Solving the Algebraic Equation \(\frac{p+q}{q+r}=\frac{r+s}{s+p}\)
This problem requires us to find a relationship between the variables \(p\), \(q\), \(r\), and \(s\) given the equation \(\frac{p+q}{q+r}=\frac{r+s}{s+p}\). We are also given that the denominators \((q+r)\) and \((s+p)\) are not equal to zero, ensuring the equation is valid.
Step-by-Step Derivation of Variable Relationships
We start with the given equation:
\(\frac{p+q}{q+r}=\frac{r+s}{s+p}\)
To simplify this equation, we can cross-multiply:
\((p+q)(s+p) = (r+s)(q+r)\)
Next, we expand both sides of the equation:
Left side expansion: \(ps + p \cdot p + qs + qp = ps + p^2 + qs + qp\)
Right side expansion: \(rq + r \cdot r + sq + sr = rq + r^2 + sq + sr\)
Now, we set the expanded expressions equal:
\(ps + p^2 + qs + qp = rq + r^2 + sq + sr\)
To find the relationship, we rearrange the terms by moving all terms to one side:
\(p^2 + ps + qp + qs - rq - r^2 - sq - sr = 0\)
Let's group the terms strategically to simplify the expression:
\((p^2 - r^2) + (ps - sr) + (qp - rq) + (qs - sq) = 0\)
Now, we factor each group:
- \(p^2 - r^2\) factors as \((p-r)(p+r)\).
- \(ps - sr\) factors as \(s(p-r)\).
- \(qp - rq\) factors as \(q(p-r)\).
- \(qs - sq\) simplifies to \(0\).
Substituting these factored forms back into the equation:
\((p-r)(p+r) + s(p-r) + q(p-r) + 0 = 0\)
We can see that \((p-r)\) is a common factor in the first three terms. Factoring out \((p-r)\):
\((p-r) [ (p+r) + s + q ] = 0\)
Simplifying the expression inside the brackets gives:
\((p-r)(p+q+r+s) = 0\)
Conclusions from the Derived Equation
The equation \((p-r)(p+q+r+s) = 0\) holds true if and only if at least one of the factors is equal to zero. This leads to two possible scenarios:
- The first factor is zero: \(p-r = 0\), which means \(p = r\).
- The second factor is zero: \(p+q+r+s = 0\).
Therefore, the relationship derived from the original equation is that either \(p = r\) or \(p+q+r+s = 0\).
Analysis of Provided Options
Let's evaluate the given options based on our findings:
- Option 1: \(p+q+r+s=0\). This is one of the two possible outcomes we derived.
- Option 2: \(p=r\). This is the second possible outcome we derived.
- Option 3: Either \(p+q+r+s=0\) or \(p=r\). This option correctly states both possible conclusions derived from the initial algebraic manipulation.
- Option 4: None of the above. Since Option 3 is mathematically sound based on our derivation, this option is incorrect.
The correct conclusion is that one of the two conditions, \(p=r\) or \(p+q+r+s=0\), must be true.
Similar Questions
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1. \(P^2 + Q^2\) varies as R.
2. PQ varies as R.
3. \(P^2 - Q^2\) varies as R.
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I. \((x^2+y^2)\) varies directly as \(y^2\).
II. \(\left(\frac{x^3}{y^2}\right)\) varies directly as \(y\).
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Important Questions from Ratio and proportion
A sum of Rs. 2,500 is distributed among X, Y and Z in the ratio 1/2 : 3/4 : 5/6 . What is the difference between the maximum share and the minimum share?
In a class, there are three groups A, B and C. If one student from group A and two students from group B are shifted to group C, then what happens to the average weight of the students of the class?
How many different sums can be formed with the denominations Rs. 50, Rs. 100, Rs. 200, Rs. 500 and Rs. 2,000 taking at least three denominations at a time?
In a 500 metres race, B starts 45 metres ahead of A, but A wins the race while B is still 35 metres behind. What is the ratio of the speeds of A to B assuming that both start at the same time?
The monthly incomes of Peter and Paul are in the ratio of 4 : 3. Their expenses are in the ratio of 3 : 2. If each saves Rs. 6,000 at the end of the month, their monthly incomes respectively are (in Rs.)
