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Question

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${\text{A}}{\text{. c = }}\dfrac{{7 - 2x}}{x},x \ne 0$

${\text{B}}{\text{. c = }}\dfrac{{8 - 2x}}{x},x \ne 0$

${\text{C}}{\text{. c = }}\dfrac{{13 - 2x}}{x},x \ne 0$

${\text{D}}{\text{. c = }}\dfrac{{19 - 2x}}{x},x \ne 0$

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Hint: Put $x = y$ in the given equation, $2x + cy = 8$, then solve the equation to find the value of c.

Complete step-by-step answer:

According to the question, we have the equation, $2x + cy = 8 - (1)$

Now, it is asked to find the value of c for which, $x = y$.

So, put $x = y$ in equation (1), we get-

$2x + cx = 8$

Solving further, take x common, we get-

$x(2 + c) = 8$

$ \Rightarrow x = \dfrac{8}{{2 + c}}$

$ \Rightarrow c = \dfrac{{8 - 2x}}{x}$, so the correct option is B.

Now, we know that the denominator should not be zero, so $\

2 + c \ne 0 \\

\Rightarrow c \ne - 2 \\

$

And, c=0,2,6 satisfies the condition, $2x + cy = 8$.

So, we can also say that c should not be equal to 2.

Note – Whenever such types of question appear, then always solve the question step by step, using the information given in the question. As mentioned in the solution, the value of c is found when $x = y$, satisfies.

Complete step-by-step answer:

According to the question, we have the equation, $2x + cy = 8 - (1)$

Now, it is asked to find the value of c for which, $x = y$.

So, put $x = y$ in equation (1), we get-

$2x + cx = 8$

Solving further, take x common, we get-

$x(2 + c) = 8$

$ \Rightarrow x = \dfrac{8}{{2 + c}}$

$ \Rightarrow c = \dfrac{{8 - 2x}}{x}$, so the correct option is B.

Now, we know that the denominator should not be zero, so $\

2 + c \ne 0 \\

\Rightarrow c \ne - 2 \\

$

And, c=0,2,6 satisfies the condition, $2x + cy = 8$.

So, we can also say that c should not be equal to 2.

Note – Whenever such types of question appear, then always solve the question step by step, using the information given in the question. As mentioned in the solution, the value of c is found when $x = y$, satisfies.

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