- Trending Categories
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Solve the following equations and verify your answer:
(i) $\frac{y-(7-8y)}{9y-(3+4y)}=\frac{2}{3}$
(ii) $\frac{6}{2x-(3-4x)}=\frac{2}{3}$
Given:
The given equations are:
(i) $\frac{y-(7-8y)}{9y-(3+4y)}=\frac{2}{3}$
(ii) $\frac{6}{2x-(3-4x)}=\frac{2}{3}$
To do:
We have to solve the given equations and verify the answers.
Solution:
To verify the answer we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.
(i) The given equation is $\frac{y-(7-8y)}{9y-(3+4y)}=\frac{2}{3}$
$\frac{y-7+8y}{9y-3-4y}=\frac{2}{3}$
$\frac{9y-7}{5y-3}=\frac{2}{3}$
On cross multiplication, we get,
$3(9y-7)=(2)(5y-3)$
$3(9y)-3(7)=2(5y)-2(3)$
$27y-21=10y-6$
On rearranging, we get,
$27y-10y=-6+21$
$17y=15$
$y=\frac{15}{17}$
Verification:
LHS $=\frac{y-(7-8y)}{9y-(3+4y)}$
$=\frac{\frac{15}{17}-(7-8(\frac{15}{17}))}{9(\frac{15}{17})-(3+4(\frac{15}{17}))}$
$=\frac{\frac{15}{17}-(7-(\frac{-8\times15}{17}))}{\frac{9\times15}{17}-(3+\frac{4\times15}{17})}$
$=\frac{\frac{15}{17}-(7-(\frac{120}{17}))}{\frac{135}{17}-(3+\frac{60}{17})}$
$=\frac{\frac{15}{17}-(\frac{7\times17-120}{17})}{\frac{135}{17}-(\frac{3\times17+60}{17})}$
$=\frac{\frac{15}{17}-(\frac{119-120}{17})}{\frac{135}{17}-(\frac{51+60}{17})}$
$=\frac{\frac{15}{17}-(\frac{-1}{17})}{\frac{135}{17}-(\frac{111}{17})}$
$=\frac{\frac{15+1}{17}}{\frac{135-111}{17}}$
$=\frac{\frac{16}{17}}{\frac{24}{17}}$
$=\frac{16}{17}\times\frac{17}{24}$
$=\frac{2}{3}$
RHS $=\frac{2}{3}$
LHS $=$ RHS
Hence verified.
(ii) The given equation is $\frac{6}{2x-(3-4x)}=\frac{2}{3}$
$\frac{6}{2x-(3-4x)}=\frac{2}{3}$
$\frac{6}{2x-3+4x}=\frac{2}{3}$
$\frac{6}{6x-3}=\frac{2}{3}$
On cross multiplication, we get,
$3(6)=2(6x-3)$
$18=2(6x)-2(3)$
$18=12x-6$
On rearranging, we get,
$12x=18+6$
$12x=24$
$x=\frac{24}{12}$
$x=2$
Verification:
LHS $=\frac{6}{2x-(3-4x)}$
$=\frac{6}{2(2)-(3-4(2))}$
$=\frac{6}{4-(3-8)}$
$=\frac{6}{4+5}$
$=\frac{6}{9}$
$=\frac{2}{3}$
RHS $=\frac{2}{3}$
LHS $=$ RHS
Hence verified.
229 Views
