Question

Solve the following system of linear equation:
${{x}^{2}}+{{y}^{2}}=41,y-x=1$

Answer 1

Hint: This problem involves two variables and hence we require equations to solve and obtain the value of each variable. First, we would try to reduce the number variables to a single variable and then we would obtain the corresponding value for that variable and by using one of the other equations we will evaluate the value of another variable.

Complete step-by-step answer:
In mathematics, the number system is the branch that deals with various types of numbers possible to form and easy to operate with different operators such as addition, multiplication and so on. Another system for variables in mathematics is an algebra system in which we have equations corresponding to some variables through which we can evaluate values of those variables.

Here, we have two equation and two variables provided in our question:

Let us assign each equation with a number so that we can easily utilize them by calling the particular number.

Let ${{x}^{2}}+{{y}^{2}}=41\ldots (1)$ be equation number 1.

Let $y-x=1\ldots (2)$ be equation number 2.

Now, by using equation (2), we replace the value of y in equation (1).
$\begin{align}
  & y=1+x \\
 & {{x}^{2}}+{{(1+x)}^{2}}=41 \\
\end{align}$

By using the expansion of ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ we get,
$\begin{align}
  & {{x}^{2}}+{{x}^{2}}+1+2x=41 \\
 & 2{{x}^{2}}+2x-40=0 \\
 & \therefore {{x}^{2}}+x-20=0 \\
\end{align}$

Now, by using the middle term splitting method of factorization we get,
$\begin{align}
  & {{x}^{2}}-4x+5x-20=0 \\
 & x(x-4)+5(x-4)=0 \\
 & (x+5)(x-4)=0 \\
 & \therefore x=-5,4 \\
\end{align}$

This implies that the possible values of x are -5 and 4.

Now, putting these values of x in equation (2) we obtain the values of y as:
$\begin{align}
  & y=1+x \\
 & y=1+(-5)=-4 \\
 & y=1+4=5 \\
\end{align}$

This implies that values of y are -4 and 5.

Hence, we have two pairs of solutions i.e., (-5, -4) and (4, 5).

Note: The key step for solving this problem is the knowledge of the algebraic system of equations. To solve any particular algebraic equation, we require the same number of equations as there are a number of variables present. In this way we can easily analyze whether we can find a solution to a given problem or it is having no solution.