Question

Solve the following pair of equations: \[6x + 5y = 7x + 3y + 1 = 2x + 12y - 2\]
A) x = 3, y = 2
B) x = 1, y = 0
C) x = 5, y = 3
D) x = 7, y = 1

Answer 1

Hint: As three equations are given, we can compare the first and the last two of them to obtain two linear equations. Using these linear equations, we can find the values of variables which will be the required solution of the given pair of equations

Complete step-by-step answer:
The given pair of equations is:
\[6x + 5y = 7x + 3y + 1 = 2x + 12y - 2\]
Here, 3 linear equations are equal to each other.
Comparing two equations at a time where one will be common in both to get two linear equations
Comparing the first two equations:
\[
  6x + 5y = 7x + 3y + 1 \\
  \Rightarrow 7x - 6x + 3y - 5y = - 1 \\
   \Rightarrow x - 2y = - 1....(1) \;
 \]
Comparing the last two equations:
\[
  7x + 3y + 1 = 2x + 12y - 2 \\
  \Rightarrow 7x - 2x + 3y - 12y = - 2 - 1 \\
   \Rightarrow 5x - 9y = - 3....(2) \;
 \]
Using these two equations, we can find the required values of x and y.
For elimination, we can multiply equation (1) with 5 and then subtracting equation (2) from the same:
 \[
  \Rightarrow 5x - 10y - 5x + 9y = - 5 + 3 \\
   \Rightarrow - y = - 2 \\
   \Rightarrow y = 2 \;
 \]
Substituting this value of y in equation (1) to find the value of x:
\[
  \Rightarrow x - 2 \times 2 = - 1{\text{ }}\left( {\because y = 2} \right) \\
  \Rightarrow x - 4 = - 1{\text{ }} \\
  \Rightarrow x = - 1 + 4 \\
   \Rightarrow x = 3 \;
 \]
Therefore, the solution of the given pair of equations is : $ x = 3 $ , $ y = 2 $ and the correct option is A).
So, the correct answer is “Option A”.

Note: As there are two variables present in the given equations with maximum power 1, these are known as linear equations in two variables.
These are the common methods we used here to solve the linear equations known as the elimination and substitution method. For elimination, we can make either of the coefficients the same so that they can be eliminated, the equation will be converted to one variable and the value of the other value of variable can be easily found. We can eliminate and substitute wherever it is easier to do, there is no such rule for the selection.