Question

One equation of a pair of dependent linear equations is $ - 5x + 7y = 2 $ . The second equation can be:
A. $ 10x + 14y + 4 = 0 $
B. $ - 10x - 14y + 4 = 0 $
C. $ - 10x + 14y + 4 = 0 $
D. $ 10x - 14y = - 4 $

Answer 1

Hint: When the same line is represented by two equations, the system of these two equations is called a dependent system and the two equations are dependent linear equations. They have an infinite number of solutions. There are many ways to identify dependent linear equations, when one of the equations is the multiple of the other equation, then the two equations are dependent linear equations. This way we can find out the correct answer.

Complete step-by-step answer:
The given equation can be rewritten as –
 $
   - 5x + 7y - 2 = 0 \\
  or \\
  5x - 7y + 2 = 0 \;
  $
Observing all the equations one by one, we can find out which equation is the multiple of the given equation.
 $
  10x + 14y + 4 = 0 \\
   \Rightarrow 2(5x + 7y + 2) = 0 \;
  $
This equation is the twice of $ 5x + 7y + 2 $ but it is not the multiple of $ 5x - 7y + 2 $ .
So, this is not the correct option.
 $
   - 10x - 14y + 4 = 0 \\
   \Rightarrow - 2(5x + 7y - 2) = 0 \;
  $
This equation is negative of the twice of $ 5x + 7y - 2 $ but it is not the multiple of $ 5x - 7y + 2 $ .
So, this is not the correct option.
 $
   - 10x + 14y + 4 = 0 \\
   \Rightarrow - 2(5x - 7y - 2) = 0 \;
  $
This equation is the negative of the twice of $ 5x - 7y - 2 $ but it is not the multiple of $ 5x - 7y + 2 $ .
So, this is not the correct option.
 $
  10x - 14y = - 4 \\
  10x - 14y + 4 = 0 \\
   \Rightarrow 2(5x - 7y + 2) = 0 \;
  $
This equation is the twice of $ 5x - 7y + 2 $ , thus it is the multiple of $ 5x - 7y + 2 $ .
So, the correct answer is “Option d”.

Note: A linear equation is the combination of constant and variable of the first order. There are three main forms of a linear equation: Slope-intercept form, Point-slope form and Standard form. In the given question, we are given the equation in the standard form that is $ Ax + By = C $ where $ A,\,B\,and\,C $ are constants.