Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{3}, \;\; x^2 + y^2 = 160$

Given system of equations:

$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{3}$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x^2 + y^2 = 160$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as $\;\;$ $\dfrac{x + y}{xy} = \dfrac{1}{3}$ $\;\;\; \cdots \; (1a)$

Equation $(2)$ can be written as $\;\;$ $\left(x + y\right)^2 - 2xy = 160$ $\;\;\; \cdots \; (2a)$

Let $\;\;$ $x + y = u$ $\;\;\; \cdots \; (3)$; $\;\;$ $x \; y = v$ $\;\;\; \cdots \; (4)$

In view of equations $(3)$ and $(4)$. equations $(1a)$ and $(2a)$ respectively become

$\dfrac{u}{v} = \dfrac{1}{3}$ $\;\;$ i.e. $\;$ $v = 3u$ $\;\;\; \cdots \; (1b)$

and $\;\;$ $u^2 - 2v = 160$ $\;\;\; \cdots \; (2b)$

Substituting for $v$ from equation $(1b)$ in equation $(2b)$ gives

$u^2 - 2\times \left(3u\right) = 160$

i.e. $\;$ $u^2 - 6u - 160 = 0$

i.e. $\;$ $\left(u - 16\right) \left(u + 10\right) = 0$

i.e. $\;$ $u = 16$ $\;\;$ or $\;\;$ $u = -10$

$\therefore \;$ We have from equation $(1b)$

when $\;\;$ $u = 16$, $\;$ $v = 3 \times 16 = 48$

and when $\;\;$ $u = -10$, $\;$ $v = 3 \times \left(-10\right) = -30$

Case 1:

Substituting $\;\;$ $u = 16, \; v = 48$ $\;\;$ in equations $(3)$ and $(4)$ gives

$x + y = 16$ $\;\;\;$ i.e. $\;$ $x = 16 - y$ $\;\;\; \cdots \; (3a)$

and $\;\;$ $x \; y = 48$

i.e. $\;$ $\left(16 - y\right) y = 48$ $\;\;\;$ [in view of equation $(3a)$]

i.e. $\;$ $y^2 - 16 y + 48 = 0$

i.e. $\;$ $\left(y - 12\right) \left(y - 4\right) = 0$

i.e. $\;$ $y = 12$ $\;\;$ or $\;\;$ $y = 4$

When $\;\;$ $y = 12$, $\;\;$ we have from equation $(3a)$, $\;\;$ $x = 16 - 12 = 4$

and when $\;\;$ $y = 4$, $\;\;$ we have from equation $(3a)$, $\;\;$ $x = 16 - 4 = 12$

Case 2:

Substituting $\;\;$ $u = -10, \; v = -30$ $\;\;$ in equations $(3)$ and $(4)$ gives

$x + y = -10$ $\;\;\;$ i.e. $\;$ $x = -y - 10$ $\;\;\; \cdots \; (3b)$

and $\;\;$ $x \; y = -30$

i.e. $\;$ $-\left(y + 10\right) y = -30$ $\;\;\;$ [in view of equation $(3b)$]

i.e. $\;$ $y^2 + 10y - 30 = 0$

i.e. $\;$ $y = \dfrac{-10 \pm \sqrt{100 + 120}}{2} = \dfrac{-10 \pm 2 \sqrt{55}}{2}$

i.e. $\;$ $y = - 5 \pm \sqrt{55}$

When $\;\;$ $y = -5 + \sqrt{55}$, $\;\;$ we have from equation $(3b)$,

$x = - \left(-5 + \sqrt{55}\right) - 10 = -5 - \sqrt{55}$

and when $\;\;$ $y = -5 - \sqrt{55}$, $\;\;$ we have from equation $(3b)$,

$x = - \left(-5 - \sqrt{55}\right) - 10 = -5 + \sqrt{55}$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(4, 12\right), \; \left(12, 4\right), \; \left(-5 - \sqrt{55}, -5 + \sqrt{55}\right), \; \left(-5 + \sqrt{55}, -5 - \sqrt{55}\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x + y + xy = 5, \;\; x^2 + y^2 + xy = 7$

Given system of equations:

$x + y + xy = 5$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x^2 + y^2 + xy = 7$ $\;\;\; \cdots \; (2)$

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

$x^2 + y^2 + 2xy + x + y = 12$

i.e. $\;$ $\left(x + y\right)^2 + \left(x + y\right) = 12$ $\;\;\; \cdots \; (3)$

Let $\;\;$ $x + y = p$ $\;\;\; \cdots \; (4)$

Then, in view of equation $(4)$, equation $(3)$ becomes

$p^2 + p - 12 = 0$

i.e. $\;$ $\left(p + 4\right) \left(p - 3\right) = 0$

i.e. $\;$ $p = - 4$ $\;\;$ or $\;\;$ $p = 3$

When $\;$ $p = - 4$, $\;$ we have from equation $(4)$

$x + y = - 4$ $\;\;$ i.e. $\;$ $x = - 4 - y$ $\;\;\; \cdots \; (5)$

In view of equation $(5)$, equation $(1)$ becomes

$- 4 - y + y - \left(4 + y\right) y = 5$

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

i.e. $\;$ $y^2 + 4y + 9 = 0$ $\;\;\; \cdots \; (6)$

Equation $(6)$ does not have a real solution since its discriminant $\;$ $\Delta = \left(4\right)^2 - \left(4 \times 1 \times 9\right) = 16 - 36 = -20 < 0$

Next, when $\;$ $p = 3$, $\;$ we have from equation $(4)$

$x + y = 3$ $\;\;$ i.e. $\;$ $x = 3 - y$ $\;\;\; \cdots \; (7)$

In view of equation $(7)$, equation $(1)$ becomes

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

i.e. $\;$ $3 + 3y - y^2 = 5$

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

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

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

Substituting the values of $y$ in equation $(1)$ gives

when $\;$ $y = 2$, $\;\;$ $x + 2 + 2x = 5$

i.e. $\;$ $3x = 3$ $\;\;$ i.e. $\;$ $x = 1$

and when $\;$ $y = 1$, $\;\;$ $x + 1 + x = 5$

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

$\therefore \;$ 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: $\;$ $4x^2 + y^2 - 2xy = 7, \;\; \left(2x - y\right) y = y$

Given system of equations:

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

Case 1:

When $\;$ $y \neq 0$, $\;$ equation $(2)$ becomes

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

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

$4x^2 + \left(2x - 1\right)^2 - 2x \left(2x - 1\right) = 7$

i.e. $\;$ $4x^2 + 4x^2 - 4x + 1 - 4x^2 + 2x = 7$

i.e. $\;$ $4x^2 -2x - 6 = 0$

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

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

i.e. $\;$ $x = \dfrac{3}{2}$ $\;\;$ or $\;\;$ $x = -1$

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

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

Case 2:

When $\;$ $y = 0$, $\;$ equation $(1)$ becomes

$4x^2 = 7$

i.e. $\;$ $x^2 = \dfrac{7}{4}$ $\implies$ $x = \pm \dfrac{\sqrt{7}}{2}$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\dfrac{3}{2}, 2\right), \; \left(-1, -3\right), \; \left(\dfrac{\sqrt{7}}{2}, 0\right), \; \left(\dfrac{- \sqrt{7}}{2}, 0\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $5 \left(x + y\right) + 2xy = -19, \;\; 15xy + 5 \left(x + y\right) = -175$

Given system of equations:

$5 \left(x + y\right) + 2xy = -19$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $15xy + 5 \left(x + y\right) = -175$ $\;\;\; \cdots \; (2)$

Let $\;$ $x + y = u$ $\;\;\; \cdots \; (3)$ $\;$ and $\;$ $x \; y = v$ $\;\;\; \cdots \; (4)$

Then equations $(1)$ and $(2)$ become

$5u + 2v = -19$ $\;\;\; \cdots \; (5)$; $\;\;$ $5u + 15v = -175$ $\;\;\; \cdots \; (6)$

Solving equations $(5)$ and $(6)$ simultaneously gives

$13v = -156$ $\implies$ $v = -12$

Substituting $\;$ $v = -12$ $\;$ in equation $(5)$ gives

$5u = -19 + 24 = 5$ $\;\;$ i.e. $\;$ $u = 1$

Substituting the values of $u$ and $v$ in equations $(3)$ and $(4)$ gives

$x + y = 1$ $\;\;$ i.e. $\;$ $x = 1 - y$ $\;\;\; \cdots \; (7)$ $\;\;$ and $\;\;$ $x \; y = -12$ $\;\;\; \cdots \; (8)$

In view of equation $(7)$ equation $(8)$ becomes

$\left(1 - y\right) y = - 12$

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

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

i.e. $\;$ $y = 4$ $\;\;$ or $\;\;$ $y = -3$

When $\;$ $y = 4$, $\;$ equation $(7)$ gives $\;\;$ $x = 1 - 4 = -3$

and when $\;$ $y = -3$, $\;$ equation $(7)$ gives $\;\;$ $x = 1 - \left(-3\right) = 4$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^2 + y^2 = 25 - 2xy, \;\; y \left(x + y\right) = 10$

Given system of equations:

$x^2 + y^2 = 25 - 2xy$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y \left(x + y\right) = 10$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as

$x^2 + y^2 + 2xy = 25$

i.e. $\;$ $\left(x + y\right)^2 = 25$

i.e. $\;$ $x = y = \pm 5$

When $\;$ $x + y = + 5$ $\;\;\; \cdots \; (3)$, $\;$ equation $(2)$ becomes

$5y = 10$ $\implies$ $y = 2$

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

When $\;$ $x + y = - 5$ $\;\;\; \cdots \; (4)$, $\;$ equation $(2)$ becomes

$-5y = 10$ $\implies$ $y = -2$

Substituting $\;$ $y = -2$ $\;$ in equation $(4)$ gives $\;\;$ $x = -5 + 2 = -3$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^2 - xy = 28, \;\; y^2 - xy = -12$

Given system of equations:

$x^2 - xy = 28$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y^2 - xy = -12$ $\;\;\; \cdots \; (2)$

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

$x^2 + y^2 - 2xy = 16$

i.e. $\;$ $\left(x - y\right)^2 = 16$ $\implies$ $x - y = \pm 4$

i.e. $\;$ $x = y + 4$ $\;\;\; \cdots \; (3)$ $\;\;$ or $\;\;$ $x = y - 4$ $\;\;\; \cdots \; (4)$

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

$\left(y + 4\right)^2 - \left(y + 4\right)y = 28$

i.e. $\;$ $y^2 + 8y + 16 - y^2 - 4y = 28$

i.e. $\;$ $4y = 12$ $\implies$ $y = 3$

Substituting $y = 3$ in equation $(3)$ gives $\;\;$ $x = 3 + 4 = 7$

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

$\left(y - 4\right)^2 - \left(y - 4\right)y = 28$

i.e. $\;$ $y^2 - 8y + 16 - y^2 + 4y = 28$

i.e. $\;$ $-4y = 12$ $\implies$ $y = -3$

Substituting $y = -3$ in equation $(4)$ gives $\;\;$ $x = -3 - 4 = -7$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(7, 3\right), \; \left(-7, -3\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^2 + xy = 15, \;\; y^2 + xy = 10$

Given system of equations:

$x^2 + xy = 15$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y^2 + xy = 10$ $\;\;\; \cdots \; (2)$

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

$x^2 + y^2 + 2xy = 25$

i.e. $\;$ $\left(x + y\right)^2 = 25$ $\implies$ $x + y = \pm 5$

i.e. $\;$ $x = 5 - y$ $\;\;\; \cdots \; (3)$ $\;\;$ or $\;\;$ $x = -5 - y$ $\;\;\; \cdots \; (4)$

When $\;$ $x = 5 - y$, $\;$ equation $(1)$ becomes

$\left(5 - y\right)^2 + \left(5 - y\right)y = 15$

i.e. $\;$ $25 + y^2 - 10y + 5y - y^2 = 15$

i.e. $\;$ $5y = 10$ $\implies$ $y = 2$

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

When $\;$ $x = -5 - y$, $\;$ equation $(1)$ becomes

$\left(-5 - y\right)^2 - \left(5 + y\right)y = 15$

i.e. $\;$ $25 + y^2 + 10y - 5y - y^2 = 15$

i.e. $\;$ $5y = -10$ $\implies$ $y = -2$

Substituting $y = -2$ in equation $(4)$ gives $\;\;$ $x = -5 + 2 = -3$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x \left(x + y\right) = 9, \;\; y \left(x + y\right) = 16$

Given system of equations:

$x \left(x + y\right) = 9$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $y \left(x + y\right) = 16$ $\;\;\; \cdots \; (2)$

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

$\dfrac{x}{y} = \dfrac{9}{16}$

i.e. $\;$ $x = \dfrac{9y}{16}$ $\;\;\; \cdots \; (3)$

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

$\dfrac{9y}{16} \left(\dfrac{9y}{16} + y\right) = 9$

i.e. $\;$ $\dfrac{81 y^2}{256} + \dfrac{9 y^2}{16} = 9$

i.e. $\;$ $\dfrac{225 y^2}{256} = 9$

i.e. $\;$ $y^2 = \dfrac{9 \times 256}{225}$ $\implies$ $y = \pm \dfrac{3 \times 16}{15} = \pm \dfrac{16}{5}$

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

$x = \pm \dfrac{9}{16} \times \dfrac{16}{5} = \pm \dfrac{9}{5}$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\dfrac{9}{5}, \dfrac{16}{5}\right), \; \left(\dfrac{-9}{5}, \dfrac{-16}{5}\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^2 + y^2 = 20, \;\; xy = 8$

Given system of equations:

$x^2 + y^2 = 20$ $\;\;\; \cdots \; (1)$

and $\;$ $xy = 8$ $\;$ i.e. $\;$ $x = \dfrac{8}{y}$ $\;\;\; \cdots \; (2)$

Then, in view of equation $(2)$ equation $(1)$ becomes

$\left(\dfrac{8}{y}\right)^2 + y^2 = 20$

i.e. $\;$ $64 + y^4 = 20 y^2$

i.e. $\;$ $y^4 - 20y^2 + 64 = 0$

i.e. $\;$ $y^4 - 16y^2 - 4y^2 + 64 = 0$

i.e. $\;$ $y^2 \left(y^2 - 4\right) - 16 \left(y^2 - 4\right) = 0$

i.e. $\;$ $\left(y^2 - 16\right) \left(y^2 - 4\right) = 0$

i.e. $\;$ $y^2 = 16$ $\;$ or $\;$ $y^2 = 4$

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

When $\;$ $y = \pm 4$, $\;$ we have from equation $(2)$,

$x = \dfrac{8}{\pm 4} = \pm 2$

When $\;$ $y = \pm 2$, $\;$ we have from equation $(2)$,

$x = \dfrac{8}{\pm 2} = \pm 4$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $xy - \dfrac{x}{y} = \dfrac{16}{3}, \;\; xy - \dfrac{y}{x} = \dfrac{9}{2}$

Given system of equations:

$xy - \dfrac{x}{y} = \dfrac{16}{3}$ $\;\;\; \cdots \; (1)$

and $\;$ $xy - \dfrac{y}{x} = \dfrac{9}{2}$ $\;\;\; \cdots \; (2)$

Let $\;$ $x \; y = u$ $\;\;\; \cdots \; (3)$ $\;$ and $\;$ $\dfrac{x}{y} = v$ $\;\;\; \cdots \; (4)$

In view of equations $(3)$ and $(4)$, equations $(1)$ and $(2)$ become

$u - v = \dfrac{16}{3}$ $\;\;\; \cdots \; (5)$

and $\;$ $u - \dfrac{1}{v} = \dfrac{9}{2}$ $\;\;\; \cdots \; (6)$

Solving equations $(5)$ and $(6)$ simultaneously gives

$v - \dfrac{1}{v} = \dfrac{9}{2} - \dfrac{16}{3} = \dfrac{-5}{6}$

i.e. $\;$ $6v^2 - 6 = -5v$

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

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

i.e. $\;$ $v = \dfrac{-3}{2}$ $\;$ or $\;$ $v = \dfrac{2}{3}$

We have from equation $(5)$, when $\;$ $v = \dfrac{-3}{2}$, $\;$ $u = \dfrac{16}{3} - \dfrac{3}{2} = \dfrac{23}{6}$

and when $\;$ $v = \dfrac{2}{3}$, $\;$ $u = \dfrac{16}{3} + \dfrac{2}{3} = 6$

Substituting $u$ and $v$ in equations $(3)$ and $(4)$ gives

when $\;$ $u = \dfrac{23}{6}$, $\;$ $v = \dfrac{-3}{2}$

we have $\;$ $\dfrac{x}{y} = \dfrac{-3}{2}$ $\;$ i.e. $\;$ $x = \dfrac{-3 y}{2}$ $\;\;\; \cdots \; (7)$

and $\;$ $xy = \dfrac{23}{6}$

i.e. $\;$ $\dfrac{-3 y^2}{2} = \dfrac{23}{6}$ $\;\;\;$ [in view of equation $(7)$]

i.e. $\;$ $y^2 = \dfrac{-23}{9}$

But square of any number cannot be negative.

$\therefore \;$ $u = \dfrac{23}{6}$, $\;$ $v = \dfrac{-3}{2}$ $\;$ are not acceptable values.

When $\;$ $u = 6$, $\;$ $v = \dfrac{2}{3}$

we have $\;$ $\dfrac{x}{y} = \dfrac{2}{3}$ $\;$ i.e. $\;$ $x = \dfrac{2 y}{3}$ $\;\;\; \cdots \; (8)$

and $\;$ $xy = 6$

i.e. $\;$ $\dfrac{2 y^2}{3} = 6$ $\;\;\;$ [in view of equation $(8)$]

i.e. $\;$ $y^2 = 9$ $\implies$ $y = \pm 3$

Substituting the value of $y$ in equation $(8)$ gives

$x = \pm \dfrac{2}{3} \times 3 = \pm 2$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $2xy - \dfrac{3x}{y} = 15, \;\; xy + \dfrac{x}{y} = 15$

Given system of equations:

$2xy - \dfrac{3x}{y} = 15$ $\;\;\; \cdots \; (1)$

and $\;$ $xy + \dfrac{x}{y} = 15$ $\;\;\; \cdots \; (2)$

Let $\;$ $x \; y = u$ $\;\;\; \cdots \; (3)$ $\;$ and $\;$ $\dfrac{x}{y} = v$ $\;\;\; \cdots \; (4)$

In view of equations $(3)$ and $(4)$, equations $(1)$ and $(2)$ become

$2u - 3v = 15$ $\;\;\; \cdots \; (5)$

and $\;$ $u + v = 15$ $\;\;\; \cdots \; (6)$

We have from equations $(5)$ and $(6)$,

$2u - 3v = u + v$

i.e. $\;$ $u = 4v$ $\;\;\; \cdots \; (7)$

Substituting the value of $u$ in equation $(6)$ gives

$4v + v = 15$ $\;$ i.e. $\;$ $5v = 15$ $\;$ i.e. $\;$ $v = 3$

When $v = 3$, we have from equation $(7)$, $\;$ $u = 4 \times 3 = 12$

Substituting the values of $u$ and $v$ in equations $(4)$ and $(3)$ gives

$\dfrac{x}{y} = 3$ $\;$ i.e. $\;$ $x = 3y$ $\;\;\; \cdots \; (8)$

and $\;\;$ $x \; y = 12$ $\;\;\; \cdots \; (9)$

In view of equation $(8)$, equation $(9)$ can be written as

$3y \cdot y = 12$

i.e. $\;$ $y^2 = 4$ $\implies$ $y = \pm 2$

Substituting the value of $y$ in equation $(8)$ gives

when $\;$ $y = + 2$, $\;$ $x = 3 \times 2 = 6$

and when $\;$ $y = -2$, $\;$ $x = 3 \times \left(-2\right) = -6$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(6, 2\right), \; \left(-6, -2\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{x + y}{x - y} + \dfrac{x - y}{x + y} = \dfrac{5}{2}, \;\; x^2 + y^2 = 20$

Given system of equations:

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

and $\;$ $x^2 + y^2 = 20$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as

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

i.e. $\;$ $x^2 + y^2 + 2xy + x^2 + y^2 - 2xy = \dfrac{5 \left(x^2 - y^2\right)}{2}$

i.e. $\;$ $2 x^2 + 2 y^2 = \dfrac{5 \left(x^2 - y^2\right)}{2}$

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

i.e. $\;$ $x^2 = 9y^2$ $\;\;\; \cdots \; (1a)$

In view of equation $(1a)$, equation $(2)$ becomes

$9 y^2 + y^2 = 20$

i.e. $\;$ $10 y^2 = 20$

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

Substituting the value of $y$ in equation $(1a)$ gives

when $\;$ $y = + \sqrt{2}$, $\;$ $x^2 = 9 \times 2 = 18$ $\implies$ $x = \pm 3 \sqrt{2}$

when $\;$ $y = - \sqrt{2}$, $\;$ $x^2 = 9 \times 2 = 18$ $\implies$ $x = \pm 3 \sqrt{2}$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{1}{y - 1} - \dfrac{1}{y + 1} = \dfrac{1}{x}, \;\; y^2 - x - 5 = 0$

Given system of equations:

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

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

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

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

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

i.e. $\;$ $2 \left(y^2 - 5\right) = y^2 - 1$

i.e. $\;$ $y^2 = 9$ $\implies$ $y = \pm 3$

Substituting $y^2 = 9$ in equation $(2)$ gives

$x = 9 - 5 = 4$

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

Algebra - System of Equations and Inequalities

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

Given system of equations:

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

and $\;$ $\dfrac{y}{x + 2y} = -3$ $\;\;\; \cdots \; (2)$

Let $\;\;$ $\dfrac{1}{x + 2y} = p$ $\;\;\; \cdots \; (3)$

Then, in view of equation $(3)$, equations $(1)$ and $(2)$ can be written respectively as

$p + y = 2$ $\;\;\; \cdots \; (4)$

and $\;$ $p \; y = -3$ $\;\;$ i.e. $\;$ $p = \dfrac{-3}{y}$ $\;\;\; \cdots \; (5)$

In view of equation $(5)$, equation $(4)$ becomes

$\dfrac{-3}{y} + y = 2$

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

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

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

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

when $\;$ $y = 3$, $\;$ $\dfrac{3}{x + 6} = -3$

i.e. $\;$ $x + 6 = -1$ $\implies$ $x = -7$

and when $\;$ $y = -1$, $\;$ $\dfrac{-1}{x - 2} = -3$

i.e. $\;$ $-1 = -3x + 6$ $\;$ i.e. $\;$ $3x = 7$ $\implies$ $x = \dfrac{7}{3}$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(-7, 3\right), \; \left(\dfrac{7}{3}, -1\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{1}{2x - y} + y = -5, \;\; \dfrac{y}{2x - y} = 6$

Given system of equations:

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

and $\;$ $\dfrac{y}{2x - y} = 6$ $\;\;\; \cdots \; (2)$

Let $\;\;$ $\dfrac{1}{2x - y} = p$ $\;\;\; \cdots \; (3)$

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

$p + y = -5$ $\;\;\; \cdots \; (4)$

and $\;$ $p \; y = 6$ $\;\;$ i.e. $\;$ $p = \dfrac{6}{y}$ $\;\;\; \cdots \; (5)$

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

$\dfrac{6}{y} + y = -5$

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

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

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

Substitute $\;$ $y = -3$ $\;$ in equation $(1)$ to get

$\dfrac{1}{2x + 3} - 3 = -5$

i.e. $\;$ $\dfrac{1}{2x + 3} = -2$

i.e. $\;$ $2x + 3 = \dfrac{-1}{2}$

i.e. $\;$ $2x = \dfrac{-7}{2}$ $\implies$ $x = \dfrac{-7}{4}$

Substitute $\;$ $y = -2$ $\;$ in equation $(1)$ to get

$\dfrac{1}{2x + 2} - 2 = -5$

i.e. $\;$ $\dfrac{1}{2x + 2} = -3$

i.e. $\;$ $2x + 2 = \dfrac{-1}{3}$

i.e. $\;$ $2x = \dfrac{-7}{3}$ $\implies$ $x = \dfrac{-7}{6}$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\dfrac{-7}{4}, -3\right), \; \left(\dfrac{-7}{6}, -2\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{1}{x + 1} + \dfrac{1}{y} = \dfrac{1}{3}, \;\; \dfrac{1}{\left(x + 1\right)^2} - \dfrac{1}{y^2} = \dfrac{1}{4}$

Given system of equations:

$\dfrac{1}{x + 1} + \dfrac{1}{y} = \dfrac{1}{3}$ $\;\;\; \cdots \; (1)$

and $\;$ $\dfrac{1}{\left(x + 1\right)^2} - \dfrac{1}{y^2} = \dfrac{1}{4}$

i.e. $\;$ $\left(\dfrac{1}{x + 1} + \dfrac{1}{y}\right) \left(\dfrac{1}{x + 1} - \dfrac{1}{y}\right) = \dfrac{1}{4}$

i.e. $\;$ $\dfrac{1}{3} \left(\dfrac{1}{x + 1} - \dfrac{1}{y}\right) = \dfrac{1}{4}$ $\;\;$ [in view of equation $(1)$]

i.e. $\;$ $\dfrac{1}{x + 1} - \dfrac{1}{y} = \dfrac{3}{4}$ $\;\;\; \cdots \; (2)$

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

$\dfrac{2}{x + 1} = \dfrac{1}{3} + \dfrac{3}{4} = \dfrac{13}{12}$

i.e. $\;$ $24 = 13x + 13$

i.e. $\;$ $13x = 11$ $\implies$ $x = \dfrac{11}{13}$

Substitute $x = \dfrac{11}{13}$ in equation $(1)$ to get

$\dfrac{1}{\dfrac{11}{13} + 1} + \dfrac{1}{y} = \dfrac{1}{3}$

i.e. $\;$ $\dfrac{13}{24} + \dfrac{1}{y} = \dfrac{1}{3}$

i.e. $\;$ $\dfrac{1}{y} = \dfrac{1}{3} - \dfrac{13}{24} = \dfrac{-5}{24}$ $\implies$ $y = \dfrac{-24}{5}$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\dfrac{11}{13}, \dfrac{-24}{5}\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{3x + y}{x - 1} - \dfrac{x - y}{2y} = 2, \;\; x - y = 4$

Given system of equations:

$\dfrac{3x + y}{x - 1} - \dfrac{x - y}{2y} = 2$ $\;\;\; \cdots \; (1)$

[Equation $(1)$ is not valid when $\;$ $x - 1 = 0$ $\;$ i.e. $\;$ $x = 1$ $\;$ or when $\;$ $y = 0$] $\;\;\; \cdots \; (1a)$

and $\;$ $x - y = 4$ $\;$ i.e. $\;$ $x = y + 4$ $\;\;\; \cdots \; (2)$

In view of equation $(2)$, equation $(1)$ becomes

$\dfrac{3 \left(y + 4\right) + y}{y + 4 - 1} - \dfrac{y + 4 - y}{2y} = 2$

i.e. $\;$ $\dfrac{4y + 12}{y + 3} - \dfrac{4}{2y} = 2$

i.e. $\;$ $\dfrac{2y + 6}{ y + 3} - \dfrac{1}{y} = 1$

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

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 $\;$ $y = -3$, $\;$ equation $(2)$ gives $\;$ $x = -3 + 4 = 1$

and when $\;$ $y = 1$, $\;$ equation $(2)$ gives $\;$ $x = 1 + 4 = 5$

But according to $(1a)$, $x \ne 1$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{x + 3}{y - 4} - \dfrac{x - 1}{y + 4} + \dfrac{16}{y^2 - 16} = 0, \;\; 11x - 3y = 1$

Given system of equations:

$\dfrac{x + 3}{y - 4} - \dfrac{x - 1}{y + 4} + \dfrac{16}{y^2 - 16} = 0$ $\;\;\; \cdots \; (1)$

and $\;$ $11x - 3y = 1$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as

$\left(x + 3\right) \left(y + 4\right) - \left(x - 1\right) \left(y - 4\right) + 16 = 0$ $\;\;\;$ (provided $\;$ $y + 4 \neq 0$ $\;$ and $\;$ $y - 4 \neq 0$ $\;$ i.e. $\;$ $y \neq -4$ $\;$ and $\;$ $y \neq 4$)

i.e. $\;$ $xy + 4x + 3y + 12 - xy + 4x + y - 4 + 16 = 0$

i.e. $\;$ $8x + 4y + 24 = 0$

i.e. $\;$ $2x + y + 6 = 0$ $\;\;\; \cdots \; (1a)$

Multiplying equation $(1a)$ with $3$ and equation $(2)$ with $1$ and adding them gives

$17x = - 17$ $\implies$ $x = -1$

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

$-2 + y + 6 = 0$ $\implies$ $y = -4$

But equation $(1)$ requires $y \neq -4$

$\therefore \;$ The given pair of equations donot have any solution i.e. the solution set is $\phi$ (null set).

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{1}{3x} - \dfrac{1}{2y} = \dfrac{1}{3}, \;\; \dfrac{1}{9x^2} - \dfrac{1}{4y^2} = \dfrac{1}{4}$

Given system of equations:

$\dfrac{1}{3x} - \dfrac{1}{2y} = \dfrac{1}{3}$ $\;\;\; \cdots \; (1)$

and $\;$ $\dfrac{1}{9x^2} - \dfrac{1}{4y^2} = \dfrac{1}{4}$ $\;\;\; \cdots \; (2)$

Let $\;$ $\dfrac{1}{3x} = u$ $\;\;\; \cdots \; (3)$ $\;$ and $\;$ $\dfrac{1}{2y} = v$ $\;\;\; \cdots \; (4)$

In view of equations $(3)$ and $(4)$, equations $(1)$ and $(2)$ respectively become

$u - v = \dfrac{1}{3}$ $\;$ i.e. $\;$ $u = v + \dfrac{1}{3}$ $\;\;\; \cdots \; (5)$

and $\;$ $u^2 - v^2 = \dfrac{1}{4}$ $\;\;\; \cdots \; (6)$

Substituting the value of $u$ from equation $(5)$ in equation $(6)$ gives

$\left(v + \dfrac{1}{3}\right)^2 - v^2 = \dfrac{1}{4}$

i.e. $\;$ $\dfrac{2v}{3} + \dfrac{1}{9} = \dfrac{1}{4}$

i.e. $\;$ $\dfrac{2v}{3} = \dfrac{5}{36}$

i.e. $\;$ $v = \dfrac{5}{24}$

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

$u = \dfrac{5}{24} + \dfrac{1}{3} = \dfrac{13}{24}$

For $\;$ $u = \dfrac{13}{24}$, $\;$ we have from equation $(3)$

$\dfrac{1}{3x} = \dfrac{13}{24}$ $\implies$ $x = \dfrac{8}{13}$

For $\;$ $v = \dfrac{5}{24}$, $\;$ we have from equation $(4)$

$\dfrac{1}{2y} = \dfrac{5}{24}$ $\implies$ $y = \dfrac{12}{5}$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\dfrac{8}{13}, \dfrac{12}{5}\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{x + y}{x - y} + \dfrac{x - y}{x + y} = \dfrac{13}{6}, \;\; x \; y = 5$

Given system of equations:

$\dfrac{x + y}{x - y} + \dfrac{x - y}{x + y} = \dfrac{13}{6}$

i.e. $\;$ $\dfrac{\left(x + y\right)^2 + \left(x - y\right)^2}{\left(x + y\right) \left(x - y\right)} = \dfrac{13}{6}$

i.e. $\;$ $\dfrac{x^2 + y^2 + 2xy + x^2 + y^2 - 2xy}{x^2 - y^2} = \dfrac{13}{6}$

i.e. $\;$ $2 \left(x^2 + y^2\right) = \dfrac{13}{6} \left(x^2 - y^2\right)$

i.e. $\;$ $12 x^2 + 12 y^2 = 13 x^2 - 13 y^2$

i.e. $\;$ $x^2 = 25 y^2$ $\;\;\; \cdots \; (1)$

and $\;$ $x \; y = 5$ $\;$ i.e. $\;$ $x = \dfrac{5}{y}$ $\;\;\; \cdots \; (2)$

In view of equation $(2)$, equation $(1)$ becomes

$\left(\dfrac{5}{y}\right)^2 = 25 y^2$

i.e. $\;$ $\dfrac{25}{y^2} = 25 y^2$ $\;$ i.e. $\;$ $y^4 = 1$ $\;\;\; \cdots \; (3)$

Let $\;$ $y^2 = p$ $\;\;\; \cdots \; (4)$

Then equation $(3)$ can be written as

$p^2 = 1$ $\implies$ $p = \pm 1$

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

when $\;$ $p = +1$, $\;$ $y^2 = 1$ $\implies$ $y = \pm 1$

and when $\;$ $p = -1$, $\;$ $y^2 = -1$

But $y^2 = -1$ is not a valid solution since square of any number cannot be negative.

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

when $\;$ $y = 1$, $\;$ $x = \dfrac{5}{1} = 5$

and when $\;$ $y = -1$, $\;$ $x = \dfrac{5}{-1} = -5$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\dfrac{x}{y} - \dfrac{y}{x} = \dfrac{5}{6}, \;\; x^2 - y^2 = 5$

Given system of equations:

$\dfrac{x}{y} - \dfrac{y}{x} = \dfrac{5}{6}$ $\;$ i.e. $\;$ $x^2 - y^2 = \dfrac{5 xy}{6}$ $\;\;\; \cdots \; (1)$

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

In view of equation $(2)$, equation $(1)$ becomes

$\dfrac{5xy}{6} = 5$

i.e. $\;$ $x \; y = 6$ $\implies$ $y = \dfrac{6}{x}$ $\;\;\; \cdots \; (3)$

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

$x^2 - \left(\dfrac{6}{x}\right)^2 = 5$

i.e. $\;$ $x^2 - \dfrac{36}{x^2} = 5$

i.e. $\;$ $x^4 - 5x^2 - 36 = 0$

i.e. $\;$ $\left(x^2 - 9\right) \left(x^2 + 4\right) = 0$

i.e. $\;$ $x^2 = 9$ $\;$ or $\;$ $x^2 = -4$

But $x^2 = -4$ is not a valid solution since square of any number cannot be negative.

Now, $\;$ $x^2 = 9$ $\implies$ $x = \pm 3$

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

when $\;$ $x = 3$, $\;$ $y = \dfrac{6}{3} = 2$

and when $\;$ $x = -3$, $\;$ $y = \dfrac{6}{-3} = -2$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^2 + y^2 + 6x + 2y = 0, \;\; x + y + 8 = 0$

Given system of equations:

$x^2 + y^2 + 6x + 2y = 0$ $\;\;\; \cdots \; (1)$

$x + y = 8$ $\;$ i.e. $\;$ $y = x - 8$ $\;\;\; \cdots \; (2)$

In view of equation $(2)$, equation $(1)$ becomes

$x^2 + \left(-x - 8\right)^2 + 6x + 2 \left(-x - 8\right) = 0$

i.e. $\;$ $x^2 + x^2 + 16x + 64 + 6x - 2x - 16 = 0$

i.e. $\;$ $2x^2 + 20x + 48 = 0$

i.e. $\;$ $\left(x + 6\right) \left(x + 4\right) = 0$

i.e. $\;$ $x = -6$ $\;$ or $\;$ $x = -4$

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

when $\;$ $x = -6$, $\;$ $y = - \left(-6\right) - 8 = -2$

and when $\;$ $x = -4$, $\;$ $y = - \left(-4\right) - 8 = -4$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(-6, -2\right), \left(-4, -4\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x + y = 7, \;\; y = \dfrac{6}{x}$

Given system of equations:

$x + y = 7$ $\;\;\; \cdots \; (1)$

$y = \dfrac{6}{x}$ $\;\;\; \cdots \; (2)$

In view of equation $(2)$, equation $(1)$ becomes

$x + \dfrac{6}{x} = 7$

i.e. $\;$ $x^2 - 7x + 6 = 0$

i.e. $\;$ $\left(x - 6\right) \left(x - 1\right) = 0$

i.e. $\;$ $x = 6$ $\;$ or $\;$ $x = 1$

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

when $\;$ $x = 6$, $\;$ $y = \dfrac{6}{6} = 1$

and when $\;$ $x = 1$, $\;$ $y = \dfrac{6}{1} = 6$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(6, 1\right), \left(1, 6\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^2 + y^2 = 41, \;\; y - x = 1$

Given system of equations:

$x^2 + y^2 = 41$ $\;\;\; \cdots \; (1)$

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

In view of equation $(2)$, equation $(1)$ becomes

$x^2 + \left(x + 1\right)^2 = 41$

i.e. $\;$ $x^2 + x^2 + 2x + 1 = 41$

i.e. $\;$ $2x^2 + 2x - 40 = 0$

i.e. $\;$ $x^2 + x - 20 = 0$

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

i.e. $\;$ $x = -5$ $\;$ or $\;$ $x = 4$

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

When $\;$ $x = -5$, $\;$ $y = -5 + 1 = -4$

and when $\;$ $x = 4$, $\;$ $y = 4 + 1 = 5$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x^2 - y^2 = 16, \;\; x + y = 8$

Given system of equations:

$x^2 - y^2 = 16$ $\;$ i.e. $\;$ $\left(x + y\right) \left(x - y\right) = 16$ $\;\;\; \cdots \; (1)$

$x + y = 8$ $\;\;\; \cdots \; (2)$

In view of equation $(2)$, equation $(1)$ becomes

$8 \left(x - y\right) = 16$

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

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

$2x = 10$ $\implies$ $x = 5$

Substituting $x = 5$ in equation $(2)$ gives

$5 + y = 8$ $\implies$ $y = 3$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $xy + 4 = 0, \;\; x + y = 3$

Given system of equations:

$xy + 4 = 0$ $\;\;\; \cdots \; (1)$

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

In view of equation $(2)$, equation $(1)$ becomes

$\left(3 - y\right) y + 4 = 0$

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

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

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

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

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

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

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $xy + x + y = 11, \;\; x^2 y + x y^2 = 30$

Given system of equations:

$xy + x + y = 11$ $\;\;\; \cdots \; (1)$

$x^2 y + x y^2 = 30$ $\;$ i.e. $\;$ $xy \left(x + y\right) = 30$ $\;\;\; \cdots \; (2)$

Let $\;$ $xy = u$ $\;\;\; \cdots \; (3)$ $\;$ and $\;$ $x + y = v$ $\;\;\; \cdots \; (4)$

In view of equations $(3)$ and $(4)$, equations $(1)$ and $(2)$ become

$u + v = 11$ $\implies$ $u = 11 - v$ $\;\;\; \cdots \; (5)$

and $\;$ $uv = 30$ $\;\;\; \cdots \; (6)$

Substituting the value of $u$ from equation $(5)$ in equation $(6)$ gives

$\left(11 - v\right) v = 30$

i.e. $\;$ $v^2 - 11v + 30 = 0$

i.e. $\;$ $\left(v - 6\right) \left(v - 5\right) = 0$

i.e. $\;$ $v = 6$ $\;$ or $\;$ $v = 5$

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

when $\;$ $v = 6$, $\;$ $u = 11 - 6 = 5$;

when $\;$ $v = 5$, $\;$ $u = 11 - 5 = 6$

Substituting $u = 5$ and $v = 6$ in equations $(3)$ and $(4)$ gives

$xy = 5$ $\;\;\; \cdots \; (7)$

and $\;$ $x + y = 6$ $\implies$ $x = 6 - y$ $\;\;\; \cdots \; (8)$

In view of equation $(8)$ equation $(7)$ becomes

$\left(6 - y\right) y = 5$

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

i.e. $\;$ $\left(y - 5\right) \left(y - 1\right) = 0$

i.e. $\;$ $y = 5$ $\;$ or $\;$ $y = 1$

$\therefore \;$ We have from equation $(8)$

when $\;$ $y = 5$, $\;$ $x = 6 - 5 = 1$

and when $\;$ $y = 1$, $\;$ $x = 6 - 1 = 5$

Next, substituting $u = 6$ and $v = 5$ in equations $(3)$ and $(4)$ gives

$xy = 6$ $\;\;\; \cdots \; (9)$

and $\;$ $x + y = 5$ $\implies$ $x = 5 - y$ $\;\;\; \cdots \; (10)$

In view of equation $(10)$ equation $(9)$ becomes

$\left(5 - y\right) y = 6$

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

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

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

$\therefore \;$ We have from equation $(10)$

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

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

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(1, 5\right), \; \left(5, 1\right), \; \left(3, 2\right), \; \left(2, 3\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x + 2y - z = 7, \;\; 2x - y + z = 2, \;\; 3x - 5y + 2z = -7$

Given system of equations:

$x + 2y - z = 7$ $\;\;\; \cdots \; (1)$

$2x - y + z = 2$ $\;\;\; \cdots \; (2)$

$3x - 5y + 2z = -7$ $\;\;\; \cdots \; (3)$

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

$3x + y = 9$ $\;\;\; \cdots \; (4)$

Multiply equation $(1)$ with $2$ and add to equation $(3)$. We get

$5x - y = 7$ $\;\;\; \cdots \; (5)$

Adding equations $(4)$ and $(5)$ gives

$8x = 16$ $\implies$ $x = 2$

Substitute the value of $x$ in equation $(4)$ to get

$6 + y = 9$ $\implies$ $y = 3$

Substituting the values of $x$ and $y$ in equation $(2)$ gives

$4 - 3 + z = 2$ $\implies$ $z = 1$

$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y, z\right) = \left\{\left(2, 3, 1\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x + y - z = 2, \;\; 2x - y + 4z = 1, \;\; -x + 6y + z = 5$

Given system of equations:

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

$2x - y + 4z = 1$ $\;\;\; \cdots \; (2)$

$-x + 6y + z = 5$ $\;\;\; \cdots \; (3)$

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

$7y = 7$ $\implies$ $y = 1$

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

$x - z = 1$ $\;\;\; \cdots \; (4)$

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

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

Subtracting equations $(4)$ and $(5)$ gives

$-3z = 0$ $\implies$ $z = 0$

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

$x = 1$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\left|x\right| + 2 \left|y\right| = 3, \;\; 5y + 7x = 2$

Given system of equations: $\;$ $\left|x\right| + 2 \left|y\right| = 3$ $\;\;\; \cdots \; (1)$; $\;$ $5y + 7x = 2$ $\;\;\; \cdots \; (2)$

Case 1:

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

Then equation $(1)$ becomes $\;\;$ $x + 2y = 3$ $\;\;\; \cdots \; (3)$

Solving equations $(2)$ and $(3)$ simultaneously gives $\;$ $x = \dfrac{-11}{9}, \; y = \dfrac{19}{9}$

But $\;$ $x = \dfrac{-11}{9}$ $\;$ is not a valid solution since it is required that $\;$ $x > 0$.

Case 2:

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

Then equation $(1)$ becomes $\;\;$ $x - 2y = 3$ $\;\;\; \cdots \; (4)$

Solving equations $(2)$ and $(4)$ simultaneously gives $\;$ $x = 1, \; y = -1$

$\left(x, y\right) = \left(1, -1\right)$ $\;$ is a valid solution since it is required that $\;$ $x > 0, \; y < 0$.

Case 3:

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

Then equation $(1)$ becomes $\;\;$ $- x + 2y = 3$ $\;\;\; \cdots \; (5)$

Solving equations $(2)$ and $(5)$ simultaneously gives $\;$ $x = \dfrac{-11}{19}, \; y = \dfrac{23}{19}$

$\left(x, y\right) = \left(\dfrac{-11}{19}, \dfrac{23}{19}\right)$ $\;$ is a valid solution since it is required that $\;$ $x < 0, \; y > 0$.

Case 4:

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

Then equation $(1)$ becomes $\;\;$ $- x - 2y = 3$ $\;\;\; \cdots \; (6)$

Solving equations $(2)$ and $(6)$ simultaneously gives $\;$ $x = \dfrac{19}{9}, \; y = \dfrac{-23}{9}$

But $\;$ $x = \dfrac{23}{9}$ $\;$ is not a valid solution since it is required that $\;$ $x < 0$.

$\therefore \;$ The solution to the given pair of equations is $\;$ $\left(x, y\right) = \left\{\left(1, -1\right), \; \left(\dfrac{-11}{19}, \dfrac{23}{19}\right) \right\}$