Question

How do you solve the following system $ x + y = 1,2x - 3y = 12 $ ?

Answer 1

Hint: In order to determine the solution of a given system of equations having two variables, use the method of elimination of term by eliminating the $ y $ term by making the mod of coefficient of $ y $ in both the equations equal. Then apply the operation of addition or subtraction between the equation to eliminate $ y $ term. Solve the result for $ x $ and put the obtained value of $ x $ in any of the equations given to get the value of $ y $ .

Complete step by step solution:
We are given pair of linear equation in two variables $ x + y = 1,2x - 3y = 12 $
 $ x + y = 1 $ ---(1)
 $ 2x - 3y = 12 $ ----(2)
In order to solve the system of equations, we have many methods like, substitution, elimination of term, and cross-multiplication.
Here we will be using an elimination method to eliminate the term having $ y $ from both the equations.
And to do so we have to first make the mod of coefficient of $ y $ in both the equation equal to each.
In the second equation the mod of coefficient of $ y $ is 3, so multiplying both side of the equation (1) with the number 3, we get
 $ 3\left( {x + y} \right) = 3\left( 1 \right) $
 $ 3x + 3y = 3 $
Now Adding the above equation with the equation (2) , we get
 $ 2x - 3y + 3x + 3y = 3 + 12 $
Combining like terms we get
 $ 5x = 15 $
Solving the equation for variable $ x $ by dividing both sides of the equation with the coefficient of $ x $ i.e. $ 5 $
\[
  \dfrac{{5x}}{5} = \dfrac{{15}}{5} \\
  x = 3 \;
 \]
Hence, we have obtained the value of \[x = 3\].
Now putting this value of $ x $ in the equation (1) to get the value of \[y\]
 $
  3 + y = 1 \\
  y = 1 - 3 \\
  y = - 2 \;
  $
Therefore, the solution of system of given equations is $ x = 3,y = - 2 $
So, the correct answer is “ $ x = 3,y = - 2 $ ”.

Note: Linear Equation in two variable: A linear equation is a equation which can be represented in the form of $ ax + by + c $ where $ x $ and $ y $ are the unknown variables and c is the number known where $ a \ne 0,b \ne 0 $ .