If \(\frac{x}{b+c}=\frac{y}{c+a}=\frac{z}{b-a}\) , then which one of the following is correct?
x - y - z = 0
Understanding the Algebraic Problem
The question gives us a relationship between three variables \(x\), \(y\), and \(z\) involving some constants \(a\), \(b\), and \(c\). The relationship is expressed as a set of equal ratios:
\(\frac{x}{b+c}=\frac{y}{c+a}=\frac{z}{b-a}\)
We need to determine which of the given options, relating \(x\), \(y\), and \(z\), is correct based on this information.
Solving the Equal Ratios Problem
A common method to solve problems involving equal ratios is to set each ratio equal to a constant, often denoted by \(k\). This method helps us express the numerators (\(x\), \(y\), \(z\)) in terms of this constant and the denominators (\(b+c\), \(c+a\), \(b-a\)).
Let's set the given ratios equal to \(k\):
\(\frac{x}{b+c}=\frac{y}{c+a}=\frac{z}{b-a} = k\)
From this, we can write expressions for \(x\), \(y\), and \(z\) in terms of \(k\), \(a\), \(b\), and \(c\):
- \(x = k \times (b+c) = k(b+c)\)
- \(y = k \times (c+a) = k(c+a)\)
- \(z = k \times (b-a) = k(b-a)\)
Now, let's test the expressions given in the options. We will substitute the expressions for \(x\), \(y\), and \(z\) into the expression \(x - y - z\) from one of the options and simplify it.
Consider the expression \(x - y - z\):
\(x - y - z = k(b+c) - k(c+a) - k(b-a)\)
We can factor out the common constant \(k\):
\(x - y - z = k[(b+c) - (c+a) - (b-a)]\)
Now, remove the parentheses inside the square brackets, being careful with the signs:
\(x - y - z = k[b+c - c-a - b+a]\)
Next, group the terms with \(a\), \(b\), and \(c\) together:
\(x - y - z = k[(b-b) + (c-c) + (-a+a)]\)
Simplify the terms within the parentheses:
- \(b-b = 0\)
- \(c-c = 0\)
- \(-a+a = 0\)
Substitute these back into the expression:
\(x - y - z = k[0 + 0 + 0]\)
\(x - y - z = k[0]\)
\(x - y - z = 0\)
Since substituting the expressions for \(x\), \(y\), and \(z\) into \(x - y - z\) resulted in 0, the relationship \(x - y - z = 0\) is correct based on the given equal ratios.
Let's quickly check why other options would not necessarily be correct:
- \(x+y+z = k(b+c) + k(c+a) + k(b-a) = k(b+c+c+a+b-a) = k(2b+2c)\). This is not necessarily 0 unless \(2b+2c=0\).
- \(x+y-z = k(b+c) + k(c+a) - k(b-a) = k(b+c+c+a-b+a) = k(2c+2a)\). This is not necessarily 0 unless \(2c+2a=0\).
- \(x+2y+3z = k(b+c) + 2k(c+a) + 3k(b-a) = k(b+c+2c+2a+3b-3a) = k(4b+3c-a)\). This is not necessarily 0.
Revision Table: Key Concepts
| Concept | Description |
|---|---|
| Equal Ratios | When two or more ratios are equal, they can be set equal to a common constant \(k\). |
| k-Method | A technique used to solve problems involving equal ratios by introducing a constant \(k\) to simplify expressions. |
| Algebraic Substitution | Replacing variables in an expression with their equivalent expressions to simplify or evaluate. |
Additional Information: Properties of Proportions
The problem uses the concept of proportions, which are statements of equality between two ratios. When multiple ratios are equal, we have an extended proportion. The k-method is derived from the fundamental property that if \(\frac{a}{b} = \frac{c}{d}\), then \(a=bk\) and \(c=dk\) for some constant \(k\).
Another useful property of equal ratios is that if \(\frac{a}{b} = \frac{c}{d} = \frac{e}{f}\), then each ratio is also equal to \(\frac{a+c+e}{b+d+f}\) (provided \(b+d+f \ne 0\)) and also equal to \(\frac{a-c-e}{b-d-f}\) (provided \(b-d-f \ne 0\)), and many other linear combinations of numerators and denominators.
In this specific problem, we are checking if a linear combination of numerators \(x, y, z\) (specifically \(x-y-z\)) divided by the corresponding linear combination of denominators \((b+c), (c+a), (b-a)\) equals the common ratio \(k\). Let's check for \(x-y-z\):
\(\frac{x-y-z}{(b+c)-(c+a)-(b-a)} = \frac{x-y-z}{b+c-c-a-b+a} = \frac{x-y-z}{0}\)
For this to be equal to \(k\) (unless \(k\) is undefined, which isn't the case here as \(x,y,z\) aren't necessarily 0), the numerator \(x-y-z\) must be 0. This confirms our earlier result obtained using the direct substitution method.
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