Question

How do you solve the system of linear equations $4x-y=6$ and $3x+y=1$?

Answer 1

Hint: Now to solve the linear equations given we will use a method of substitution. Consider any one equation and try to write y in terms of x. Now substitute the value of y in another equation and hence we get a linear equation in x. Solve the equation to find the value of x. Substitute the value of x in any equation and hence find the value of y. Hence we have the values of x and y.

Complete step by step solution:
Now consider the given equation $4x-y=6$ and $3x+y=1$ .
Now the given equations are linear equations in two variables x and y.
We know that a linear equation in two variables represents a straight line in the XY plane.
Now we want to find the solution of the linear equation. Hence we want to find the values of x and y such that it satisfies both the equation.
Now let us understand what this means geometrically. Now all the values of x and y which satisfies an equation lies on the line as point (x, y). Now since we want to find a point which satisfies both the equations we can say that geometrically the point will lie on a graph of both lines. Hence we want to find the intersection point of the lines.
Now we will solve the equation simultaneously by substitution method.
Consider the equation $3x+y=1$ . On rearranging the terms of the equation we get$y=1-3x$
Now let us substitute this value of y in the equation $4x-y=6$ . Hence we get,
$\begin{align}
  & \Rightarrow 4x-\left( 1-3x \right)=6 \\
 & \Rightarrow 4x-1+3x=6 \\
 & \Rightarrow 7x=6+1 \\
 & \Rightarrow 7x=7 \\
\end{align}$
Now dividing the whole equation by 7 we get x = 1.
Now let us substitute the value of x obtained in $y=1-3x$ . Hence we get,
$\begin{align}
  & \Rightarrow y=1-3 \\
 & \Rightarrow y=-2 \\
\end{align}$.
Hence the solution of the given equation is x = 1 and y = -2.

Note: Now note that we can also solve the equation by drawing the graph of the equation. First take any two points which satisfy the equation and draw the graph of the line representing the equation. Similarly draw the line for the second equation. The intersection point of the lines is the solution of the equation.