SOLUTION: Anjali and joe improved their yards by planting daylilies and ivy. They bought their supplies from the same store. Anjali spent $233 on 13 daylilies and 14 pots of ivy. Joe spent $

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Question 1103078: Anjali and joe improved their yards by planting daylilies and ivy. They bought their supplies from the same store. Anjali spent $233 on 13 daylilies and 14 pots of ivy. Joe spent $109 on 5 daylilies and 7 pots of ivy. find the cost of one daylily and the cost of one pot of ivy
Found 2 solutions by addingup, ikleyn:
Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
13d + 14i = 233
5d + 7i = 109 multiply all times -2 and add to above
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
13d + 14i = 233
+
-10d - 14i = -218
------------------
3d 0i = 15
3d = 15
d = 5 the daylilies are 5 each. Now let's solve for ivy:
------------------
13(5) + 14i = 233
65 + 14i = 233
14i = 168
i = 12 the ivy are 12 per pot.
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In case you are wondering, you would get the same answer using the other equation to find i:
5d + 7i = 109
5(5) + 7i = 109
25 + 7i = 109
7i = 84
i = 84/7 = 12
:
Happy learning!

Answer by ikleyn(52756)   (Show Source): You can put this solution on YOUR website!
.
From the condition, you have the system of 2 equations in 2 unknowns

13*d + 14*p = 233,    (1)    (Anjali' spending)
 5*d +  7*p = 109.    (2)    (Joe' spending)


To solve the system, multiply equation (2) by 2 (both sides). The modified system is

13*d + 14*p = 233,    (3)   
10*d + 14*p = 218.    (4) 


Now subtract eq(4) from eq(3) (both sides).  The terms "14*p" in both equations will cancel each other, and 
you will get a single equation for only one unknown "d" (it is how the Elimination method works):

3d = 233 - 218  = 15  ====>  d =   = 5.


Thus you found that the pot of ivy costs $5.


Then from eq(2),   7p = 109 - 5*5 = 84  and  d =  = 12.


Answer.  One daylily costs $12  and  one pot of ivy costs $5.


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On solving systems oi two linear equations in two unknowns and related word problems see the lessons
    - Solution of a linear system of two equations in two unknowns by the Substitution method
    - Solution of a linear system of two equations in two unknowns by the Elimination method
    - Solution of a linear system of two equations in two unknowns using determinant
    - Geometric interpretation of a linear system of two equations in two unknowns
    - Solving word problems using linear systems of two equations in two unknowns

    - Oranges and grapefruits
    - Using systems of equations to solve problems on tickets
    - Three methods for solving standard (typical) problems on tickets
    - Using systems of equations to solve problems on shares
    - Using systems of equations to solve problems on investment
    - Two mechanics work on a car
    - The Robinson family and the Sanders family each used their sprinklers last summer
    - Roses and vilolets
    - Counting calories and grams of fat in combined food
    - HOW TO algebreze and solve this problem on 2 equations in 2 unknowns
    - One unusual problem to solve using system of two equations
    - Solving mentally word problems on two equations in two unknowns
    - Solving systems of non-linear equations by reducing to linear ones
    - System of equations helps to solve a problem for the Thanksgiving day
    - Using system of two equations to solve the problem for the day of April, 1
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Systems of two linear equations in two unknowns".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.



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