Reduce the following equations into slope -intercept form and find their slopes and the $ y $ -intercepts.
A. $ x + 7y = 0 $ ,
B. $ 6x + 3y - 5 = 0 $ ,
C. $ y = 0 $ .
Answer
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Hint: First to find the slope-intercept form the formula for the slope-intercept form is the value of $ y $ - axis is equal to product of slope and $ x $ -axis and intercept. So need to write all the above equations in the slope intercept form.
Complete step-by-step answer:
Given:
For the first problem we have the equation is $ x + 7y = 0 $ .
For the second problem we have the equation is $ 6x + 3y - 5 = 0 $ .
For the third problem we have the equation is $ y = 0 $ .
The general formula for the slope – intercept form is $ y = mx + c $ .
Where $ m $ is the slope and $ c $ is the intercept.
i)
The given equation is $ x + 7y = 0 $ .
Now, we will add $ - 7y $ both sides to the above equation we obtain,
$ x + 7y - 7y = 0 - 7y $
Since $ 7y $ and $ - 7y $ get cancelled we obtain,
$ x = - 7y $
Multiply $ - 1 $ both sides to the above equation,
$ 7y = - x $
Then, multiply both sides by $ \dfrac{1}{7} $ we obtain,
$ y = - \dfrac{x}{7} + 0 $
The slope for the above equation is $ m = \dfrac{{ - 1}}{7} $ .
The intercept for the equation is $ 0 $ .
The slope -intercept form for the equation $ x + 7y = 0 $ is $ y = - \dfrac{x}{7} + 0 $ .
ii)
The given equation is $ 6x + 3y - 5 = 0 $ .
Adding $ - 3y $ both sides to the above equation we obtain,
$ 6x + 3y - 3y - 5 = - 3y $
Since $ 3y $ nd $ - 3y $ get cancelled we obtain,
$ 6x - 5 = - 3y $
Multiply $ - 1 $ both sides to the above equation,
$ 3y = - 6x + 5 $
Then, multiply both sides by $ \dfrac{1}{3} $ we obtain,
$ y = - \dfrac{{6x}}{3} + \dfrac{5}{3} $
The slope for the above equation is $ m = - \dfrac{6}{3} $ .
The intercept for the equation is $ \dfrac{5}{3} $ .
The slope - intercept form for the equation $ 6x + 3y - 5 = 0 $ is $ y = - \dfrac{{6x}}{3} + \dfrac{5}{3} $ .
iii)
The slope -intercept form is $ y = 0 $ .
Where, $ m = 0 $ and intercept is $ 0 $ .
Note: If $ x = 0 $ then in the slope-intercept form the slope will be $ 1 $ and intercept is $ 0 $ . If the slope is not given, then determine the slope and intercept and substitute in the general formula.
Complete step-by-step answer:
Given:
For the first problem we have the equation is $ x + 7y = 0 $ .
For the second problem we have the equation is $ 6x + 3y - 5 = 0 $ .
For the third problem we have the equation is $ y = 0 $ .
The general formula for the slope – intercept form is $ y = mx + c $ .
Where $ m $ is the slope and $ c $ is the intercept.
i)
The given equation is $ x + 7y = 0 $ .
Now, we will add $ - 7y $ both sides to the above equation we obtain,
$ x + 7y - 7y = 0 - 7y $
Since $ 7y $ and $ - 7y $ get cancelled we obtain,
$ x = - 7y $
Multiply $ - 1 $ both sides to the above equation,
$ 7y = - x $
Then, multiply both sides by $ \dfrac{1}{7} $ we obtain,
$ y = - \dfrac{x}{7} + 0 $
The slope for the above equation is $ m = \dfrac{{ - 1}}{7} $ .
The intercept for the equation is $ 0 $ .
The slope -intercept form for the equation $ x + 7y = 0 $ is $ y = - \dfrac{x}{7} + 0 $ .
ii)
The given equation is $ 6x + 3y - 5 = 0 $ .
Adding $ - 3y $ both sides to the above equation we obtain,
$ 6x + 3y - 3y - 5 = - 3y $
Since $ 3y $ nd $ - 3y $ get cancelled we obtain,
$ 6x - 5 = - 3y $
Multiply $ - 1 $ both sides to the above equation,
$ 3y = - 6x + 5 $
Then, multiply both sides by $ \dfrac{1}{3} $ we obtain,
$ y = - \dfrac{{6x}}{3} + \dfrac{5}{3} $
The slope for the above equation is $ m = - \dfrac{6}{3} $ .
The intercept for the equation is $ \dfrac{5}{3} $ .
The slope - intercept form for the equation $ 6x + 3y - 5 = 0 $ is $ y = - \dfrac{{6x}}{3} + \dfrac{5}{3} $ .
iii)
The slope -intercept form is $ y = 0 $ .
Where, $ m = 0 $ and intercept is $ 0 $ .
Note: If $ x = 0 $ then in the slope-intercept form the slope will be $ 1 $ and intercept is $ 0 $ . If the slope is not given, then determine the slope and intercept and substitute in the general formula.
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