Question

How do you solve the equations $x+3y=-1$ and $y=x+1$ .

Answer 1

Hint: Now we are given with two linear equations. To solve the linear equation we will first arrange both the equations such that all the variables are on one side. Now if necessary we will multiply the equations with scalar such that one variable is eliminated after addition or subtraction of the equation. Hence we will get a linear equation in one variable and we will solve the equation to find the solution. Now we will substitute the value obtained in any equation to find the value of another variable. Hence we will have the solution to the equation.

Complete step-by-step solution:
Now consider the given equations $x+3y=-1$ and $y=x+1$ .
These are linear equations in two variables. Now to find the solution of the two equations means we want to find the values of x and y such that they satisfy both the equations.
Similarly if we try to understand the concept geometrically we know that linear equations represent straight lines on graphs. Hence the solution to two linear equations is nothing but the intersection point of the two lines.
Now let $x+3y=1.............\left( 1 \right)$
Now consider $y=x+1$
Taking x on LHS of the equation we get,
$\Rightarrow -x+y=1............\left( 2 \right)$
Now let us add equation (1) and equation (2), Hence we get,
$\begin{align}
  & x+3y-x+y=1+1 \\
 & \Rightarrow 4y=2 \\
 & \Rightarrow y=\dfrac{1}{2} \\
\end{align}$
Now substituting $y=\dfrac{1}{2}$ in equation (2) we get,
$\begin{align}
  & -x+\dfrac{1}{2}=1 \\
 & \Rightarrow x=\dfrac{-1}{2} \\
\end{align}$
Hence the solution of the equation is $x=\dfrac{-1}{2}$ and $y=\dfrac{1}{2}$.

Note: Note that if linear equations need not always have a solution. For example if the lines represent parallel lines then the point of intersection does not exist and hence there is no solution to the given equation. In this case both the equations are linearly dependent. Which means both can be expressed as scalar × other equations.