Question

Solving Systems...I'm Stuck?

I'm unsure how to solve these systems..what's the best formula?
Also, if the picture doesn't come up, here's the first problem:

A local computer company sells three types of laptop computers to three nearby stores. The number of laptops ordered by each store and the amount owed to the company for the order is shown in the following table:

Store Laptop A Laptop B Laptop C Amount Owed ($)
Wal-Mart 10 8 6 21,200
Sears 7 9 5 18,700
Target 8 4 3 13,000

Write a system of equations to represent the above information and determine the unit price of each type of laptop computer. Define your variables! Show and explain all work!

There's other ones but I figure if I can see how this is done I can work throught others

Answer 1

`x = 800, y = 900, z = 1000`

Explanation:

Actually, This is a question for heck of a hard work.

You can use Matrices to solve this system of equation in a quicker and way way more simpler way, but unfortunately, I'm just a novice and I don't know about matrices that much. (T-T).

First, the Simplest Task, Assuming the Variables.

Let The Price of a laptop of Kind A be `$x`.

Similarly, The price of each laptop of Kind B and Kind C are `$y` and `$z` respectively.

So, For the Wal-Mart Company :

As your table tells, They ordered `10` laptops of Kind A, `8` laptops of Kind B, and `6` laptops of Kind C; and the total price of order is `$21,200`.

So, The equation for Wal-Mart Company :

`10x + 8y + 6z = 21,200` ........................................................(i)

Similarly,

The equation for Sears :

`7x + 9y + 5z = 18,700` ...........................................................(ii)

The equation for Target :

`8x + 4y + 3z = 13,000` ...........................................................(iii)

So, Now comes the arduous task.

WARNING! BRACE YOURSELF FOR THE CONFUSING WORK!

First, Multiply The (iii) equation with `2`.

So, It will be :

`16x + 8y + 6z = 26,000`..........................................................(iv)

Now, Subtract eq(iv) from eq(i). [WOW! I got The `x`, yeah!!!!]

`color(white)(xxx)10x + cancel(8y) + cancel(6z) - 16x - cancel(8y) - cancel(6z) = 21,200 - 26000`

`rArr -6x = -4800`

`rArr x = (cancel(-)4800)/(cancel(-)6) = 800` [OMG $800!!!!!]

Now, Substitute `x = 800` in eq(i) and eq(ii)..

`color(white)(xx)800*10 + 8y + 6z = 21,200`

`rArr 8y + 6z = 21,200 - 8,000`

`rArr 8y + 6z = 13,200`.....................................(v)

And, `7 * 800 + 9y + 5z = 18,700`

`rArr 5600 + 9y + 5z = 18,700`

`rArr 9y + 5z = 18,700 - 5600`

`rArr 9y + 5z = 13,100`...................................(vi)

Now, Let's Multiply eq(v) with `5` and eq(vi) with `6`.

So, `40y + 30z = 66,000`....................................(vii)

And, `54y + 30z = 78,600`.....................................(viii)

Now, Subtract eq(vii) from eq(viii).

So, `54y + cancel(30z) - 40y - cancel(30z) = 78,600 - 66,000`

`rArr 14y = 12,600`

`rArr y = 900` [Got the Y!!!!!!!!!! Yeah!!!!!!!!! Wooohoo!]

Now, Substitute `y = 900` in eq(vii).

`color(white)(xx)900 * 40 + 30z = 66,000`

`rArr 36000 + 30z = 66,000`

`rArr 30z = 30000`

`rArr z = 1000`

So, Finally Got Everything.

I pray to God that this helps.