Question

Solve the following systems of equations-
$
 x + 2y = \dfrac{3}{2} ,
 2x + y = \dfrac{3}{2}
$

Answer 1

Hint- We will use the method of elimination by equating the coefficients in order to solve this question. After doing so, we will also use subtraction in the solution below to find out the value of x and y.

Complete step-by-step answer:
Mark the equations given in the question as equation 1 and equation 2 respectively, we will get:
$ \Rightarrow x + 2y = \dfrac{3}{2}$ (equation 1)
$ \Rightarrow 2x + y = \dfrac{3}{2}$ (equation 2)
Now, in order to use the method of elimination by making the coefficients equal, we will multiply the equation 1 by 2 so that the coefficients of x in both the equations will be equal, we will get-
$
   \Rightarrow \left( {x + 2y = \dfrac{3}{2}} \right) \times 2 \\
    \\
   \Rightarrow 2x + 4y = 3 \\
$
Mark the equation as equation 3, we get-
$ 2x + 4y = 3$ (equation 3)
Now, subtracting equation 3 from equation 2 in order to get the value of y-
$
   \Rightarrow 2x + 4y - 2x - y = 3 - \dfrac{3}{2} \\
    \\
   \Rightarrow 3y = \dfrac{3}{2} \\
    \\
   \Rightarrow y = \dfrac{1}{2} \\
$
Substituting this value of y in equation 2 or equation 1 in order to find the value of x-
Consider equation 1-
$
   \Rightarrow x + 2y = \dfrac{3}{2} \\
    \\
   \Rightarrow x + 2\dfrac{1}{2} = \dfrac{3}{2} \\
    \\
   \Rightarrow x + 1 = \dfrac{3}{2} \\
    \\
   \Rightarrow x = \dfrac{3}{2} - 1 \\
    \\
   \Rightarrow x = \dfrac{1}{2} \\
$
Thus, the value of $x = \dfrac{1}{2},y = \dfrac{1}{2}$.

Note: Such questions are very easy to solve once we use the method of elimination by equating the coefficients and the method of substitution. Remember to equation the coefficients if we use the method of elimination.