Question

A candy man manufacture wishes to mix two candies as a sales promotion when candy sells for $3per pound in other cells for $1.75 per pound the manufacture which is to have 1000 lbs of a mixture and stir the mixture for $2.35 per lbs how many pounds of each candy should be used in the mixture

Answer 1

Answer:

$\displaystyle x=480\ ,\ y=520$

Step-by-step explanation:

System Of Two Linear Equations

A system of two linear equations is given as

$\displaystyle \left\{\begin{matrix}ax+by=c\\ dx+ey=f\end{matrix}\right.$

We must find the values of x and y who make both equations comply, i.e. they become identities

The question talks about one candy who sells for $3 per pound and others who sells for $1.75 per pound. Let's call x and y the pounds of each candy that must be used in a mixture with the conditions that:

• 1000 lbs of the mixture will be produced
• They will be sold for $2.35 per lb, i.e. for a total of $2,350

We form the system with both conditions

$\displaystyle \left\{\begin{matrix}x+y=1000\\ 3x+1,75y=2,350\end{matrix}\right.$

Multiplying the second equation by 4

$\displaystyle \left\{\begin{matrix}x+y=1,000\\ 12x+7y=9,400\end{matrix}\right.$

Multiplying the first equation by -7

$\displaystyle \left\{\begin{matrix}-7x-7y=-7,000\\ 12x+7y=9,400\end{matrix}\right.$

Adding both equations

$\displaystyle 5x=2,400$

$\displaystyle x=480$

Using the relation

$\displaystyle x+y=1,000$

We solve for y

$\displaystyle y=1,000-480$

$\displaystyle y=520$

The solution is

$\displaystyle x=480\ ,\ y=520$