If \(\frac{2}{x}+\frac{3}{y}=13\) and \(\frac{3}{x}-\frac{2}{y}=4\), what is the value of \(\frac{1}{x}\)?
Answer and explanation
Correct answer: \(\frac{38}{13}\)
Let \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\). The equations become \(2u+3v=13\) and \(3u-2v=4\). Multiply the first equation by 2 and the second by 3 to get \(4u+6v=26\) and \(9u-6v=12\). Adding them gives \(13u=38\), so \(u=\frac{38}{13}\). Hence, \(\frac{1}{x}=\frac{38}{13}\). The value \(\frac{31}{13}\) is for \(v=\frac{1}{y}\), not for \(\frac{1}{x}\). Exam tip: For reciprocal terms, substitute \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\) first to form linear equations.
Frequently asked questions
What is the correct answer to this question?
\(\frac{38}{13}\)
Why is this the correct answer?
Let \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\). The equations become \(2u+3v=13\) and \(3u-2v=4\). Multiply the first equation by 2 and the second by 3 to get \(4u+6v=26\) and \(9u-6v=12\). Adding them gives \(13u=38\), so \(u=\frac{38}{13}\). Hence, \(\frac{1}{x}=\frac{38}{13}\). The value \(\frac{31}{13}\) is for \(v=\frac{1}{y}\), not for \(\frac{1}{x}\). Exam tip: For reciprocal terms, substitute \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\) first to form linear equations.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Pair of Linear Equations in Two Variables. Topic: Algebraic methods: Substitution method and Elimination method..
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