Question

Write an equation of a line with slope $3$ and $y-\text{intersect}$ $6?$

Answer 1

Where we have to write an equation of a given line in the slope intercept form. The equation of line can be written as $y=mx+b.$
Whereas $m$ be the slope $b$ is the $y-\text{intercept}.$
Here the coefficient given in the question will be the values written in the form. Also it may be considered as a parameter of the equation. But they do not contain any of the variables.

Complete step by step solution:
Here given, the equation of line with the slope is $3$ and $y-\text{intercept}$ is $6.$
Now, we have to write the given values in the form of
$\Rightarrow y=mx+c$
Where, as $m$ is the slope and $b$ is the $y-\text{intercept}.$
Now plug the given values in the given form.
$m=3$ and $b=6$
$\therefore y=3x+6$
Thus, the required question will be $y=3x+6.$

Additional information:
The another way is to write slope intercept form is the standard form of equation and the standard form of equation is written as $Ax+By+C.$ As you can also change the slope intercept form is in the standard form for better understanding we take the example as, $y=\dfrac{-3x}{2}+3.$ now isolate the $y-\text{intercept}$ and add $\dfrac{3x}{2}$ then the equation we get, $\dfrac{3x}{2}+y=3.$ But as we have standard form fraction part does not consider their so, we have to solve it, the equation we get $2\left( \dfrac{3x}{2}+y \right)=3\left( 2 \right)$
$\therefore 3x+2y=6.$ Then the given equation is considered as in a standard form equation.

Note: The standard form of a linear equation is one as follows: $\text{A}x+\text{B}y=\text{C}.$ There are some restriction which you need to remember that is A and B cannot be zero and A and B both are integers and A is positive number. In the standard form no fraction nor decimal accepts in the equation. For example we take $\dfrac{1}{3}x+\dfrac{1}{4}y=4.$ We can say that the equation is not in the standard from and another examples as $4x+3y=8$ then the given equation is in the standard form.