To solve this problem, we'll let "x" represent the number of pounds of the high quality coffee bean that costs $6.00 per pound, and "y" represent the number of pounds of the cheaper coffee bean that costs $3.25 per pound.
Sarah wants the final blend to weigh 160 pounds, so we can write the first equation that represents the total weight of the blend:
1) \( x + y = 160 \)
She also wants the final blend to sell for $4.97 per pound. Therefore, the total cost of the 160 pounds at $4.97 per pound would be:
\( 160 \times 4.97 \)
Now let's write the second equation that represents the total cost of the blend using the individual costs of each type of bean:
2) \( 6x + 3.25y = 160 \times 4.97 \)
Let's calculate the total cost of 160 pounds at $4.97 per pound:
$\( 160 \times 4.97 = 795.20 \)
So, the second equation using this total cost is:
\( 6x + 3.25y = 795.20 \)
Now we have a system of two equations with two variables:
1) \( x + y = 160 \)
2) \( 6x + 3.25y = 795.20 \)
We will solve this system by expressing one variable in terms of the other from the first equation and then substituting it into the second equation.
From equation 1), we can express "y" as:
\( y = 160 - x \)
Now we substitute this expression for "y" into equation 2):
\( 6x + 3.25(160 - x) = 795.20 \)
Now let's distribute and simplify:
\( 6x + 520 - 3.25x = 795.20 \)
Combine like terms:
\( (6 - 3.25)x + 520 = 795.20 \)
\( 2.75x + 520 = 795.20 \)
Now subtract 520 from both sides:
\( 2.75x = 795.20 - 520 \)
\( 2.75x = 275.20 \)
Now we can solve for "x" by dividing both sides by 2.75:
\( x = \frac{275.20}{2.75} \)
\( x ≈ 100 \)
So, Sarah should use approximately 100 pounds of the high quality coffee beans.
Now we need to find "y". From the equation \( y = 160 - x \), we substitute the value of "x":
\( y = 160 - 100 \)
\( y = 60 \)
Therefore, Sarah should blend approximately 100 pounds of high quality coffee beans with 60 pounds of the cheaper coffee beans. This will give her a 160-pound blend that sells for $4.97 per pound.