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$\dfrac{x}{2} + 3y = 11;x + 5y = 20$

A. $x = 20,y = 2$

B. $x = 15,y = 2$

C. $x = 5,y = 2$

D. $x = 10,y = 2$

Answer

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Hint: To solve the given equations, we should focus on the coefficients of x and y and try to make one of the coefficients equal so that it gets cancelled in the later steps, find one of the values of x or y and then equate this value to one of the equation and find the value of the second variable that is x or y.

Complete step-by-step answer:

Given,

$\dfrac{x}{2} + 3y = 11$

Multiply 2 in the entire equation, so as to make the simplification easier,

$ \Rightarrow x + 6y = 22$ …..(i)

and,

$x + 5y = 20$ …..(ii)

Since, coefficients of x in both equations are equal.

On subtracting (ii) from (i), we get,

$ \Rightarrow x + 6y - \left( {x + 5y} \right) = 22 - 20$

$ \Rightarrow x + 6y - x - 5y = 2$

$ \Rightarrow y = 2$

Substitute the value of $y = 2$ in equation (i),

$x + 6\left( 2 \right) = 22$

$ \Rightarrow x = 10$

Therefore, $x = 10,y = 2$.

Note: The first step is to make x or y equal in both the equations, so we can cancel out one of the terms and calculate the second variable, putting this value in one of the equations we can obtain the value of the second variable as well.

Complete step-by-step answer:

Given,

$\dfrac{x}{2} + 3y = 11$

Multiply 2 in the entire equation, so as to make the simplification easier,

$ \Rightarrow x + 6y = 22$ …..(i)

and,

$x + 5y = 20$ …..(ii)

Since, coefficients of x in both equations are equal.

On subtracting (ii) from (i), we get,

$ \Rightarrow x + 6y - \left( {x + 5y} \right) = 22 - 20$

$ \Rightarrow x + 6y - x - 5y = 2$

$ \Rightarrow y = 2$

Substitute the value of $y = 2$ in equation (i),

$x + 6\left( 2 \right) = 22$

$ \Rightarrow x = 10$

Therefore, $x = 10,y = 2$.

Note: The first step is to make x or y equal in both the equations, so we can cancel out one of the terms and calculate the second variable, putting this value in one of the equations we can obtain the value of the second variable as well.

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