equation $(4)$ becomes

$p^2 - 7p - 8 = 0$

i.e. $\;$ $\left(p - 8\right) \left(p + 1\right) = 0$

i.e. $\;$ $p = 8$ $\;\;$ or $\;\;$ $p = -1$

Substituting the value of $p$ in equation $(5)$ gives

when $\;$ $p = 8$, $\;$ $y^3 = 8$ $\;\;$ i.e. $\;$ $y = 2$

and when $\;$ $p = -1$, $\;$ $y^3 = -1$ $\;\;$ i.e. $\;$ $y = -1$

Substituting the value of $y$ in equation $(3)$ gives

when $\;\;$ $y = 2$, $\;\;$ $x = \dfrac{-2}{2} = -1$

and when $\;\;$ $y = -1$, $\;\;$ $x = \dfrac{-2}{-1} = 2$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(-1, 2\right), \; \left(2, -1\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^4 + y^4 = 82, \;\; xy = 3$

Given system of equations:

$x^4 + y^4 = 82$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $xy = 3$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as

$\left(x^2\right)^2 + \left(y^2\right)^2 = 82$

i.e. $\;$ $\left(x^2 + y^2\right)^2 - 2 \left(xy\right)^2 = 82$

i.e. $\;$ $\left(x^2 + y^2\right)^2 - 18 = 82$ $\;\;\;$ [by equation $(2)$]

i.e. $\;$ $\left(x^2 + y^2\right)^2 = 100$

i.e. $\;$ $x^2 + y^2 = \pm 10$

1. When $\;$ $x^2 + y^2 = 10$
   
   i.e. $\;$ $\left(x + y\right)^2 - 2xy = 10$
   
   i.e. $\;$ $\left(x + y\right)^2 - 6 = 10$ $\;\;\;$ [by equation $(2)$]
   
   i.e. $\;$ $\left(x + y\right)^2 = 16$
   
   i.e. $\;$ $x + y = \pm 4$



1. When $\;$ $x + y = + 4$ $\;\;\;$ i.e. $\;$ $x = 4 - y$ $\;\;\; \cdots \; (3)$
   
   we have from equation $(2)$,
   
   $\left(4 - y\right) y = 3$
   
   i.e. $\;$ $y^2 - 4y + 3 = 0$
   
   i.e. $\;$ $\left(y - 3\right) \left(y - 1\right) = 0$
   
   i.e. $\;$ $y = 3$ $\;\;$ or $\;\;$ $y = 1$
   
   Substituting the values of of $y$ in equation $(3)$ gives



1. when $\;$ $y = 3$, $\;\;$ $x = 4 - 3 = 1$



2. when $\;$ $y = 1$, $\;\;$ $x = 4 - 1 = 3$



2. When $\;$ $x + y = - 4$ $\;\;\;$ i.e. $\;$ $x = -4 - y$ $\;\;\; \cdots \; (4)$
   
   we have from equation $(2)$,
   
   $- \left(4 + y\right) y = 3$
   
   i.e. $\;$ $y^2 + 4y + 3 = 0$
   
   i.e. $\;$ $\left(y + 3\right) \left(y + 1\right) = 0$
   
   i.e. $\;$ $y = - 3$ $\;\;$ or $\;\;$ $y = -1$
   
   Substituting the values of of $y$ in equation $(4)$ gives



1. when $\;$ $y = -3$, $\;\;$ $x = -4 - \left(-3\right) = -1$



2. when $\;$ $y = -1$, $\;\;$ $x = -4 - \left(-1\right) = -3$



2. When $\;$ $x^2 + y^2 = -10$
   
   i.e. $\;$ $\left(x + y\right)^2 - 2xy = -10$
   
   i.e. $\;$ $\left(x + y\right)^2 - 6 = -10$ $\;\;\;$ [by equation $(2)$]
   
   i.e. $\;$ $\left(x + y\right)^2 = -16$
   
   But square of any number cannot be negative.
   
   $\therefore \;$ $x^2 + y^2 = -10$ $\;$ is not a valid solution.

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{x^2}{y} + \dfrac{y^2}{x} = 18, \;\; x + y = 12$

Given system of equations:

$\dfrac{x^2}{y} + \dfrac{y^2}{x} = 18$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x + y = 12$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as

$x^3 + y^3 = 18 xy$

i.e. $\;$ $\left(x + y\right)^3 - 3xy \left(x + y\right) = 18xy$

i.e. $\;$ $12^3 - 3 x y \times 12 = 18xy$ $\;\;\;$ [by equation $(2)$]

i.e. $\;$ $1728 - 36xy = 18xy$

i.e. $\;$ $54 xy = 1728$ $\;\;\;$ i.e. $\;$ $xy = 32$ $\implies$ $x = \dfrac{32}{y}$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(2)$ becomes

$\dfrac{32}{y} + y = 12$

i.e. $\;$ $y^2 - 12y + 32 = 0$

i.e. $\;$ $\left(y - 8\right) \left(y - 4\right) = 0$

i.e. $\;$ $y = 8$ $\;\;$ or $\;\;$ $y = 4$

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

when $\;$ $y = 8$, $\;$ $x = 12 - 8 = 4$

and when $\;$ $y = 4$, $\;$ $x = 12 - 4 = 8$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(4, 8\right), \; \left(8, 4\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^3 + y^3 = 7, \;\; xy \left(x + y\right) = -2$

Given system of equations:

$x^3 + y^3 = 7$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $xy \left(x + y\right) = -2$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as

$\left(x + y\right)^3 - 3xy \left(x + y\right) = 7$ $\;\;\; \cdots \; (1a)$

Let $\;$ $x + y = u$ $\;\;\; \cdots \; (3a)$, $\;\;$ $xy = v$ $\;\;\; \cdots \; (3b)$

Then, in view of equations $(3a)$ and $(3b)$, equations $(2)$ and $(1a)$ respectively become

$uv = -2$ $\;\;$ i.e. $\;$ $u = \dfrac{-2}{v}$ $\;\;\; \cdots \; (4)$

and $\;\;$ $u^3 - 3uv = 7$ $\;\;\; \cdots \; (5)$

In view of equation $(4)$, equation $(5)$ can be written as

$\left(\dfrac{-2}{v}\right)^3 - 3 \times \left(\dfrac{-2}{v}\right) \times v = 7$

i.e. $\;$ $\dfrac{-8}{v^3} + 6 = 7$

i.e. $\;$ $\dfrac{-8}{v^3} = 1$ $\implies$ $v^3 = -8$ $\implies$ $v = -2$

Substituting $\;$ $v = -2$ $\;$ in equation $(4)$ gives $\;\;$ $u = \dfrac{-2}{-2} = 1$

Substituting the values of $u$ and $v$ in equations $(3a)$ and $(3b)$ gives

$x + y = 1$ $\;\;\; \cdots \; (6a)$

and $\;\;$ $xy = -2$ $\;\;\;$ i.e. $\;$ $x = \dfrac{-2}{y}$ $\;\;\; \cdots \; (6b)$

In view of equation $(6b)$, equation $(6a)$ becomes

$\dfrac{-2}{y} + y = 1$

i.e. $\;$ $y^2 - y - 2 = 0$

i.e. $\;$ $\left(y - 2\right) \left(y + 1\right) = 0$

i.e. $\;$ $y = 2$ $\;\;$ or $\;\;$ $y = -1$

Substituting the value of $y$ in equation $(6b)$ gives

when $\;\;$ $y = 2$, $\;\;$ $x = \dfrac{-2}{2} = -1$

and when $\;\;$ $y = -1$, $\;\;$ $x = \dfrac{-2}{-1} = 2$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(-1, 2\right), \; \left(2, -1\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^3 + y^3 = 35, \;\; x + y = 5$

Given system of equations:

$x^3 + y^3 = 35$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x + y = 5$ $\;\;\; \cdots \; (2)$

We have from equation $(2)$, $\;\;\;$ $x = 5 - y$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(1)$ becomes

$\left(5 - y\right)^3 + y^3 = 35$

i.e. $\;$ $125 - 75y + 15 y^2 - y^3 + y^3 = 35$

i.e. $\;$ $15 y^2 - 75 y + 90 = 0$

i.e. $\;$ $y^2 - 5y + 6 = 0$

i.e. $\;$ $\left(y - 3\right) \left(y - 2\right) = 0$

i.e. $\;$ $y = 3$ $\;\;$ or $\;\;$ $y = 2$

$\therefore \;$ We have from equation $(3)$,

when $\;$ $y = 3$, $\;$ $x = 5 - 3 = 2$

and when $\;$ $y = 2$, $\;$ $x = 5 - 2 = 3$

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(2, 3\right), \; \left(3, 2\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $3x^2 + xy - 2x + y - 5 = 0, \;\; 2x^2 - xy -3x - y - 5 = 0$

Given system of equations:

$3x^2 + xy - 2x + y - 5 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $2x^2 - xy -3x - y - 5 = 0$ $\;\;\; \cdots \; (2)$

Adding equations $(1)$ and $(2)$ gives

$5x^2 - 5x - 10 = 0$

i.e. $\;$ $x^2 - x - 2 = 0$

i.e. $\;$ $\left(x - 2\right) \left(x + 1\right) = 0$

i.e. $\;$ $x = 2$ $\;\;$ or $\;\;$ $x = -1$

Substituting $\;$ $x = 2$ $\;$ in equation $(1)$ gives

$12 + 2y - 4 + y - 5 = 0$

i.e. $\;$$3y = -3$ $\implies$ $y = -1$

Substituting $\;$ $x = -1$ $\;$ in equation $(1)$ gives

$3 - y + 2 + y - 5 = 0$

i.e. $\;$ $0 = 0$

$\implies$ $x = -1$ is satisfied for all $y \in R$.

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(2, -1\right), \; \left(-1, p\right) \mid p \in R \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $xy + 3y^2 - x + 4y - 7 = 0, \;\; 2xy + y^2 - 2x - 2y + 1 = 0$

Given system of equations:

$xy + 3y^2 - x + 4y - 7 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and

$2xy + y^2 - 2x - 2y + 1 = 0$ $\;\;\; \cdots \; (2)$

Multiply equation $(1)$ with $2$ to get

$2xy + 6y^2 - 2x + 8y - 14 = 0$ $\;\;\; \cdots \; (3)$

Subtracting equation $(2)$ from equation $(3)$ gives

$5y^2 + 10y - 15 = 0$

i.e. $\;$ $y^2 + 2y - 3 = 0$

i.e. $\;$ $\left(y + 3\right) \left(y - 1\right) = 0$

i.e. $\;$ $y = -3$ $\;\;$ or $\;\;$ $y = + 1$

When $\;$ $y = -3$, $\;$ we have from equation $(1)$

$-3x + 27 -x - 12 - 7 = 0$

i.e. $\;$ $4x = 8$ $\implies$ $x = 2$

When $\;$ $y = +1$, $\;$ we have from equation $(1)$

$x + 3 - x + 4 - 7 = 0$

i.e. $\;$ $0 = 0$

i.e. $\;$ $y = +1$ $\;$ is satisfied for all values of $\;$ $x \in R$.

$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(2, -3\right), \; \left(p, +1\right) \mid p \in R \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $2y^2 + xy -x^2 = 0, \;\; x^2 - xy - y^2 + 3x + 7y + 3 = 0$

Given system of equations:

$2y^2 + xy -x^2 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x^2 - xy - y^2 + 3x + 7y + 3 = 0$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as

$2y^2 + 2xy -xy - x^2 = 0$

i.e. $\;$ $2y \left(y + x\right) - x \left(y + x\right) = 0$

i.e. $\;$ $\left(2y - x\right) \left(y + x\right) = 0$

i.e. $\;$ $x = 2y$ $\;\;$ or $\;\;$ $x = -y$

Case 1:

When $\;$ $x = 2y$ $\;\;\; \cdots \; (3)$, $\;$ equation $(2)$ becomes

$\left(2y\right)^2 - 2y \times y - y^2 + 3 \times 2y + 7y + 3 = 0$

i.e. $\;$ $4y^2 - 2y^2 - y^2 + 6y + 7y + 3 = 0$

i.e. $\;$ $y^2 + 13y + 3 = 0$

i.e. $\;$ $y = \dfrac{-13 \pm \sqrt{13^2 - 4 \times 1 \times 3}}{2} = \dfrac{-13 \pm \sqrt{157}}{2}$

When $\;$ $y = \dfrac{-13 + \sqrt{157}}{2}$, $\;$ we have from equation $(3)$,

$x = 2 \times \left(\dfrac{-13 + \sqrt{157}}{2}\right) = -13 + \sqrt{157}$

and when $\;$ $y = \dfrac{-13 - \sqrt{157}}{2}$, $\;$ we have from equation $(3)$,

$x = 2 \times \left(\dfrac{-13 - \sqrt{157}}{2}\right) = -13 - \sqrt{157}$

Case 2:

When $\;$ $x = -y$ $\;\;\; \cdots \; (4)$, $\;$ equation $(2)$ becomes

$\left(-y\right)^2 - \left(-y\right) y - y^2 + 3 \left(-y\right) + 7y + 3 = 0$

i.e. $\;$ $y^2 + y^2 - y^2 - 3y + 7y + 3 = 0$

i.e. $\;$ $y^2 + 4y + 3 = 0$

i.e. $\;$ $\left(y + 3\right) \left(y + 1\right) = 0$

i.e. $\;$ $y = -3$ $\;\;$ or $\;\;$ $y = -1$

When $\;$ $y = -3$, $\;$ we have from equation $(4)$, $\;\;$ $x = - \left(-3\right) = 3$

and when $\;$ $y = -1$, $\;$ we have from equation $(4)$, $\;\;$ $x = - \left(-1\right) = 1$

$\therefore \;$ The solution to the given system of equations is

$\left(x, y\right) = \left(-13 + \sqrt{157}, \dfrac{-13 + \sqrt{157}}{2}\right), \; \left(-13 - \sqrt{157}, \dfrac{-13 - \sqrt{157}}{2}\right)$,
$\hspace{2cm} \left(3, -3\right), \; \left(1, -1\right)$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $2x^2 + y^2 + 3xy = 12, \;\; 2 \left(x + y\right)^2 - y^2 = 14$

Given system of equations:

$2x^2 + y^2 + 3xy = 12$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $2 \left(x + y\right)^2 - y^2 = 14$ $\;\;\; \cdots \; (2)$

Equation $(2)$ can be written as

$2x^2 + 4xy + 2y^2 - y^2 = 14$

i.e. $\;$ $2x^2 + 4xy + y^2 = 14$ $\;\;\; \cdots \; (3)$

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

$x \; y = 2$ $\;\;$ i.e. $\;$ $x = \dfrac{2}{y}$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$, equation $(1)$ becomes

$2 \times \left(\dfrac{2}{y}\right)^2 + y^2 + 3 \times \dfrac{2}{y} \times y = 12$

i.e. $\;$ $\dfrac{8}{y^2} + y^2 + 6 = 12$

i.e. $\;$ $y^4 - 6y^2 + 8 = 0$

i.e. $\;$ $\left(y^2 - 2\right) \left(y^2 - 4\right) = 0$

i.e. $\;$ $y^2 = 2$ $\;\;$ or $\;\;$ $y^2 = 4$

i.e. $\;$ $y = \pm \sqrt{2}$ $\;\;$ or $\;\;$ $y = \pm 2$

When $\;$ $y = \pm \sqrt{2}$, $\;$ we have from equation $(4)$, $\;$ $x = \dfrac{2}{\pm \sqrt{2}} = \pm \sqrt{2}$

When $\;$ $y = \pm 2$, $\;$ we have from equation $(4)$, $\;$ $x = \dfrac{2}{\pm 2} = \pm 1$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\sqrt{2}, \sqrt{2}\right), \; \left(- \sqrt{2}, - \sqrt{2}\right), \; \left(1, 2\right), \; \left(-1, -2\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^2 - xy + y^2 = 21, \;\; y^2 - 2xy + 15 = 0$

Given system of equations:

$x^2 - xy + y^2 = 21$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y^2 - 2xy + 15 = 0$ $\;\;\; \cdots \; (2)$

Least Common Multiple (LCM) of $15$ and $21$ is $105$.

Multiplying equation $(1)$ with $5$ gives $\;\;$ $5x^2 - 5xy + 5y^2 = 105$ $\;\;\; \cdots \; (3)$

Multiplying equation $(2)$ with $7$ gives $\;\;$ $-14xy + 7y^2 = -105$ $\;\;\; \cdots \; (4)$

Adding equations $(3)$ and $(4)$ gives

$5x^2 - 19xy + 12y^2 = 0$

i.e. $\;$ $5x^2 - 15xy - 4xy + 12y^2 = 0$

i.e. $\;$ $5x \left(x - 3y\right) - 4y \left(x - 3y\right) = 0$

i.e. $\;$ $\left(5x - 4y\right) \left(x - 3y\right) = 0$

i.e. $\;$ $x = \dfrac{4 y}{5}$ $\;\;$ or $\;\;$ $x = 3y$

Case 1:

When $\;$ $x = \dfrac{4y}{5}$ $\;\;\; \cdots \; (5)$, $\;$ equation $(1)$ becomes

$\left(\dfrac{4y}{5}\right)^2 - \left(\dfrac{4y}{5}\right) y + y^2 = 21$

i.e. $\;$ $\dfrac{16 \; y^2}{25} - \dfrac{4 \; y^2}{5} + y^2 = 21$

i.e. $\;$ $\left(16 - 20 + 25\right) y^2 = 21 \times 25$

i.e. $\;$ $21 \; y^2 = 21 \times 25$

i.e. $\;$ $y^2 = 25$ $\implies$ $y = \pm 5$

Substituting $\;$ $y = \pm 5$ $\;$ in equation $(5)$ gives

$x = \pm \dfrac{4}{5} \times 5 = \pm 4$

Case 2:

When $\;$ $x = 3y$ $\;\;\; \cdots \; (6)$, $\;$ equation $(1)$ becomes

$\left(3y\right)^2 - \left(3y\right) y + y^2 = 21$

i.e. $\;$ $9y^2 - 3y^2 + y^2 = 21$

i.e. $\;$ $7y^2 = 21$

i.e. $\;$ $y^2 = 3$ $\implies$ $y = \pm \sqrt{3}$

Substituting $\;$ $y = \pm \sqrt{3}$ $\;$ in equation $(6)$ gives

$x = \pm 3 \sqrt{3}$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(4, 5\right), \; \left(-4, -5\right), \; \left(3 \sqrt{3}, \sqrt{3}\right), \; \left(- 3\sqrt{3}, - \sqrt{3}\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $2y^2 - 4xy + 3x^2 = 17, \;\; y^2 - x^2 = 16$

Given system of equations:

$2y^2 - 4xy + 3x^2 = 17$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y^2 - x^2 = 16$ $\;\;\; \cdots \; (2)$

Subtracting equations $(1)$ and $(2)$ gives

$y^2 - 4xy + 4x^2 = 1$

i.e. $\;$ $\left(y - 2x\right)^2 = 1$

i.e. $\;$ $y - 2x = \pm 1$ $\;\;$ i.e. $\;$ $y = 2x \pm 1$ $\;\;\; \cdots \; (3)$

Substituting the value of $y$ from equation $(3)$ in equation $(2)$

Case 1:

When $\;$ $y = 2x + 1$ $\;\;\; \cdots \; (3a)$ $\;$ we have $\;$ $\left(2x + 1\right)^2 - x^2 = 16$

i.e. $\;$ $4x^2 + 4x + 1 - x^2 = 16$

i.e. $\;$ $3x^2 + 4x - 15 = 0$

i.e. $\;$ $\left(3x - 5\right) \left(x + 3\right) = 0$

i.e. $\;$ $x = \dfrac{5}{3}$, $\;\;$ or $\;\;$ $x = -3$

When $\;$ $x = \dfrac{5}{3}$, $\;$ we have from equation $(3a)$, $\;$ $y = \left[2 \times \dfrac{5}{3}\right] + 1 = \dfrac{13}{3}$

and when $\;$ $x = -3$, $\;$ we have from equation $(3a)$, $\;$ $y = \left[2 \times \left(-3\right)\right] + 1 = -5$

Case 2:

When $\;$ $y = 2x - 1$ $\;\;\; \cdots \; (3b)$ $\;$ we have $\;$ $\left(2x - 1\right)^2 - x^2 = 16$

i.e. $\;$ $4x^2 - 4x + 1 - x^2 = 16$

i.e. $\;$ $3x^2 - 4x - 15 = 0$

i.e. $\;$ $\left(3x + 5\right) \left(x - 3\right) = 0$

i.e. $\;$ $x = \dfrac{-5}{3}$, $\;\;$ or $\;\;$ $x = 3$

When $\;$ $x = \dfrac{-5}{3}$, $\;$ we have from equation $(3b)$, $\;$ $y = \left[2 \times \left(\dfrac{-5}{3}\right)\right] - 1 = \dfrac{-13}{3}$

and when $\;$ $x = 3$, $\;$ we have from equation $(3b)$, $\;$ $y = \left[2 \times 3\right] - 1 = 5$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(-3, -5\right), \; \left(\dfrac{-5}{3}, \dfrac{-13}{3}\right), \; \left(\dfrac{5}{3}, \dfrac{13}{3}\right), \; \left(3, 5\right) \right\}$