SOLUTION: Donna and Kayleigh both go to the same high school. Donna lives 21 miles from the school.Kayleigh lives 6 miles from Donna. Part A. Write an absolute value equation to represent t

Algebra ->  Absolute-value -> SOLUTION: Donna and Kayleigh both go to the same high school. Donna lives 21 miles from the school.Kayleigh lives 6 miles from Donna. Part A. Write an absolute value equation to represent t      Log On


   



Question 1206363: Donna and Kayleigh both go to the same high school. Donna lives 21 miles from the school.Kayleigh lives 6 miles from Donna.
Part A. Write an absolute value equation to represent the location of Kayleigh’s house in relation to the high school.
Part B. How far could Kayleigh live from her school?

Found 3 solutions by ikleyn, Theo, greenestamps:
Answer by ikleyn(52847) About Me  (Show Source):
You can put this solution on YOUR website!
.

The problem does not say that all locations are along one straight line.


Therefore, based on common sense, we may assume that the locations are on the 2D 
Earth surface, which we may assume flat.


If so, then the problem is worded incorrectly.


In this case, the question should be posed about an inequality, not about an equation.


An inequality for the problem is  

      15 <= |H - K| <= 27  miles.


Here 15 = 21-6;  27 = 21 + 6.


It is the answer to question (A).


The answer to question (B) is  that Kayleigh can live from 15 to 27 miles from her school.

Solved.


From reading this post, I see a fog in the head of the "math composer".

Also, I see that this "math composer" is not able to write Math problem in unambiguous mode.


At this forum, I see it every day and, to say more, several times per day.

But this so called "math composer" should not struggle of my words.
In the positive mode, it simply means that for him (or for her) there is a room to grow.



Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
donna lives 21 miles from the school.
kayleigh lives 6 miles from donna.
the absolute value inequality that models this would be abs(x - 21) <= 6
this says that the maximum distance from the school to kayleigh's house will be 21 + 6 = 27 miles and the minimum distance from the school to kayleigh's house will be 21 - 6 = 15 miles.

the absolute value inequality is solved by breaking it up into two separate parts.
the first part assumes x - 21 is positive and the second part assumes x - 21 is negative.

first part is (x-21) <= 6.
add 21 to both sides of that inequality to get x <= 27

second part is -(x-21) <= 6
multiply both sides of that by -1 to get (x-21) >= -6.
that's because multiplying both sides of an inequality by a negative number reverses the inequality.
solve for x to get x >= 15.

the smallest distance from kayleigh to the school is 15.
the largest distance from kayleigh to the school is 27.

here's a reference on absolute value equalities and inequalities.

https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra/02%3A_Graphing_Functions_and_Inequalities/206%3A_Solving_Absolute_Value_Equations_and_Inequalities

let me know if you have any questions.
theo

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Donna and Kayleigh both go to the same high school. Donna lives 21 miles from the school. Kayleigh lives 6 miles from Donna.

Kayleigh lives somewhere on a circle with radius 6 miles with center at Donna's house. So the distance from Kayleigh's house to the school can be anything from some minimum to some maximum. It should be obvious that the minimum and maximum distances of Kayleigh's house from the school are if the school and the two girls' houses lie in a straight line.

Part A. Write an absolute value equation to represent the location of Kayleigh’s house in relation to the high school.

That is poorly written. What is in fact needed (I think!) is an absolute value INEQUALITY that can be solved to find the range of possible distances from Kayleigh's house to the school.

That inequality says that the difference between the distance x of Kayleigh's house to the school and the distance of Donna's house from the school (21 miles) is at most 6 miles:

abs%28x-21%29%3C=6

Part B. How far could Kayleigh live from her school?

Informally, the distance from Kayleigh's house to the school is at least 21-6 = 15 miles and at most 21+6 = 27 miles.

Formally....

abs%28x-21%29%3C=6
-6%3C=x-21%3C=6
-6%2B21%3C=x%3C=6%2B21
15%3C=x%3C=27