Oranges and grapefruits
Problem 1
On Monday you bought 11 oranges and 4 grapefruit for a total of $7.40. Later that day, you bought 3 oranges and 5 grapefruits
for a total of $4.95. Find the price of each orange and each grapefruit.
Solution
Let "n" be the number of oranges, and let "g" be the number of grapefruits.
So, you have this system of two equations in two unknowns
11n + 4g = 740, (1)
3n + 5g = 495. (2)
For this system the Elimination method suits very well, so I will apply it.
Multiply first equation by 5 (both sides) and multiply second equation by 4.
In this way you get the system of EQUIVALENT equations
55n + 20g = 3700, (3)
12n + 20g = 1980. (4)
Now the coefficients at "g" are the same in both equations. It was my goal to do it, and I reached this goal.
Now subtract equation (4) from equation (3). The terms "20g" will cancel each other, and you will get SINGLE equation for ONLY ONE unknown "n":
55n - 12n = 3700 - 1980, or
43n = 1720.
From the last equation you easily get n =
= 40.
Thus we just found the price of one ORANGE. It is 40 cents.
Now substitute the found value of n= 40 into the either equation (1) or (2) to find "g". I will substitute into eq(2):
3*40 + 5g = 495.
It gives you
120 + 5g = 495 ====> 5g = 495 - 120 = 375 ====> g =
= 75 cents.
Answer. One orange costs 40 cents. One grapefruit costs 75 cents.
Solved.
Lessons to learn from this solution
1. The key step is to write correctly the system of equations for the given problem.
It is not difficult. The major precaution you should make for this step is to read the condition carefully.
2. The second step is to look into the system and to decide which method of solution to choose.
There are different methods: Elimination or/and Substitution.
Actually, there are other methods too (using determinants, or the Cramer's rule).
But if you only started Algebra-I, you, presumably, know only Elimination or Substitution methods.
It is totally enough to solve any problem of your curriculum.
3. And the last step is to implement the method and the solution ACCURATELY.
4. Did I say "the last" ? - No, it is not the last step.
The last step is to CHECK the solution. For it, simply substitute the found values into your original equations:
11*40 + 4*75 = 740. ! Correct !, and
3*40 + 5*75 = 495. ! Correct !
5. And now the solution is completed.
My other lessons in this site on solving systems of two linear equations in two unknowns (Algebra-I curriculum) are
- Solution of the linear system of two equations in two unknowns by the Substitution method
- Solution of the linear system of two equations in two unknowns by the Elimination method
- Solution of the linear system of two equations in two unknowns using determinant
- Geometric interpretation of the linear system of two equations in two unknowns
- Useful tricks when solving systems of 2 equations in 2 unknowns by the Substitution method
- Solving word problems using linear systems of two equations in two unknowns
- Word problems that lead to a simple system of two equations in two unknowns
- 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
- A theater group made appearances in two cities
- Exchange problems solved using systems of linear equations
- Typical word problems on systems of 2 equations in 2 unknowns
- HOW TO algebreze and solve these problems on 2 equations in 2 unknowns
- One unusual problem to solve using system of two equations
- Non-standard problem with a tricky setup
- Sometimes one equation is enough to find two unknowns in a unique way
- Solving mentally word problems on two equations in two unknowns
- Solving systems of non-linear equations by reducing to linear ones
- Solving word problems for 3 unknowns by reducing to equations in 2 unknowns
- 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
- OVERVIEW of lessons on solving systems of two linear equations in two unknowns
Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.