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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

User Junfei
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1 Answer

4 votes

Answer:


\displaystyle x=480\ ,\ y=520

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

User Krethika
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