Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
489 views
in Linear Equations by (41.5k points)
closed by

Solve the following equations:

1. 9y + 5 = 15y – 1 

2. 3x + 4 = 5(x – 2) 

3. 3(t – 3) = 5(2t – 1) 

4. 5(p – 3) = 3(p – 2) 

5. 5(z + 3) = 4(2z + 1)

1 Answer

+1 vote
by (40.1k points)
selected by
 
Best answer

1. 9y + 5 = 15y – 1 

⇒ 9y – 15y = – 1 – 5 

⇒ -6y = -6

⇒ y = \(\frac{-6}{-6}\)

∴ y = 1

2. 3x + 4 = 5(x – 2) 

⇒ 3x + 4 =5x – 10 

⇒ 3x – 5x= – 10 – 4 

⇒ – 2x = – 14 

:. x = 71

3. 3(t – 3) = 5(2t – 1) 

⇒ 3t – 9 = 10t – 5 

⇒ 3t – 10t = – 5+ 9 

⇒ – 7t = 4 

∴ t = \(\frac{-4}{7}\)

4. 5 (p – 3) = 3 (p – 2) 

⇒ 5p – 15 = 3p – 6 

⇒ 5p – 3p = -6 + 15 

⇒ 2p = 9 

∴ p = \(\frac92\)

5. 5(z + 3) = 4(2z + 1) 

⇒ 5z + 15 = 8z + 4 

⇒ 5z – 8z = 4 – 15 

⇒ – 3z = – 11 

⇒ z = \(\frac{-11}3\)

∴ z = \(\frac{-11}3\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...