Algebraic Equations

Find the quotient and remainder when $(24a^2+48a+2)$ as divided by $(6a+12)$ by using the method of long division of polynomials.

Divide $6{\mathrm{x}}^4-30{\mathrm{x}}^2+24$ by $2{\mathrm{x}}^2-8$:

Solve the following systems of equations by using the method of substitution:

$\frac{2\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=2$
$\frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{b}}=4$

Solve the following systems of equations by using the method of substitution:

$$\begin{gathered} 2\mathrm{x}+3\mathrm{y}=9 \\ 3\mathrm{x}+4\mathrm{y}=5 \end{gathered}$$

For the following system of equations determine the value of$\mathrm{k}$for which the given system has no solution:

$3 \mathrm{x}-4 \mathrm{y}+7=0, \mathrm{k} \mathrm{x}+3 \mathrm{y}-5=0$

If $\left(x+\frac{1}{x}\right)=2,$ then what is the value of $\left(x^2+\frac{1}{x^2}\right)?$

If $x=(b-c)(a-d), y=(c-a)(b-d), z=(a-b)(c-d),$ $x^3+y^3+z^3$ then what is $x^3+y^3+z^3$ equal to?

If, $\frac{a+b}{\sqrt{ab}}=\frac{2}{1}$ , then the value of $(a-b)$ is:

What is the value of $a(a + b^2 + c) + b^2(a^2 + b^2 + c^2) - c(a + b^2)$ , when $a = 1, b = - 3 and c = -2$? 

If $a+\frac{1}{b}=1$ and $b+\frac{1}{c}=1,$ then $c+\frac{1}{a}=?$

If $a^3=1+7,3^3=1+7+b$ and $4^3=1+7+c$ here a, b, c are positive integers, then the value of $a+b+c$ is
 

If$x=\frac{\sqrt{3}+1}{\sqrt{3}-1}$ and $y=\frac{\sqrt{3}-1}{\sqrt{3}+1}$ then value of $x^2{y}^2$ is

If $x+y=2$ and $x^2+y^2=4,$ then the value of $x^3+y^3$ is-

If $a+\frac{1}{b}=1$ and $b+\frac{1}{c}=1,$ then $c+\frac{1}{a}=?$

If a and b are rational numbers and $a+4\sqrt{3}=2+b(\sqrt{3}+1)$ then $a-b=?$

If $\mathrm{x}+\frac{4}{\mathrm{x}}=4;\mathrm{x}\ne 0$  then, find the value of ${\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^3}$.

If $3\left(a^2+b^2+c^2\right)=(a+b+c{)}^2$, then the relation between $\mathrm{a}, \mathrm{b}, \mathrm{c}$ is:

If $\frac{a}{b}=\frac{5}{7}$, then find the value of $\left(\frac{a}{b}+\frac{b-a}{b+a}\right)$.

If $x=2+\sqrt{3},$ then the value of $\sqrt{x}+\frac{1}{\sqrt{x}}$ is:

If $a+\frac{1}{a}=1$ then the value of $a^3$ is: