Question

# Solve the following pair of equations. 5xâˆ’1+1yâˆ’2=2 6xâˆ’1âˆ’3yâˆ’2=1 Here, x â‰  1 and y â‰  2

A

x = 4, y = 5

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B

x = 5, y = 3

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C

x=13, y=13

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D

x = 7, y = 11

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Solution

## The correct option is A x = 4, y = 5 If we substitute 1x−1 as p and 1y−2 as q in the given equations, we get the equations as 5p+q=2 .......(1) 6q−3q=1 .......(2) Now, we can solve the pair of equations by method of elimination. On multiplying the first equation by 3 and then adding it to (2), we get 15p+3q=6 6p−3q=1 _____________ p=13 Now by substituting the value of p in equation (2) we get q=13 Now, p=1x−1 ⇒1x−1=13 ⇒x=4 Similarly we assumed q=1y−2 ⇒1y−2=13 ⇒y−2=3 ⇒y=5 ∴x=4 and y=5

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The text passage contains a mathematical question in the form of a pair of equations to solve, and it provides a detailed solution leading to the correct answer, as well as several answer options.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Question

# Solve the following pair of equations. 5xâˆ’1+1yâˆ’2=2 6xâˆ’1âˆ’3yâˆ’2=1 Here, x â‰  1 and y â‰  2

A

x = 4, y = 5

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

x = 5, y = 3

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C

x=13, y=13

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D

x = 7, y = 11

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Solution

## The correct option is A x = 4, y = 5 If we substitute 1x−1 as p and 1y−2 as q in the given equations, we get the equations as 5p+q=2 .......(1) 6q−3q=1 .......(2) Now, we can solve the pair of equations by method of elimination. On multiplying the first equation by 3 and then adding it to (2), we get 15p+3q=6 6p−3q=1 _____________ p=13 Now by substituting the value of p in equation (2) we get q=13 Now, p=1x−1 ⇒1x−1=13 ⇒x=4 Similarly we assumed q=1y−2 ⇒1y−2=13 ⇒y−2=3 ⇒y=5 ∴x=4 and y=5

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0
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The text passage contains a mathematical question in the form of a pair of equations to solve, and it provides a detailed solution leading to the correct answer, as well as several answer options.

TYPE: SINGLE_QUESTION_WITH_ANSWER