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+--- Thread: College Algebra "Graph Function Module" help! (/Thread-College-Algebra-Graph-Function-Module-help)
College Algebra "Graph Function Module" help! - purp!chickpot822 - 07-20-2015
Question 36 referring to identifying equations that are perpendicular.
The video is cut off from the bottom so I cannot see and barely hear what the narrator is explaining.
What I do not understand is how 15/4 ends up being just -5/4 after she "adds y=-(-5) to both sides of the equation given
that y-(-5) = -3/4 (x - 5).
Any help whatsoever will be greatly appreciated!
Thank you!
College Algebra "Graph Function Module" help! - gingerbeefE - 07-20-2015
Think there is a different thread for this but maybe I can explain
Y-(-5)=-3/4 (x-5)
Y+5= -3/4x +15/4 [ distributed -3/4 to x and -5]
-5 -20/4(this is = to -5)[subtracting -5 from both side]
Y=-3/4x -5/4
so to answer your question, she is not adding 5 she should be adding -5 to both sides
15/4 - 5 (or -20/4)= -5/4
College Algebra "Graph Function Module" help! - purp!chickpot822 - 07-21-2015
Thank you so much! I am going crazy with this algebra - being removed from it for over 30 years - it's like I am a first grader!
Thank you again.