What is the value of (x) from (\frac{x}{3}+\frac{y}{5}=4) and (x-y=6)?
Answer and explanation
Correct answer: \(x=\frac{39}{4}\)
From the second equation, \(x=y+6\). Substituting this into the first equation gives \(\frac{y+6}{3}+\frac{y}{5}=4\). Multiplying by 15, \(5y+30+3y=60\), so \(8y=30\) and \(y=\frac{15}{4}\). Hence, \(x=\frac{15}{4}+6=\frac{39}{4}\). If \(x=\frac{37}{4}\), then \(x-y=6\) gives \(y=\frac{13}{4}\), which does not satisfy the first equation. Exam tip: after substitution in equations containing fractions, multiply by the LCM to clear the denominators.
Frequently asked questions
What is the correct answer to this question?
\(x=\frac{39}{4}\)
Why is this the correct answer?
From the second equation, \(x=y+6\). Substituting this into the first equation gives \(\frac{y+6}{3}+\frac{y}{5}=4\). Multiplying by 15, \(5y+30+3y=60\), so \(8y=30\) and \(y=\frac{15}{4}\). Hence, \(x=\frac{15}{4}+6=\frac{39}{4}\). If \(x=\frac{37}{4}\), then \(x-y=6\) gives \(y=\frac{13}{4}\), which does not satisfy the first equation. Exam tip: after substitution in equations containing fractions, multiply by the LCM to clear the denominators.
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..
