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Mike has coffee worth $4 per pound that he wishes to mix with 20 pounds of coffee worth $7 per pound to get a mixture that can be sold for $5 per pound. How many pounds of the cheaper coffee should he use?

Respuesta :

[tex]\bf \begin{array}{lccclll} &\stackrel{lbs}{amount}&\stackrel{per~lb}{price}&\stackrel{amount}{price}\\ &------&------&------\\ \textit{\$4/lb coffee}&x&4&4x\\ \textit{\$7/lb coffee}&20&7&140\\ ------&------&------&------\\ mixture&y&5&5y \end{array}[/tex]

so, we know the mixture is the sum of both types of coffee, thus x + 20 = y, and 4x + 140 = 5y.

[tex]\bf \begin{cases} x+20=\boxed{y}\\ 4x+140=5y\\ ----------\\ 4x+140=5\left( \boxed{x+20} \right) \end{cases} \\\\\\ 4x+140=5x+100\implies 140-100=5x-4x\implies 40=x[/tex]
facundo3141592

By solving a linear equation, we will see that Mike must use 40 pounds of the cheaper coffee.

How many pounds of the cheaper coffee should he use?

First, let's define the variable

  • x = pounds of the $4 coffee used.

We know that we will use 20 pounds of the $7 coffee, then the total mass of the mix will be (20 + x) pounds, and the price of this mix must be $5.

Now, the costs must be the same in both sides of the equation, so we can write:

x*$4 + 20*$7 = (20 + x)*$5

Now we can solve this linear equation for x:

x*$4 + $140 = $100 + x*$5

$140 - $100 = x*$5 - x*$4

$40 = $1*x

$40/$1 = 40 = x

So he should use 40 pounds of the cheaper coffee.

If you want to learn more about linear equations, you can read:

https://brainly.com/question/1884491

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