Question

One equation of a pair of dependent linear equations is –3x + 7y =4. The second equation can be

Answer 1

Step-by-step explanation:

Here,

1st equation : -3x + 7y = 4

2nd equation : -3x + 7y = 4

or, 7y = 4 + 3x

or , y= ( 4 + 3x) / 7 ......2nd equation

Answer 2

Answer:

Second equation can be $&12 x-28 y+16=0$

Step-by-step explanation:

Given: Linear equation

To find: The second equation

Solution:

Hint:

• An equation is said to be linear when it makes a straight line when it is graphed. This type of straight-line equation can be written in the form of $y=mx+c$.
• In the given question we have to check it by the given options one by one as we know that any dependent equation gives an independent equation when it is divided by 2 or we can say that any independent equation gives a dependent equation when it is multiplied by 2.
• So in the question, we have two procedures to solve it: we can check it by given options or directly multiply it by 2 and get a dependent equation.
• So first we will check it with options and then directly multiply it by 2 to get the correct answer.

Condition for dependent linear equations -

$\frac{a_{1} }{a_{2} }=\frac{b_{1} }{b_{2} }=\frac{c_{1} }{c_{2} } \ldots (i)$

Given the equation of the line is, $-3 x+7 y-4=0$;

Comparing with $a x+b y+c=0$;

Неге, $a_{1}=-3,\\\\ b_{1}=7,\\\\ c_{1}=-4$;

For the second equation, let's assume $a_{2} x+b_{2} y+c_{2}=0 ;$

From Eq. (i), $-\frac{3}{a_{2}} =\frac{7}{b_{2}} =-\frac{4}{c_{2}} =\frac{1}{k}$

Where, $k$ is an arbitrary constant.

Putting $k=-\frac{1}{4}$then

$a_{2}=12, \\b_{2}=-28,\\ c_{2}=16$

$\therefore$ The required equation of line becomes

$\begin{aligned}&a_{2} x+b_{2} y+c_{2}=0 \\&12 x-28 y+16=0\end{aligned}$