The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
0 votes
4 views

Let two systems of linear equations be defined as follows:

$\begin{array}{lll} &  x+y  & =1 \\  P: & 3x+3y & =3 \\ &  5x+5y & =5 \end{array}$ and $\begin{array}{lll} & x+y & =3 \\ Q: & 2x+2y & =4 \\  & 5x+5y & =12 \end{array}$. Then,

  1. $P$ and $Q$ are inconsistent
  2. $P$ and $Q$ are consistent
  3. $P$ is consistent but $Q$ is inconsistent
  4. None of the above
in Others by Boss (16.3k points)
edited by | 4 views
0

1 Answer

0 votes
P

$x+y = 1$ ....$eq(i)$

$3x+3y = 3$....$eq(ii)$

$5x+5y = 5$...$eq(iii)$

As we can see

$2*eq(i) + eq(ii) = eq(iii)$ this rule is satisfied by both the LHS and RHS parts of the equations.

$\implies$ they can have infinite solution

$\implies$ They are having consistent solutions.

 

Q

$x+y = 3$ ....$eq(i)$

$2x+2y = 4$....$eq(ii)$

$5x+5y = 12$...$eq(iii)$

As we can see

$eq(i) + 2*eq(ii) = eq(iii)$ this rule is satisfied by the LHS part and not by the RHS parts of the equations.(i.e. $3+4*2\neq 12$)

$\implies$ They have no solution

$\implies$ They are having inconsistent solution.

 

$\therefore$ Option $C.$ is the correct answer
by Boss (19.1k points)

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,309 questions
55,743 answers
192,226 comments
90,495 users