Question

The solution of the system of equations \[\dfrac{{2x + 5y}}{{xy}} = 6\] and \[\dfrac{{4x - 5y}}{{xy}} + 3 = 0\] (where \[x \ne 0,y \ne 0\]), respectively is__________.
A. 1, 2
B. 0, 0
C. -1, 2
D. 1, -2

Answer 1

Hint: The algebraic expression should be any one of the forms such as addition, subtraction, multiplication and division. The given equation is a linear equation as there are two constant variables x and y involved and to solve this equation, Isolate the variable term using the addition property of equality, combine all the like terms and then simplify the terms to get the value of x and y.

Complete step by step solution:
The given equation is
\[\dfrac{{2x + 5y}}{{xy}} = 6\] …………….. (1)
\[\Rightarrow\dfrac{{4x - 5y}}{{xy}} + 3 = 0\]………………. (2)
Equation 1 can be written as
\[\dfrac{{2x + 5y}}{{xy}} = 6\]
\[\Rightarrow 2x + 5y = 6xy\] …………… (3)
Equation 2 can be written as
\[\dfrac{{4x - 5y}}{{xy}} + 3 = 0\]
\[\Rightarrow\dfrac{{4x - 5y}}{{xy}} = - 3\]
\[\Rightarrow 4x - 5y = - 3xy\] …………. (4)
Add equations 3 and 4 as
\[2x + 5y = 6xy\]
\[\Rightarrow 4x - 5y = - 3xy\]
\[ \Rightarrow \]\[6x = 3xy\] ……………. (5)
Hence, from equation 5 the value of y is
\[y = 2\]
Now, substitute the value of y in equation 3 to get the value of x as
\[2x + 5y = 6xy\]
\[\Rightarrow 2x + 5\left( 2 \right) = 6x\left( 2 \right)\]
Solve for x i.e.,
\[2x + 10 = 12x\]
\[\Rightarrow 2x - 12x = - 10\]
\[ \Rightarrow - 10x = - 10\]
\[ \Rightarrow \]\[x = \dfrac{{10}}{{10}} = 1\]
Therefore, the values of $x$ and $y$ are 1 and 2.

Hence, option A is the right answer.

Note: We solve algebraic equations by isolating the variable with a coefficient of 1. Multiplying or dividing both sides of an equation by 0 is carefully avoided. Dividing by 0 is undefined and multiplying both sides by 0 results in the equation 0 = 0. If given a linear equation of the form \[ax + b = c\], then we can solve it in two steps. First, use the appropriate equality property of addition or subtraction to isolate the variable term. Next, isolate the variable using the equality property of multiplication or division.