Question

How do you solve the system of linear equations \[y = 2x + 3\],$ - 2x + y = 1$ ?

Answer 1

Hint: To solve this question we have to write a linear equation using the slope-intercept Form. After that you can identify the slope, m. This can be done by calculating the slope between two known points of the line using the slope formula. Then Find the y-intercept. This can be done by substituting the slope and the coordinates of a point (x, y) on the line in the slope-intercept formula and then solve for b. Once you've got both m and b you can just put them in the equation at their respective position.

Complete step-by-step solution:
The given system of linear equation is
$ \Rightarrow y = 2x + 3.........\left( 1 \right)$
$ \Rightarrow - 2x + y = 1.........\left( 2 \right)$
Now we have to solve the equation and find the solution
We have consider two equation in Slope-intercept form
$y = mx + b$
Where $m$ is slope and $b$ is $y$- intercept
From $\left( 1 \right)$ equation
$ \Rightarrow y = 2x + 3$
We get
$m = 2$
From $\left( 2 \right)$ equation
$ \Rightarrow - 2x + y = 1$
We get
$y = 2x + 1$
$\therefore m = 2$
From this two equation, we came to know that the slope of two equations are same, therefore this
System of equations has no point of intersection.
Hence this equation has no solution.

Note: The slope-intercept is the most “popular” form of a straight line. Many students find this is useful because of its simplicity. One can easily describe the characteristics of the straight line even without seeing its graph because of the slope and $y$-intercept which can be easily identified
$1)$ The slope $m$ measures how steep the line is with respect to the horizontal. Given two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ found in the line, the slope is computed as
$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
$2)$ The $y$ intercept -$b$ is the point where the line crosses the $y$ axis. That is the basic characteristic of the $y$ -intercept.
$3)$ In many cases the value of $b$ is not as easily read. In those cases, or if you're uncertain whether the line actually crosses the $y$ axis in this particular point you can calculate $b$ by solving the equation for b $b$ and then substituting $x$ and $y$ with one of your two points.