Algebra - Exponential Equations

Solve the equation: $\;$ $3^{2x - 3} - 9^{x - 1} + 27^{\frac{2x}{3}} = 675$

Given equation: $\;\;$ $3^{2x - 3} - 9^{x - 1} + 27^{\frac{2x}{3}} = 675$

i.e. $\;$ $3^{2x} \times 3^{-3} - \left(3^2\right)^{x - 1} + \left(3^3\right)^{\frac{2x}{3}} = 25 \times 27$

i.e. $\;$ $3^{2x} \times 3^{-3} - 3^{2x} \times 3^{-2} + 3^{2x} = 25 \times 3^3$

i.e. $\;$ $3^{2x} \left(\dfrac{1}{3^3} - \dfrac{1}{3^2} + 1\right) = 25 \times 3^3$

i.e. $\;$ $3^{2x - 3} \left(\dfrac{1 - 3 + 27}{27}\right) = 25$

i.e. $\;$ $3^{2x - 3} \times \dfrac{25}{27} = 25$

i.e. $\;$ $3^{2x - 3} = 27 = 3^3$

$\implies$ $2x - 3 = 3$

i.e. $\;$ $2x = 6$ $\implies$ $x = 3$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = 3$

Algebra - Exponential Equations

Solve the equation: $\;$ $5^{2x} - 7^x - 5^{2x} \times 35 + 7^x \times 35 = 0$

Given equation: $\;\;$ $5^{2x} - 7^x - 5^{2x} \times 35 + 7^x \times 35 = 0$

i.e. $\;$ $5^{2x} - 7^x = 5^{2x} \times 35 - 7^x \times 35$

i.e. $\;$ $5^{2x} - 7^x = 35 \left(5^{2x} - 7^x\right)$ $\;\;\; \cdots \; (1)$

In equation $(1)$, if $\;$ $5^{2x} - 7^x \neq 0$, $\;$ then dividing both sides of equation $(1)$ by $5^{2x} - 7^x$ gives

$1 = 35$ $\;\;$ which is not possible.

$\implies$ In equation $(1)$, $\;\;$ $5^{2x} - 7^x = 0$

i.e. $\;$ $\left(5^2\right)^x = 7^x$

i.e. $\;$ $25^x = 7^x$

i.e. $\;$ $\dfrac{25^x}{7^x} = 1$

i.e. $\;$ $\left(\dfrac{25}{7}\right)^x = 1$

i.e. $\;$ $\left(\dfrac{25}{7}\right)^x = \left(\dfrac{25}{7}\right)^0$

$\implies$ $x = 0$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = 0$

Algebra - Exponential Equations

Solve the equation: $\;$ $5^{x-1} = 10^x \times 2^{-x} \times 5^{x + 1}$

Given equation: $\;\;$ $5^{x-1} = 10^x \times 2^{-x} \times 5^{x + 1}$

i.e. $\;$ $5^{x - 1} = \left(5 \times 2\right)^x \times 2^{-x} \times 5^{x + 1}$

i.e. $\;$ $5^{x - 1} = 5^x \times 2^x \times 2^{-x} \times 5^{x + 1}$

i.e. $\;$ $5^{x - 1} = 5^{x + x + 1} \times 2^{x - x}$

i.e. $\;$ $5^{x - 1} = 5^{2x + 1} \times 2^0$

i.e. $\;$ $5^{x - 1} = 5^{2x + 1} \times 1$

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

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

$\therefore \;$ The solution to the given equation is $\;\;$ $x = -2$

Algebra - Exponential Equations

Solve the equation: $\;$ $\left(\dfrac{5}{12}\right)^x \times \left(\dfrac{6}{5}\right)^{x - 1} = \left(0.3\right)^{-1}$

Given equation: $\;\;$ $\left(\dfrac{5}{12}\right)^x \times \left(\dfrac{6}{5}\right)^{x - 1} = \left(0.3\right)^{-1}$

i.e. $\;$ $\left(\dfrac{5}{3 \times 2^2}\right)^x \times \left(\dfrac{3 \times 2}{5}\right)^{x - 1} = \left(\dfrac{3}{10}\right)^{-1}$

i.e. $\;$ $\dfrac{5^x}{3^x \times 2^{2x}} \times \dfrac{3^{x-1} \times 2^{x-1}}{5^{x-1}} = \left(\dfrac{3}{5 \times 2}\right)^{-1}$

i.e. $\;$ $5^{x - x + 1} \times 3^{x - 1 - x} \times 2^{x - 1 - 2x} = \dfrac{3^{-1}}{5^{-1} \times 2^{-1}}$

i.e. $\;$ $\dfrac{5^1 \times 3^{-1} \times 2^{-x - 1} \times 5^{-1} \times 2^{-1}}{3^{-1}} = 1$

i.e. $\;$ $5^{1 - 1} \times 3^{-1 + 1} \times 2^{-x - 1 - 1} = 1$

i.e. $\;$ $5^0 \times 3^0 \times 2^{-x - 2} = 1$

i.e. $\;$ $1 \times 1 \times 2^{-x - 2} = 2^0$

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

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

Algebra - Exponential Equations

Solve the equation: $\;$ $\dfrac{2^{x - 1} \times 4^{1 - x}}{8^{x - 1}} = 64$

Given equation: $\;\;$ $\dfrac{2^{x - 1} \times 4^{1 - x}}{8^{x - 1}} = 64$ $\;\;\; \cdots \; (1)$

i.e. $\;$ $\dfrac{2^{x - 1} \times \left(2^2\right)^{1 - x}}{\left(2^3\right)^{x - 1}} = 2^6$

i.e. $\;$ $2^{\left(x - 1 + 2 - 2x - 3x + 3\right)} = 2^6$

i.e. $\;$ $2^{-4x + 4} = 2^6$

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

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

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

Algebra - Exponential Equations

Solve the equation: $\;$ $4^{\left(x - \sqrt{x^2 - 5}\right)} - 12 \times 2^{\left(x - 1 - \sqrt{x^2 - 5}\right)} + 8 = 0$

Given equation: $\;\;$ $4^{\left(x - \sqrt{x^2 - 5}\right)} - 12 \times 2^{\left(x - 1 - \sqrt{x^2 - 5}\right)} + 8 = 0$ $\;\;\; \cdots \; (1)$

i.e. $\;$ $\left(2^2\right)^{\left(x - \sqrt{x^2 - 5}\right)} - \dfrac{12}{2} \times 2^{\left(x - \sqrt{x^2 - 5}\right)} + 8 = 0$

i.e. $\;$ $\left[2^{\left(x - \sqrt{x^2 - 5}\right)}\right]^2 - 6 \times 2^{\left(x - \sqrt{x^2 - 5}\right)} + 8 = 0$ $\;\;\; \cdots \; (1a)$

Let $\;$ $2^{\left(x - \sqrt{x^2 - 5}\right)} = p$ $\;\;\; \cdots \; (2)$

Then, equation $(1a)$ becomes,

$p^2 - 6p + 8 = 0$

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

i.e. $\;$ $p = 4$ $\;\;$ or $\;\;$ $p = 2$

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

when $\;$ $p = 4$, $\;$ $2^{\left(x - \sqrt{x^2 - 5}\right)} = 4 = 2^2$

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

i.e. $\;$ $\sqrt{x^2 - 5} = x - 2$ $\;\;\; \cdots \; (3)$

Squaring equation $(3)$ gives

$x^2 - 5 = x^2 - 4x + 4$

i.e. $\;$ $4x = 9$ $\implies$ $x = \dfrac{9}{4}$

when $\;$ $p = 2$, $\;$ $2^{\left(x - \sqrt{x^2 - 5}\right)} = 2 = 2^1$

i.e. $\;$ $x - \sqrt{x^2 - 5} = 1$

i.e. $\;$ $\sqrt{x^2 - 5} = x - 1$ $\;\;\; \cdots \; (4)$

Squaring equation $(4)$ gives

$x^2 - 5 = x^2 - 2x + 1$

i.e. $\;$ $2x = 6$ $\implies$ $x = 3$

$\therefore \;$ The solution to the given equation are $\;\;$ $x = \dfrac{9}{4}$, $\;$ $x = 3$

Algebra - Exponential Equations

Solve the equation: $\;$ $4^x - 10 \times 2^{x - 1} = 24$

Given equation: $\;\;$ $4^x - 10 \times 2^{x - 1} = 24$ $\;\;\; \cdots \; (1)$

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

Then, $\;\;$ $4^x = \left(2^2\right)^x = \left(2^x\right)^2 = p^2$ $\;\;\; \cdots \; (3)$

Equation $(1)$ can be written as

$4^x - 10 \times \dfrac{2^x}{2} = 24$

i.e. $\;$ $4^x - 5 \times 2^x = 24$

i.e. $\;$ $p^2 - 5p - 24 = 0$ $\;\;\;$ [in view of equations $(2)$ and $(3)$]

i.e. $\;$ $\left(p - 8\right) \left(p + 3\right) = 0$

i.e. $\;$ $p = 8$ $\;\;$ or $\;\;$ $p = -3$

When $\;$ $p = 8$, $\;$ we have from equation $(2)$, $\;$ $2^x = 8 = 2^3$ $\implies$ $x = 3$

When $\;$ $p = -3$, $\;$ we have from equation $(2)$, $\;$ $2^x = -3$ $\implies$ $x = \log_2 \left(-3\right)$

But logarithim of a negative number is not defined.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = 3$.

Algebra - Exponential Equations

Solve the equation: $\;$ $5^x - 24 = \dfrac{25}{5^x}$

Given equation: $\;\;$ $5^x - 24 = \dfrac{25}{5^x}$ $\;\;\; \cdots \; (1)$

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

$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes

$p^2 - 24p - 25 = 0$

i.e. $\;$ $\left(p - 25\right) \left(p + 1\right) = 0$

i.e. $\;$ $p = 25$ $\;\;$ or $\;\;$ $p = -1$

When $\;$ $p = 25$, $\;$ we have from equation $(2)$,

$5^x = 25 = 5^2$ $\implies$ $x = 2$

When $\;$ $p = -1$, $\;$ we have from equation $(2)$,

$5^x = -1$ $\implies$ $x = \log_5 \left(-1\right)$

But logarithom of a negative number is not defined.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = 2$.

Algebra - Exponential Equations

Solve the equation: $\;$ $4^{\frac{2}{x}} - 5 \times 4^{\frac{1}{x}} + 4 = 0$

Given equation: $\;\;$ $4^{\frac{2}{x}} - 5 \times 4^{\frac{1}{x}} + 4 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $4^{\frac{1}{x}} = p$ $\;\;\; \cdots \; (2)$

Then $\;\;$ $4^{\frac{2}{x}} = \left[\left(4\right)^{\frac{1}{x}}\right]^2 = p^2$ $\;\;\; \cdots \; (3)$

$\therefore \;$ In view of equations $(2)$ and $(3)$, equation $(1)$ 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 $\;$ $p = 4$, $\;$ we have from equation $(2)$,

$4^{\frac{1}{x}} = 4 = 4^1$ $\implies$ $\dfrac{1}{x} = 1$ $\implies$ $x = 1$

When $\;$ $p = 1$, $\;$ we have from equation $(2)$,

$4^{\frac{1}{x}} = 1 = 4^0$ $\implies$ $\dfrac{1}{x} = 0$ $\implies$ $x = \dfrac{1}{0}$ $\;$ which is not possible.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = 1$.

Algebra - Exponential Equations

Solve the equation: $\;$ $2^{\frac{3}{\log_2 x}} = \dfrac{1}{64}$

Given equation: $\;\;$ $2^{\frac{3}{\log_2 x}} = \dfrac{1}{64}$

i.e. $\;$ $2^{\frac{3}{\log_2 x}} = 2^{-6}$

i.e. $\;$ $\dfrac{3}{\log_2 x} = -6$

i.e. $\;$ $-2 = \dfrac{1}{\log_2 x}$

i.e. $\;$ $\log_2 x = \dfrac{-1}{2}$

i.e. $\;$ $x = 2^{\frac{-1}{2}} = \dfrac{1}{\sqrt{2}}$

Algebra - Exponential Equations

Solve the equation: $\;$ $\left(\dfrac{5}{3}\right)^{x + 1} \times \left(\dfrac{9}{25}\right)^{x^2 + 2x - 11} = \left(\dfrac{5}{3}\right)^9$

Given equation: $\;\;$ $\left(\dfrac{5}{3}\right)^{x + 1} \times \left(\dfrac{9}{25}\right)^{x^2 + 2x - 11} = \left(\dfrac{5}{3}\right)^9$

i.e. $\;$ $\left(\dfrac{5}{3}\right)^{x + 1} \times \left[\left(\dfrac{3}{5}\right)^2\right]^{x^2 + 2x - 11} = \left(\dfrac{5}{3}\right)^9$

i.e. $\;$ $\dfrac{5^{x + 1}}{3^{x + 1}} \times \dfrac{3^{2x^2 + 4x - 22}}{5^{2x^2 + 4x - 22}} = \left(\dfrac{5}{3}\right)^9$

i.e. $\;$ $\dfrac{5^{x + 1 - 2x^2 - 4x + 22}}{3^{x + 1 - 2x^2 - 4x + 22}} = \left(\dfrac{5}{3}\right)^9$

i.e. $\;$ $\left(\dfrac{5}{3}\right)^{-2x^2 - 3x + 23} = \left(\dfrac{5}{3}\right)^9$

i.e. $\;$ $-2x^2 - 3x + 23 = 9$

i.e. $\;$ $2x^2 + 3x - 14 = 0$

i.e. $\;$ $\left(x - 2\right) \left(2x + 7\right) = 0$

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\sqrt[3]{x} + \sqrt[3]{y} = 3$, $\;$ $\sqrt[3]{x^2} - \sqrt[3]{xy} + \sqrt[3]{y^2} = 3$

Given system of equations:

$\sqrt[3]{x} + \sqrt[3]{y} = 3$ $\;\;\; \cdots \; (1)$, $\;\;$ $\sqrt[3]{x^2} - \sqrt[3]{xy} + \sqrt[3]{y^2} = 3$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as

$x^{\frac{1}{3}} + y^{\frac{1}{3}} = 3$ $\;\;$ i.e. $\;\;$ $x^{\frac{1}{3}} = 3 - y^{\frac{1}{3}}$ $\;\;\; \cdots \; (3)$

Equation $(2)$ can be written as

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

In view of equation $(3)$, equation $(4)$ becomes

$\left(3 - y^{\frac{1}{3}}\right)^2 - \left(3 - y^{\frac{1}{3}}\right) y^{\frac{1}{3}} + y^{\frac{2}{3}} = 3$

i.e. $\;$ $9 + y^{\frac{2}{3}} - 6 y^{\frac{1}{3}} - 3 y^{\frac{1}{3}} + y^{\frac{2}{3}} + y^{\frac{2}{3}} = 3$

i.e. $\;$ $3 y^{\frac{2}{3}} - 9 y^{\frac{1}{3}} + 6 = 0$

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

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

i.e. $\;$ $y^{\frac{1}{3}} = 2$ $\;\;\;$ or $\;\;\;$ $y^{\frac{1}{3}} = 1$

i.e. $\;$ $y = 8$ $\;\;\;$ or $\;\;\;$ $y = 1$

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

when $\;$ $y = 8$, $\;$ $x^{\frac{1}{3}} = 3 - 8^{\frac{1}{3}} = 3 - 2 = 1$ $\implies$ $x = 1$

when $\;$ $y = 1$, $\;$ $x^{\frac{1}{3}} = 3 - 1^{\frac{1}{3}} = 3 - 1 = 2$ $\implies$ $x = 8$

$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(1, 8\right), \; \left(8, 1\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\sqrt{\dfrac{20 y}{x}} = \sqrt{x + y} + \sqrt{x - y}$, $\;$ $\sqrt{\dfrac{16 x}{5 y}} = \sqrt{x + y} - \sqrt{x - y}$

Given system of equations:

$\sqrt{\dfrac{20 y}{x}} = \sqrt{x + y} + \sqrt{x - y}$ $\;\;\; \cdots \; (1)$, $\;\;$ $\sqrt{\dfrac{16x}{5y}} = \sqrt{x + y} - \sqrt{x - y}$ $\;\;\; \cdots \; (2)$

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

$\sqrt{\dfrac{20 y}{x} \times \dfrac{16 x}{5 y}} = \left(\sqrt{x + y} + \sqrt{x - y}\right) \left(\sqrt{x + y} - \sqrt{x - y}\right)$

i.e. $\;$ $2 \times 4 = x + y - \sqrt{x^2 - y^2} + \sqrt{x^2 - y^2} - x + y$

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

Substituting $y = 4$ in equation $(1)$ gives

$\sqrt{\dfrac{20 \times 4}{x}} = \sqrt{x + 4} + \sqrt{x - 4}$

i.e. $\;$ $\dfrac{4 \sqrt{5}}{\sqrt{x}} = \sqrt{x + 4} + \sqrt{x - 4}$ $\;\;\; \cdots \; (3)$

Squaring equation $(3)$ gives

$\dfrac{80}{x} = x + 4 + x - 4 + 2 \sqrt{x^2 - 16}$

i.e. $\;$ $\dfrac{40}{x} - x = \sqrt{x^2 + 16}$ $\;\;\; \cdots \; (4)$

Squaring equation $(4)$ gives

$\dfrac{1600}{x^2} + x^2 - 80 = x^2 - 16$

i.e. $\;$ $\dfrac{1600}{x^2} = 64$

i.e. $\;$ $x^2 = \dfrac{1600}{64} = 25$ $\implies$ $x = \pm 5$

When $\;$ $x = -5, \; y = 4$, $\;$ $\sqrt{x + y} = -1$ $\;$ i.e. an imaginary number.

$\therefore \;$ The solution $\;$ $\left(x, y\right) = \left(-5, 4\right)$ is discarded.

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

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x \sqrt{y} + y \sqrt{x} = 6$, $\;$ $x^2 y + y^2 x = 20$

Given system of equations:

$x \sqrt{y} + y \sqrt{x} = 6$ $\;\;\; \cdots \; (1)$, $\;\;$ $x^2 y + y^2 x = 20$ $\;\;\; \cdots \; (2)$

Squaring equation $(1)$ gives

$\left(x \sqrt{y} + y \sqrt{x} \right)^2 = 6^2$

i.e. $\;$ $x^2 y + y^2 x + 2 xy \sqrt{xy} = 36$

i.e. $\;$ $20 + 2x^{\frac{3}{2}} y^{\frac{3}{2}} = 36$ $\;\;\;$ [in view of equation $(2)$]

i.e. $\;$ $x^{\frac{3}{2}} y^{\frac{3}{2}} = 8 = 2^3$

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

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

Substituting $\;$ $x = \dfrac{4}{y}$ $\;$ in equation $(1)$ gives

$\dfrac{4}{y} \times \sqrt{y} + y \times \dfrac{2}{\sqrt{y}} = 6$

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

i.e. $\;$ $y - 3 \sqrt{y} + 2 = 0$

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

i.e. $\;$ $\sqrt{y} = 2$ $\;\;\;$ or $\;\;\;$ $\sqrt{y} = 1$

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

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

when $\;$ $y = 4$, $\;$ $x = \dfrac{4}{4} = 1$

when $\;$ $y = 1$, $\;$ $x = \dfrac{4}{1} = 4$

$\therefore \;$ 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: $\;$ $\sqrt{x} + \sqrt{y} = 10$, $\;$ $\sqrt[4]{x} + \sqrt[4]{y} = 4$

Given system of equations:

$\sqrt{x} + \sqrt{y} = 10$ $\;\;\; \cdots \; (1)$, $\;\;$ $\sqrt[4]{x} + \sqrt[4]{y} = 4$ $\;\;\; \cdots \; (2)$

Let $\;\;$ $\sqrt[4]{x} = p$ $\;\;\; \cdots \; (3a)$, $\;$ $\sqrt[4]{y} = q$ $\;\;\; \cdots \; (3b)$

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

$p^2 + q^2 = 10$ $\;\;\; \cdots \; (1a)$ $\;\;$ and $\;\;$ $p + q = 4$ $\;\;\; \cdots \; (2a)$

We have from equation $(2a)$, $\;\;$ $p = 4 - q$ $\;\;\; \cdots \; (4)$

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

$\left(4 - q\right)^2 + q^2 = 10$

i.e. $\;$ $16 - 8q + q^2 + q^2 = 10$

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

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

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

i.e. $\;$ $q = 3$ $\;\;\;$ or $\;\;\;$ $q = 1$

When $\;$ $q = 3$, $\;$ we have from equation $(4)$, $\;$ $p = 4 - 3 = 1$

When $\;$ $q = 1$, $\;$ we have from equation $(5)$, $\;$ $p = 4 - 1 = 3$

For $\;$ $p = 1, \; q = 3$, $\;$ we have from equations $(3a)$ and $(3b)$

$\sqrt[4]{x} = 1$ $\implies$ $x = 1$ $\;\;$ and $\;\;$ $\sqrt[4]{y} = 3$ $\implies$ $y = 81$

For $\;$ $p = 3, \; q = 1$, $\;$ we have from equations $(3a)$ and $(3b)$

$\sqrt[4]{x} = 3$ $\implies$ $x = 81$ $\;\;$ and $\;\;$ $\sqrt[4]{y} = 1$ $\implies$ $y = 1$

$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(1, 81\right), \left(81, 1\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x + y = 9$, $\;$ $x^{\frac{1}{3}} + y^{\frac{1}{3}} = 3$

Given system of equations:

$x + y = 9$ $\;\;\; \cdots \; (1)$, $\;\;$ $x^{\frac{1}{3}} + y^{\frac{1}{3}} = 3$ $\;\;\; \cdots \; (2)$

Let $\;\;$ $x^{\frac{1}{3}} = p$ $\;\;\; \cdots \; (3)$, $\;$ $y^{\frac{1}{3}} = q$ $\;\;\; \cdots \; (4)$

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

$p^3 + q^3 = 9$ $\;\;\; \cdots \; (1a)$ $\;\;$ and $\;\;$ $p + q = 3$ $\;\;\; \cdots \; (2a)$

We have from equation $(2a)$, $\;\;$ $p = 3 - q$ $\;\;\; \cdots \; (5)$

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

$\left(3 - q\right)^3 + q^3 = 9$

i.e. $\;$ $27 - 27q + 9q^2 - q^3 + q^3 = 9$

i.e. $\;$ $9q^2 - 27q + 18 = 0$

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

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

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

When $\;$ $q = 2$, $\;$ we have from equation $(5)$, $\;$ $p = 3 - 2 = 1$

When $\;$ $q = 1$, $\;$ we have from equation $(5)$, $\;$ $p = 3 - 1 = 2$

When $\;$ $p = 1$ $\;$ and $\;$ $q = 2$,

we have from equation $(3)$, $\;$ $x^{\frac{1}{3}} = 1$ $\implies$ $x = 1$

and from equation $(4)$, $\;$ $y^{\frac{1}{3}} = 2$ $\implies$ $y = 8$

When $\;$ $p = 2$ $\;$ and $\;$ $q = 1$,

we have from equation $(3)$, $\;$ $x^{\frac{1}{3}} = 2$ $\implies$ $x = 8$

and from equation $(4)$, $\;$ $y^{\frac{1}{3}} = 1$ $\implies$ $y = 1$

$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(1, 8\right), \left(8, 1\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\sqrt[3]{x} - \sqrt[3]{y} = 2$, $\;$ $xy = 27$

Given system of equations:

$\sqrt[3]{x} - \sqrt[3]{y} = 2$ $\;\;\; \cdots \; (1)$, $\;\;$ $xy = 27$ $\;\;\; \cdots \; (2)$

We have from equation $(2)$, $\;\;\;$ $x = \dfrac{27}{y}$ $\;\;\; \cdots \; (3)$

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

$\sqrt[3]{\dfrac{27}{y}} - \sqrt[3]{y} = 2$

i.e. $\;$ $\dfrac{3}{\sqrt[3]{y}} - \sqrt[3]{y} = 2$ $\;\;\; \cdots \; (4)$

Let $\;\;\;$ $\sqrt[3]{y} = p$ $\;\;\; \cdots \; (5)$

Then equation $(4)$ becomes

$\dfrac{3}{p} - p = 2$

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

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

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

Case 1:

When $\;$ $p = -3$, $\;$ we have from equation $(5)$

$\sqrt[3]{y} = -3$ $\implies$ $y = -27$

Substituting $\;$ $y = -27$ $\;$ in equation $(2)$ gives

$-27x = 27$ $\implies$ $x = -1$

Case 2:

When $\;$ $p = 1$, $\;$ we have from equation $(5)$

$\sqrt[3]{y} = 1$ $\implies$ $y = 1$

Substituting $\;$ $y = 1$ $\;$ in equation $(2)$ gives $\;\;\;$ $x = 27$

$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(27, 1\right), \left(-1, -27\right) \right\}$

Algebra - System of Equations and Inequalities

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

Given system of equations:

$\sqrt{\dfrac{x}{y}} - \sqrt{\dfrac{y}{x}} = \dfrac{3}{2}$ $\;\;\; \cdots \; (1)$, $\;\;$ $x + y + xy = 9$ $\;\;\; \cdots \; (2)$

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

$p - \dfrac{1}{p} = \dfrac{3}{2}$

i.e. $\;$ $p^2 - 1 = \dfrac{3}{2} p$

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

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

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
   
   $4y + y + 4y^2 = 9$
   
   i.e. $\;$ $4y^2 + 5y - 9 = 0$
   
   i.e. $\;$ $\left(y - 1\right) \left(4y + 9\right) = 0$
   
   i.e. $\;$ $y = 1$ $\;\;$ or $\;\;$ $y = \dfrac{-9}{4}$
   
   Substituting the value of $y$ in equation $(4a)$ gives
   
   
   1. when $\;$ $y = 1$, $\;$ $x = 4 \times 1 = 4$
   
   
   
   2. when $\;$ $y = \dfrac{-9}{4}$, $\;$ $x = 4 \times \left(\dfrac{-9}{4}\right) = -9$
   
   
   
2. when $\;$ $p = \dfrac{-1}{2}$, $\;\;\;$ then $\;\;$ $\sqrt{\dfrac{x}{y}} = \dfrac{-1}{2}$
   
   But square root of any number cannot be negative.
   
   $\implies$ $p = \dfrac{-1}{2}$ $\;$ is not an acceptable solution.
   
   

$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(-9, \dfrac{-9}{4}\right), \left(4, 1\right) \right\}$