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If a 2 − by − cz = 0, ax − b 2  + cz = 0 and ax+ by − c 2  = 0, then what is the value of  \(\rm\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}\)  will be

This question was previously asked in
CDS I 2016 English Previous Year Paper (14-Feb-2016)
The correct answer is

1

Analyzing the Given System of Equations

We are given a system of three linear equations involving variables \(x\), \(y\), and \(z\), and constants \(a\), \(b\), and \(c\). The equations are:

  • \(\rm a^2 - by - cz = 0\)
  • \(\rm ax - b^2 + cz = 0\)
  • \(\rm ax + by - c^2 = 0\)

We can rewrite these equations to isolate the constant terms on one side:

  • Equation 1: \(\rm by + cz = a^2\)
  • Equation 2: \(\rm ax + cz = b^2\)
  • Equation 3: \(\rm ax + by = c^2\)

Our goal is to find the value of the expression \(\rm\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}\).

Solving the System of Equations for x, y, and z

To find the value of the expression, we first need to determine the values of \(x\), \(y\), and \(z\) in terms of \(a\), \(b\), and \(c\) by solving the system of equations. We can use methods like elimination or substitution.

Solving for x Variable

Let's add Equation 2 and Equation 3:

\(\rm (ax + cz) + (ax + by) = b^2 + c^2\)

\(\rm 2ax + by + cz = b^2 + c^2\)

Notice that the term \(\rm by + cz\) appears in this combined equation. From Equation 1, we know that \(\rm by + cz = a^2\). Substitute this into the equation:

\(\rm 2ax + a^2 = b^2 + c^2\)

Now, solve for \(x\):

\(\rm 2ax = b^2 + c^2 - a^2\)

\(\rm x = \frac{b^2 + c^2 - a^2}{2a}\)

Solving for y Variable

Similarly, let's add Equation 1 and Equation 3:

\(\rm (by + cz) + (ax + by) = a^2 + c^2\)

\(\rm ax + 2by + cz = a^2 + c^2\)

From Equation 2, we know that \(\rm ax + cz = b^2\). Substitute this into the equation:

\(\rm (ax + cz) + 2by = a^2 + c^2\)

\(\rm b^2 + 2by = a^2 + c^2\)

Now, solve for \(y\):

\(\rm 2by = a^2 + c^2 - b^2\)

\(\rm y = \frac{a^2 + c^2 - b^2}{2b}\)

Solving for z Variable

Finally, let's add Equation 1 and Equation 2:

\(\rm (by + cz) + (ax + cz) = a^2 + b^2\)

\(\rm ax + by + 2cz = a^2 + b^2\)

From Equation 3, we know that \(\rm ax + by = c^2\). Substitute this into the equation:

\(\rm (ax + by) + 2cz = a^2 + b^2\)

\(\rm c^2 + 2cz = a^2 + b^2\)

Now, solve for \(z\):

\(\rm 2cz = a^2 + b^2 - c^2\)

\(\rm z = \frac{a^2 + b^2 - c^2}{2c}\)

So we have found the values of \(x\), \(y\), and \(z\):

  • \(\rm x = \frac{b^2 + c^2 - a^2}{2a}\)
  • \(\rm y = \frac{a^2 + c^2 - b^2}{2b}\)
  • \(\rm z = \frac{a^2 + b^2 - c^2}{2c}\)

Calculating the Expression Terms Value

Now we need to calculate each term of the expression \(\rm\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}\) using the values we found for \(x\), \(y\), and \(z\).

Calculating Term 1: x/(a+x)

First, let's find \(a+x\):

\(\rm a+x = a + \frac{b^2 + c^2 - a^2}{2a}\)

To add these, we find a common denominator:

\(\rm a+x = \frac{2a \cdot a}{2a} + \frac{b^2 + c^2 - a^2}{2a} = \frac{2a^2 + b^2 + c^2 - a^2}{2a} = \frac{a^2 + b^2 + c^2}{2a}\)

Now, calculate \(\rm\frac{x}{a+x}\):

\(\rm\frac{x}{a+x} = \frac{\frac{b^2 + c^2 - a^2}{2a}}{\frac{a^2 + b^2 + c^2}{2a}}\)

We can cancel out the \(\rm 2a\) in the numerator and denominator:

\(\rm\frac{x}{a+x} = \frac{b^2 + c^2 - a^2}{a^2 + b^2 + c^2}\)

Calculating Term 2: y/(b+y)

Next, let's find \(b+y\):

\(\rm b+y = b + \frac{a^2 + c^2 - b^2}{2b}\)

Finding a common denominator:

\(\rm b+y = \frac{2b \cdot b}{2b} + \frac{a^2 + c^2 - b^2}{2b} = \frac{2b^2 + a^2 + c^2 - b^2}{2b} = \frac{a^2 + b^2 + c^2}{2b}\)

Now, calculate \(\rm\frac{y}{b+y}\):

\(\rm\frac{y}{b+y} = \frac{\frac{a^2 + c^2 - b^2}{2b}}{\frac{a^2 + b^2 + c^2}{2b}}\)

Cancel out the \(\rm 2b\):

\(\rm\frac{y}{b+y} = \frac{a^2 + c^2 - b^2}{a^2 + b^2 + c^2}\)

Calculating Term 3: z/(c+z)

Finally, let's find \(c+z\):

\(\rm c+z = c + \frac{a^2 + b^2 - c^2}{2c}\)

Finding a common denominator:

\(\rm c+z = \frac{2c \cdot c}{2c} + \frac{a^2 + b^2 - c^2}{2c} = \frac{2c^2 + a^2 + b^2 - c^2}{2c} = \frac{a^2 + b^2 + c^2}{2c}\)

Now, calculate \(\rm\frac{z}{c+z}\):

\(\rm\frac{z}{c+z} = \frac{\frac{a^2 + b^2 - c^2}{2c}}{\frac{a^2 + b^2 + c^2}{2c}}\)

Cancel out the \(\rm 2c\):

\(\rm\frac{z}{c+z} = \frac{a^2 + b^2 - c^2}{a^2 + b^2 + c^2}\)

Summing the Expression Terms Value

Now that we have the value for each term, we can add them together:

\(\rm\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z} = \frac{b^2 + c^2 - a^2}{a^2 + b^2 + c^2} + \frac{a^2 + c^2 - b^2}{a^2 + b^2 + c^2} + \frac{a^2 + b^2 - c^2}{a^2 + b^2 + c^2}\)

Since all terms have the same denominator \(\rm a^2 + b^2 + c^2\), we can add the numerators directly:

\(\rm = \frac{(b^2 + c^2 - a^2) + (a^2 + c^2 - b^2) + (a^2 + b^2 - c^2)}{a^2 + b^2 + c^2}\)

Combine the terms in the numerator:

\(\rm = \frac{b^2 + c^2 - a^2 + a^2 + c^2 - b^2 + a^2 + b^2 - c^2}{a^2 + b^2 + c^2}\)

Group like terms in the numerator:

\(\rm = \frac{(-a^2 + a^2 + a^2) + (-b^2 + b^2 + b^2) + (c^2 + c^2 - c^2)}{a^2 + b^2 + c^2}\)

\(\rm = \frac{a^2 + b^2 + c^2}{a^2 + b^2 + c^2}\)

Assuming \(\rm a^2 + b^2 + c^2 \neq 0\), the fraction simplifies to:

\(\rm = 1\)

Conclusion on Expression Value

Based on the calculations, the value of the expression \(\rm\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}\) is 1.

Revision Table: Key Steps Reviewed

Here is a summary of the key steps taken to solve the problem:

Step Description Result
1 Identify the given system of linear equations. \(\rm by + cz = a^2\)
\(\rm ax + cz = b^2\)
\(\rm ax + by = c^2\)
2 Solve the system to find expressions for \(x\), \(y\), and \(z\). \(\rm x = \frac{b^2 + c^2 - a^2}{2a}\)
\(\rm y = \frac{a^2 + c^2 - b^2}{2b}\)
\(\rm z = \frac{a^2 + b^2 - c^2}{2c}\)
3 Calculate each term of the expression \(\rm\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}\). \(\rm\frac{x}{a+x} = \frac{b^2 + c^2 - a^2}{a^2 + b^2 + c^2}\)
\(\rm\frac{y}{b+y} = \frac{a^2 + c^2 - b^2}{a^2 + b^2 + c^2}\)
\(\rm\frac{z}{c+z} = \frac{a^2 + b^2 - c^2}{a^2 + b^2 + c^2}\)
4 Sum the calculated terms. \(\rm\frac{a^2 + b^2 + c^2}{a^2 + b^2 + c^2}\)
5 Simplify the sum. 1

Additional Information: Systems of Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. The goal is often to find the values of the variables that satisfy all equations simultaneously.

Common methods for solving systems of linear equations include:

  • Substitution Method: Solve one equation for one variable and substitute that expression into the other equations.
  • Elimination Method: Multiply equations by constants so that when they are added or subtracted, one or more variables are eliminated. This problem primarily used an elimination approach by adding equations.
  • Matrix Methods: For larger systems, matrix methods like Gaussian elimination or Cramer's rule can be used.

In this specific problem, the symmetric structure of the resulting values for \(x\), \(y\), and \(z\) in terms of \(a\), \(b\), and \(c\) led to a significant simplification when calculating the sum of the terms in the expression.

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Important Questions from Quadratic Equation

  1. For what values of k, the roots of 9x 2 + 8kx + 16 = 0 are real and equal?

  2. If \(\rm \left( \frac{x}{x+1} \right)^2 -5 \left( \frac{x}{x+1} \right) +6=0 \) , then the value of  \(\rm \left( 1+\frac{1}{x} \right) \)  is equal to :
  3. The nature of the roots of the equation 4x 2 - 2x - 3 = 0.

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