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myrzilka [38]
3 years ago
11

Solve for x if 10/4x-3 = 2​

Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
6 0

Answer:

Kindly see attached picture

hope it helped you:)

Marysya12 [62]3 years ago
3 0

Answer:

We need to solve this equation:

\frac{10}{4x}-3=2

We pass -3 to the right summing:

\frac{10}{4x} = 2 + 3

\frac{10}{4x}=5

We pass the 4x that is dividing to the right multiplying:

10 = 5*4x

10=20x

We pass the 20 of the right to the left dividing:

\frac{10}{20}=x

x=\frac{1}{2}

Now, we know x=\frac{1}{2}, so we can verify:

\frac{10}{4*\frac{1}{2}}-3 = 2

\frac{10}{2}-3=2

5 - 3 = 2 TRUE

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The Fine Line Pen Company makes two types of ballpoint pens: a silver model and a gold model. The silver model requires 1 minute
tresset_1 [31]

Answer:

Optimal production = 600 gold pens

Revenue  = 600*7 = $4200 gold pens

Step-by-step explanation:

The Fine Line Pen Company makes two types of ballpoint pens: a silver model and a gold model.

A. The silver model requires 1 minute in a grinder and 3 minutes in a bonder.

B. The gold model requires 3 minutes in a grinder and 4 minutes in a bonder.

Because of maintenance procedures,

C. the grinder can be operated no more than 30 hours per week and

D. the bonder no more than 50 hours per week.

The company makes

E. $5 on each silver pen and

F. $7 on each gold pen.

How many of each type of pen should be produced and sold each week to maximize profits?

Solution:

We will solve the problem graphically, with number of silver pens, x, on the x axis, and number of gold pens, y, on the y axis, i.e.

1. From A and C, the maximum number of silver pens

x <= 30*60 / 1 = 1800 and

x <= 50*60 /3 = 1000  ....................(1)   bonder governs

2. from A & D, the maximum number of gold pens

y <= 30*60 / 3 = 600 .....................(2) grinder governs

y <= 50*60 / 4 = 750

3. From D,

x + 3y <= 30*60 = 1800  ...................(limit of grinder) ..... (3)

3x + 4y <= 50*60 = 3000 .................(limit of bonder)  .......(4)

Need to maximize profit,

Z(x,y) = 5x+7y, represented by parallel lines y = -5x/7 + k such that all constraints of (3) and (4) are satisfied.

The maximum is obtained when Z passes through (360,480), i.e. at intersection of constraints (3) and (4).  Using slope intercept form,

(y-480) = -(5/7)(x-360)

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(y-600) = -(5/7)(x-0), or

y = 600 - (5/7)x   [the black line]

As we can see all point on the black (in the first quadrant) satisfy the constraints, so it is a feasible solution, and is the optimal solution, with a revenue of

Revenue  = 600*7 = 4200 gold pens

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