Algebraic Expressions

Given, $\dfrac{1}{2}x=5-\dfrac{1}{4}y$

Given, $\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$
$\Rightarrow \displaystyle \frac{2+y}{y-7}=\frac{4}{5}$

Given x/y = 3/5, let x = 3k and y = 5k. Substitute these into (2x+3y)/(5x+7y): (2(3k) + 3(5k)) / (5(3k) + 7(5k)) = (6k + 15k) / (15k + 35k) = 21k / 50k = 21/50.

Substitute y = 1 into the equation: 2(1) - 3(1) - a = 3. This simplifies to 2 - 3 - a = 3, which is -1 - a = 3. Solving for a gives -a = 4, so a = -4.

Sum the terms: (xy) + (x + y) + (y + xy). Grouping like terms gives (xy + xy) + x + (y + y), which results in 2xy + x + 2y.

Factoring the numerator: y^2 + 7y + 10 = (y + 5)(y + 2). Dividing by (y + 5) leaves (y + 2).

Substitute x = 2 and y = 3 into x^x + y^x: 2^2 + 3^2 = 4 + 9 = 13.

Let the required expression be X. Then 3x + y + X = 2x + 3y, so X = 2x + 3y - 3x - y = -x + 2y. This gives us what must be added to transform the first expression into the second.

$\sqrt{2} =1.414$

$\sqrt{17} =4.123$

Hence,

$x = \sqrt{2}+1 =1.414+1 = 2.414$

$ y= \sqrt{17} -\sqrt{2} =4.123- 1.414 = 2.709$

Hence , $x<y$

$x = \sqrt{7}-\sqrt{5}$

$= 2.64 – 2.23= 0.41$

$y=  \sqrt{13}-\sqrt{11}$

$= 3.60- 3.31 = 0.29$

$Hence, x> y$

(A) Given, $x=\sqrt{2}+1$ and $y=\sqrt{17}-\sqrt{2}$

$\dfrac{x}{y}=\dfrac{\sqrt{2}+1}{\sqrt{17}-\sqrt{2}}\times \dfrac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}$

$=\dfrac{(\sqrt{2}+1)(\sqrt{17}+\sqrt{2})}{17-2}$

$=\dfrac{(\sqrt{2}+1)(\sqrt{17}+\sqrt{2})}{15}< 1$

$\therefore x< y$

A constant term is a value that do not have variable.
So, in this quadratic equation, −3−3 is a  constant term.

Given $x=2, y=3$

Here, the constant in the given equation is $7$ as it contains no variable.