What is equal to?
Simplifying Algebraic Expression with x, y, z
The problem asks us to simplify the following complex algebraic expression:
\[ \frac{\frac{x}{x-y}+\frac{y}{y-z}+\frac{z}{z-x}} {\frac{x+y}{x-y}+\frac{y+z}{y-z}+\frac{z+x}{z-x}+3} \]Let's denote the numerator as \( N \) and the denominator as \( D \).
Numerator \( N \):
\[ N = \frac{x}{x-y}+\frac{y}{y-z}+\frac{z}{z-x} \]Denominator \( D \):
\[ D = \frac{x+y}{x-y}+\frac{y+z}{y-z}+\frac{z+x}{z-x}+3 \]Our goal is to simplify the fraction \( \frac{N}{D} \).
Analyzing the Denominator Structure
Let's examine the terms in the denominator \( D \). We can rewrite the denominator by adding 1 to each of the first three terms:
- Consider the first term and add 1: \[ \frac{x+y}{x-y} + 1 = \frac{x+y + (x-y)}{x-y} = \frac{x+y+x-y}{x-y} = \frac{2x}{x-y} \]
- Consider the second term and add 1: \[ \frac{y+z}{y-z} + 1 = \frac{y+z + (y-z)}{y-z} = \frac{y+z+y-z}{y-z} = \frac{2y}{y-z} \]
- Consider the third term and add 1: \[ \frac{z+x}{z-x} + 1 = \frac{z+x + (z-x)}{z-x} = \frac{z+x+z-x}{z-x} = \frac{2z}{z-x} \]
Now, let's rewrite the denominator \( D \) using these results. The original denominator is:
\[ D = \left(\frac{x+y}{x-y}\right) + \left(\frac{y+z}{y-z}\right) + \left(\frac{z+x}{z-x}\right) + 3 \]We can group the terms like this:
\[ D = \left(\frac{x+y}{x-y} + 1\right) + \left(\frac{y+z}{y-z} + 1\right) + \left(\frac{z+x}{z-x} + 1\right) \]Substituting the results from above:
\[ D = \left(\frac{2x}{x-y}\right) + \left(\frac{2y}{y-z}\right) + \left(\frac{2z}{z-x}\right) \]We can factor out a 2:
\[ D = 2 \left( \frac{x}{x-y} + \frac{y}{y-z} + \frac{z}{z-x} \right) \]Final Simplification
We recognize that the expression inside the parentheses is exactly the numerator \( N \).
\[ D = 2 \times N \]Now we can substitute this back into the original fraction \( \frac{N}{D} \):
\[ \frac{N}{D} = \frac{N}{2N} \]Assuming \( N \neq 0 \), we can cancel \( N \) from the numerator and the denominator:
\[ \frac{N}{2N} = \frac{1}{2} \]Therefore, the simplified value of the given algebraic expression is \( \frac{1}{2} \).
Similar Questions
Let \(k\) be a positive integer. What is the quotient when \(x^{8k+3}+x^{8k+6}+x^{8k+9} + x^{8k+12}\) is divided by \((1+x^3)(1+x^6)\) ?
If 2s = a + b + c, then what is s(s-a)(s-b)(s-c) [ (1 / s-a) + (1 / s-b) + (1 / s-c) - (1 / s)] equal to?
- If \((x + 1)(x + p)(x^2 + p^2) = x^4 - 1\), then what is the value of \(p\)?
- If \((2+\sqrt{3})^x + (2-\sqrt{3})^x = 2\), then what is \((2+\sqrt{3})^x-(2-\sqrt{3})^x\) equal to?
What is
\(\frac{(a+b)^2}{(c-a)(c+a+b)} + \frac{(a+b)c}{c^2 + bc-a^2 - ab}\) - \(\frac{(a+2b + c)}{2(c-a)}\), \(a \neq b\), \(b \neq c\), \(c \neq a\)
equal to?
Which of the following is/are the factor(s) of \((3x + y)^2 + (3x+y)(x+5y)-20(x+5y)^2\)?
I. \((4x+13y)\)
II. \((x + 19y)\)
Select the correct answer using the code given below.
- If $p = \sqrt[3]{a + \sqrt{a^2 + b^3}} + \sqrt[3]{a - \sqrt{a^2 + b^3}}$, then what is $p^3 + 3bp$ equal to ?
If \(x^3 + \frac{1}{x^3} = \frac{65}{8}\) and \(y^3 + \frac{1}{y^3} = \frac{730}{27}\), then which one of the following is a value of \(xy\)?
- What is HCF of
\(a^4+2a^3+3a^2+2a+1\) and \(a^6-2a^3+1\)? Let x, y, z be variables such that (x + y + z) = k, where k is a constant. If (x + z - y) × (x - z + y) is proportional to yz, then (y + z - x) is proportional to:
Important Questions from Algebra
- Simplify $x(9x - 8) + 7(x^2 - 3) + 18$.
Simplify the following.
$4\left((\frac{3}{4})x^2 - 26x + 17\right) - 3(x^2 + 4x - 14)$- Simplify the given expression.
$x(3x - 5) + 5(x^2 – 3) + 16$ Simplify: $3\left((\frac{7}{3})x^2 - 25x + 12\right) - 7(x^2 + 9x - 13)$.
Simplify the following.
$3\left((\frac{6}{3})x^2 - 21x + 18\right) - 6(x^2 + 5x - 18)$
