SOLUTION: Use two equations in two variables to solve the application. See Example 6. (Objective 1)
A merchant wants to mix peanuts worth $3 per pound with jelly beans worth $1.50 per pou
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A merchant wants to mix peanuts worth $3 per pound with jelly beans worth $1.50 per pou
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Question 1152316: Use two equations in two variables to solve the application. See Example 6. (Objective 1)
A merchant wants to mix peanuts worth $3 per pound with jelly beans worth $1.50 per pound to make 30 pounds of a mixture worth $2.40 per pound. How many pounds of each should he use? Answer by ikleyn(52847) (Show Source):
Let x = pounds of peanuts
y = pounds of jelly beans.
Then you have these two equations
x + y = 30 (1) (pounds of the new mixture)
3x + 1.50y = 2.40*30 (2) (the total cost of the mixture in both sides)
From equation (1), express y = 30 - x and substitute it into equation (2), replacing y. You will get
3x + 1.50*(30-x) = 72.
From the last equation.
3x + 1.50*30 - 1.50x = 72
3x - 1.50x = 72 - 45
1.50x = 27
x = = 18.
ANSWER. 18 pounds of peanuts and the rest (30-18) = 12 pounds of jelly beans.
Solved.
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It is a standard and typical mixture word problem.
You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.
A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.
Read them and become an expert in solution mixture word problems.
