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Question 245815: I need help. Thank
1.Translate the problem into a pair of linear equations in two vaiables. Solve the equations using either elimination or substitution. State your answer for the specified variable.
2. Ellen wishes to mix candy worth $1.80 per pound with candy worth $2.40 per pound to form 48 pounds of a mixture worth $2.00 per pound. How many pounds of the more expensive candy should she use? . Given the pair of linear equations in two variables:
•Find the x- and y-intercepts (if any) for each line. Show your work.
•Plot those intercepts, and graph the two lines. Submit your graph through the Dropbox.
•Apply elimination or substitution to find the coordinates of the point of intersection (if there are no solutions or infinite solutions, state this). Show your work.
15x + 11y = -17
8x - 6y = 74
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! I will solve the candy problem. Start with what you know.
x = candy @ $1.80/lb
y = candy @ $2.40/lb
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She wants to produce 48 lbs valued at $2.00/lb.
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So we have x(1.80) + y(2.40) = 48(2.00) = 96.
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But with only one equation, having two unknowns is not solvable. But since we know the total, we define x in terms of y or y in terms of x.
x lb = 48 lb - y lb
y lb = 48 lb = x lb
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Rewriting the equation we have:
1.8x + 2.4y = 96
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Substitute y = 48-x
1.8x + 2.4(48-x) = 96
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Multiply thru
1.8x + 115.2 - 2.4x = 96
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Collect terms
-0.6x + 115.2 = 96
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Subtract 115.2 from both sides
-0.6x = 96 - 115.2 = -19.2
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Multiply both sides by -10
6x = 192
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Divide both sides by 6
x = 32
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Looking back to our setup, we know that y = 48-x = 48-32 = 16.
So we have 32 lb @ $1.80/lb = $57.60
And we have 16 lb @ $2.40/lb = $38.40
Thus we have a total of 48 lb and a total cost of $57.60 + $38.40 = $96.00.
96/48 = $2/lb.
48 lb @ $2/lb = $96.
So our solution checks.
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