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How do you write $6y = 3x + 4\;or\;y - x = 0$ in slope intercept form?

Last updated date: 05th Mar 2024
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Hint: We will compare these two equations with the general form of slope intercept equation and then we will convert these equations accordingly. Finally we get the required answer.

Formula used: The general equation of slope intercept form of any equation is $y = mx + c$, where $m$ is the slope of the equation and $c$ is an arbitrary constant.
Slope of any equation states the nature of the equation.
Slope of an equation is defined by the ratio of rise and run made by the straight line.
Suppose coordinates joining the line are $({x_1},{y_1})\;and\;({x_2},{y_2})$.
So, the raise made by the line would be $({y_2} - {y_1})$ and the run made by the line would become $({x_2} - {x_1})$.
In such cases slope of the equation will be $= \dfrac{{({y_2} - {y_1})}}{{({x_2} - {x_1})}}$.

Complete Step by Step Solution:
The given equations are $6y = 3x + 4\;and\;y - x = 0.$
The first equation is: $6y = 3x + 4$.
So, divide the both sides of the equation by $6$, we get:
$\Rightarrow y = \dfrac{{3x + 4}}{6}$.
Now, by splitting up the variables on R.H.S, we get:
$\Rightarrow y = \dfrac{{3x}}{6} + \dfrac{4}{6}$.
Now, by completing the division, we get:
$\Rightarrow y = \dfrac{x}{2} + \dfrac{2}{3}$, where $m = \dfrac{1}{2}\;and\;c = \dfrac{2}{3}.$
The second equation is: $y - x = 0.$
Taking the term ‘$x$’ on the R.H.S, we get:
$y = x$.
So, the above equation states that the line goes through the coordinate$(0,0)$ and the slope of the equation is also $1$.

$\therefore$The slope intercepts form of $6y = 3x + 4$ is $y = \dfrac{x}{2} + \dfrac{2}{3}$, where $m = \dfrac{1}{2}\;and\;c = \dfrac{2}{3}$, and the slope intercept form of $y - x = 0$ is $y = x$, where $m = 1\;and\;c = 0.$

Note: The general equation of slope intercept form of any equation is $y = mx + c$, where $m$ is the slope of the equation and $c$ is an arbitrary constant.
For, $y = mx + c$:
If we have a negative slope, the line is decreasing or falling from left to right, and passing through the point $(0,c)$.
On the other hand, if we have a positive slope, the line is increasing or rising from left to right, and passing through the point $(0,c)$.
For, $y = mx - c$
If we have a negative slope, the line is decreasing or falling from left to right, and passing through the point $(0, - c)$.
On the other hand, if we have a positive slope, the line is increasing or rising from left to right, and passing through the point $(0, - c)$.