Question

Solve the following simultaneous equation by the method of elimination:
$\begin{align}
  & 2a-b=0 \\
 & 5a+b=14 \\
\end{align}$

Answer 1

Hint: Given, to solve the question in elimination method. The system of equations is solved in elimination method by altering one of the equations to match with one of the variables of the second equation and substitute that variable to completely eliminate one variable and solve for the variable remaining. However, you know the relation between them by which you eliminate the first one. By that relation find the remaining variable’s value. This is called an elimination method because you are finding the value of the first variable by eliminating the second one.
Complete step-by-step answer:
Definition of system of equations: If simultaneously we have more than one equation, then the set of those equations is called the system of equations. We can project systems of equations as lines, planes, etc. depending on the number of variables.
If we have 2 variables: Then the system of equations is analogous to straight lines.
If we have 3 variables: Then the system of equations is analogous to the planes.
First, we need to alter one or both equations to match the variable of the other equation.
Given equations in the question all given by as follows:
$\begin{align}
  & 2a-b=0..........(i) \\
 & 5a+b=14........(ii) \\
\end{align}$
By observation we can say that coefficients of b all matching without any alteration by adding both the equation, we get
$\begin{align}
  & \left( 2a-b \right)+\left( 5a+b \right)=0+14 \\
 & \\
\end{align}$
By simplifying the above equation, we get the following:
$\begin{align}
  & \left( 2a+5a \right)+\left( b-b \right)=14 \\
 & \\
\end{align}$
By cancelling the common terms in above equation, we get:
$7a=14$
By dividing with 7 on both sides of equation, we get:
$a=2.......(iii)$
By substituting equation(iii) in equation (i), we get:
$2\left( 2 \right)-b=0$
By adding the term ‘b’ on both sides of equation, we get:
$b=4..........(iv)$
By equation (iii) and equation (iv), we get values of a, b as:
a = 2
b = 4
By above we can say:
$\left( a,b \right)=\left( 2,4 \right)$
Therefore, by elimination method we obtained values of $\left( a,b \right)$ as $\left( 2,4 \right)$.
Note: The observation of b’s coefficient being same made this case easier if not that direct you must find least common multiple and turn each coefficient into the least common multiple and then follow the same procedure.