Question

If k is a rational constant not equal to 11 , which of the following graphs could represent the equation y+5=k(x+y)+5y+5=k(x+y)+5 y+5=k(x+y)+5 y, plus, 5, equals, k, (, x, plus, y, ), plus, 5 ?

Answer 1

Answer:

Slope: y is a straight line through the origin with a slope of $(\frac{k}{1-k})$

All the options (A) (B) and (D) are valid and represent the equation$y+5 = k (x+y) + 5$ unless otherwise any other condition is provided.

Correction in Problem Statement:

If k is a rational constant not equal to 1 , which of the following graphs could represent the equation  y+5 = k (x+y) + 5.

I have uploaded a picture of answer choices.

Step-by-step explanation:

To solve this problem , we do not need visual representation of the graphs.

Simply, as we are given the equation in form y = mx +c :

y+5 = k (x+y) + 5

y+5-5 = k(x+y)

y = k(x+y)

y = kx+ky

y-ky=kx

y(1-k)= kx

$y= \frac{kx}{1-k}$

$y= (\frac{k}{1-k})x$

Since , we are given that k is not equal to 1, it leads to many possibilities depending on the value of k but not vertical, which means that:

y is a straight line through the origin with a slope of $(\frac{k}{1-k})$

Option (C) is not possible as the straight line does not pass through origin.

Other than that, All options (A), (B) and (D) are valid based on the current information provided with the question.

Option (A) has slope of 3, Option (B) has a slope of $\frac{-1}{3}$, Option (D) has a slope of -3

So various values of k for these options will be as  follows;

(A) $(\frac{k}{1-k})=3$

k=3-3k

$k=\frac{3}{4}$

(B)$(\frac{k}{1-k})=\frac{-1}{3}$

k= -0.5

(D)$(\frac{k}{1-k})=-3$

$k=\frac{3}{2}$

All the options (A) (B) and (D) are valid and represent the equation$y+5 = k (x+y) + 5$ unless otherwise any other condition is provided.