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Ilia_Sergeevich [38]
3 years ago
7

Trey's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Trey $4.80 per pound, and type B

coffee costs $5.85 per pound. This month's blend used three times as many pounds of type B coffee as type A, for a total cost of $603.45 . How many pounds of type A coffee were used?
Mathematics
1 answer:
boyakko [2]3 years ago
4 0

Let x = the number of pounds of Type A coffee.

 

We know that this month's blend used 3 times as many pounds of type B coffee as type A coffee.

Total pounds of coffee =  x + 3x

 

The total cost is:

(cost for Type A)(x) + (cost for Type B)(3x) = $603.45

$4.80x + $5.85(3x) = $603.45

$4.80x + $17.55x = $603.45

$22.35x = $603.45

x = 603.45/22.35 = 27 pounds


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Find all solutions of the given system of equations (If the system is infinite many solution, express your answer in terms of x)
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Answer:

(a) The system of the equations \left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right. has no solution.

(b) The system of the equations \left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right. has many solutions y=\frac{2x}{3}-\frac{5}{3}

Step-by-step explanation:

(a) To find the solutions of the following system of equations \left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right. you must:

Multiply 2x-3y=3 by 2:

\begin{bmatrix}4x-6y=6\\ 4x-6y=3\end{bmatrix}

Subtract the equations

4x-6y=3\\-\\4x-6y=6\\------\\0=-3

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(b) To find the solutions of the system \left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right. you must:

Isolate x for 4x-6y=10

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Substitute x=\frac{5+3y}{2} into the second equation

16\cdot \frac{5+3y}{2}-24y=40\\8\left(3y+5\right)-24y=40\\24y+40-24y=40\\40=40

The system has many solutions.

Isolate y for 4x-6y=10

y=\frac{2x}{3}-\frac{5}{3}

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