If x 2= y + z, y 2= z + x and z 2= x + y, then what is the value of \(\frac{1}{{x\; + \;1}} + \frac{1}{{y\; + \;1\;}} + \frac{1}{{z\; + \;1}}\;?\)
1
Solving the System of Symmetric Equations
We are given a system of three symmetric equations:
- \(x^2 = y + z\)
- \(y^2 = z + x\)
- \(z^2 = x + y\)
We need to find the value of the expression \( \frac{1}{{x\; + \;1}} + \frac{1}{{y\; + \;1\;}} + \frac{1}{{z\; + \;1}}\; \).
Analyzing the Equations and Deriving a Key Relationship
Let's look at the first equation: \(x^2 = y + z\). To relate this to the terms in the expression we need to evaluate, let's try adding \(x\) to both sides:
\(x^2 + x = x + y + z\)
Factoring the left side gives:
\(x(x+1) = x + y + z\)
Similarly, from the other two equations, by adding \(y\) and \(z\) respectively to both sides, we get:
\(y^2 + y = y + z + x \implies y(y+1) = x + y + z\)
\(z^2 + z = z + x + y \implies z(z+1) = x + y + z\)
This shows that \(x(x+1) = y(y+1) = z(z+1) = x+y+z\). Let \(S = x+y+z\). Then we have:
- \(x(x+1) = S\)
- \(y(y+1) = S\)
- \(z(z+1) = S\)
Evaluating the Expression \(\frac{1}{{x\; + \;1}} + \frac{1}{{y\; + \;1\;}} + \frac{1}{{z\; + \;1}}\)
Consider the term \( \frac{1}{{x\; + \;1}} \). From the relation \(x(x+1) = S\), if \(S \neq 0\) and \(x \neq -1\), we can divide by \(S\) and \((x+1)\) to get:
\(\frac{x(x+1)}{S(x+1)} = \frac{S}{S(x+1)}\)
\(\frac{x}{S} = \frac{1}{x+1}\)
This gives us a way to rewrite each term in the expression:
- \( \frac{1}{{x\; + \;1}} = \frac{x}{S} \) (provided \(S \neq 0\) and \(x \neq -1\))
- \( \frac{1}{{y\; + \;1}} = \frac{y}{S} \) (provided \(S \neq 0\) and \(y \neq -1\))
- \( \frac{1}{{z\; + \;1}} = \frac{z}{S} \) (provided \(S \neq 0\) and \(z \neq -1\))
Now, let's sum these terms:
\(\frac{1}{{x\; + \;1}} + \frac{1}{{y\; + \;1\;}} + \frac{1}{{z\; + \;1}} = \frac{x}{S} + \frac{y}{S} + \frac{z}{S}\)
\(= \frac{x+y+z}{S}\)
Since \(S = x+y+z\), if \(S \neq 0\), the sum is:
\(= \frac{S}{S} = 1\)
Handling Special Cases: \(S=0\) and Denominators Being Zero
Case 1: \(x+y+z = S = 0\)
If \(x+y+z = 0\), then the relations \(x(x+1)=S\), \(y(y+1)=S\), \(z(z+1)=S\) become:
- \(x(x+1) = 0 \implies x=0\) or \(x=-1\)
- \(y(y+1) = 0 \implies y=0\) or \(y=-1\)
- \(z(z+1) = 0 \implies z=0\) or \(z=-1\)
So, each variable must be either 0 or -1. We also require \(x+y+z = 0\). The possible combinations from \(\{0, -1\}\) that sum to zero are only \((0,0,0)\).
Let's check if \((0,0,0)\) is a valid solution to the original system:
- \(x^2 = y+z \implies 0^2 = 0+0 \implies 0 = 0\) (True)
- \(y^2 = z+x \implies 0^2 = 0+0 \implies 0 = 0\) (True)
- \(z^2 = x+y \implies 0^2 = 0+0 \implies 0 = 0\) (True)
So, \((0,0,0)\) is a valid solution. For this solution, the value of the expression is:
\(\frac{1}{0+1} + \frac{1}{0+1} + \frac{1}{0+1} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 1 + 1 + 1 = 3\)
The derivation \(\frac{1}{x+1} = \frac{x}{S}\) is not valid when \(S=0\).
Case 2: Denominators are Zero (\(x=-1\) or \(y=-1\) or \(z=-1\))
Suppose \(x=-1\). Substitute this into the original equations:
- \(x^2 = y+z \implies (-1)^2 = y+z \implies 1 = y+z\)
- \(y^2 = z+x \implies y^2 = z-1\)
- \(z^2 = x+y \implies z^2 = -1+y\)
From \(1 = y+z\), we get \(z = 1-y\). Substitute this into the second equation:
\(y^2 = (1-y) - 1 \implies y^2 = -y \implies y^2 + y = 0 \implies y(y+1) = 0\)
This gives \(y=0\) or \(y=-1\).
If \(y=0\), then \(z = 1-y = 1-0 = 1\). Let's check the third equation with \((x,y,z) = (-1, 0, 1)\):
\(z^2 = y-1 \implies 1^2 = 0-1 \implies 1 = -1\) (False)
If \(y=-1\), then \(z = 1-y = 1-(-1) = 2\). Let's check the third equation with \((x,y,z) = (-1, -1, 2)\):
\(z^2 = y-1 \implies 2^2 = -1-1 \implies 4 = -2\) (False)
Since assuming \(x=-1\) leads to no valid solutions, the denominators \(x+1, y+1, z+1\) are never zero for any real solution of the system.
Conclusion on Solutions and Expression Value
We found that all real solutions must satisfy \(x(x+1)=y(y+1)=z(z+1)\). This implies that at least two of the variables must be equal (since the quadratic \(t^2+t-S=0\) has at most two distinct roots). Considering cases like \(x=y\), we find the real solutions are \((0,0,0)\) and \((2,2,2)\).
- For the solution \((0,0,0)\), the value of the expression is 3.
- For the solution \((2,2,2)\), \(x+y+z = 2+2+2 = 6 \neq 0\). The derivation \( \frac{1}{{x\; + \;1}} + \frac{1}{{y\; + \;1\;}} + \frac{1}{{z\; + \;1}} = 1 \) is valid. Let's check: \( \frac{1}{2+1} + \frac{1}{2+1} + \frac{1}{2+1} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \).
The question asks for "the value", implying a single value for the expression. Since 1 is one of the options and 3 is not, it is implied that the intended solutions are those for which \(x+y+z \neq 0\), which yield the value 1.
| Solution \((x,y,z)\) | \(x+y+z\) | Check Equations | Value of \( \frac{1}{{x\; + \;1}} + \frac{1}{{y\; + \;1\;}} + \frac{1}{{z\; + \;1}}\; \) |
|---|---|---|---|
| \((0,0,0)\) | 0 | \(0^2=0+0\) (True) | \(1/1+1/1+1/1 = 3\) |
| \((2,2,2)\) | 6 | \(2^2=2+2 \implies 4=4\) (True) | \(1/3+1/3+1/3 = 1\) |
| \((-1,0,1)\) | 0 | \(1^2 = 0-1 \implies 1=-1\) (False) | (Undefined due to invalid solution) |
The value of the expression is 1 for the solution \((2,2,2)\), which is a valid solution to the system of equations.
Revision Table: Key Steps in Solving Symmetric Equations
| Step | Description | Purpose |
|---|---|---|
| 1 | Add \(x, y, z\) to respective equations. | Derive \(x(x+1) = y(y+1) = z(z+1) = x+y+z\). |
| 2 | Let \(S = x+y+z\). | Simplify notation and identify common value. |
| 3 | Rewrite expression terms using the derived relation. | Show \( \frac{1}{x+1} = \frac{x}{S} \) (for \(S \neq 0\)). |
| 4 | Sum the rewritten terms. | Show sum is \(\frac{x+y+z}{S}\) which is 1 (for \(S \neq 0\)). |
| 5 | Analyze the case \(S=0\). | Find solution \((0,0,0)\) and calculate expression value (3). |
| 6 | Analyze cases where denominators are zero. | Show \(x, y, z \neq -1\) for real solutions. |
| 7 | Compare calculated values with options. | Determine the intended value of the expression. |
Additional Information: Symmetric Systems and Solutions
A system of equations is called symmetric if interchanging any two variables results in the same system. The given system \(x^2=y+z, y^2=z+x, z^2=x+y\) is symmetric in \(x,y,z\).
Symmetric systems often have solutions where variables are equal (like \(x=y=z\)). We found two such solutions: \((0,0,0)\) and \((2,2,2)\). We also showed that any real solution must have at least two variables equal.
When solving algebraic problems derived from equations, it is crucial to check for special cases like division by zero or cases where the derived simplification steps might not be valid. In this problem, the case \(x+y+z=0\) was a special case for the derivation \(\frac{1}{x+1} = \frac{x}{x+y+z}\), and it yielded a different value for the expression.
The context of multiple-choice questions often implies that there is a unique answer among the options for all relevant solutions of the system. When multiple solutions yield different values for the expression, the intended solution set usually excludes trivial or degenerate cases, or the question might be implicitly limited to a specific domain (like positive numbers, though not stated here).
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