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k0ka [10]
3 years ago
12

Two step Equation a.3x-8=31

Mathematics
2 answers:
Arturiano [62]3 years ago
6 0
3x - 8 = 31

1) 3x = 31 + 8 (add 8 to both sides)
2) x = 39/3 = 13 (divide both sides by 3)

Hope this is what you're lookin for!
stira [4]3 years ago
5 0
<span>Simplifying 3x + -8 = 31 Reorder the terms: -8 + 3x = 31 Solving -8 + 3x = 31 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '8' to each side of the equation. -8 + 8 + 3x = 31 + 8 Combine like terms: -8 + 8 = 0 0 + 3x = 31 + 8 3x = 31 + 8 Combine like terms: 31 + 8 = 39 3x = 39 Divide each side by '3'. x = 13 Simplifying x = 13</span>
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A coordinate plane graph is shown. Point C is at negative 4 comma 3. Point D is at 1 comma 0. A segment connects the two points.
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Is y= 1/2x -4 a linear equation
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Answer:

Yes

Step-by-step explanation:

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Southern Oil Company produces two grades of gasoline: regular and premium. The profit contributions are $0.30 per gallon for reg
Contact [7]

Answer:

a) MAX--> PC (R,P) = 0,3R+ 0,5P

b) <u>Optimal solution</u>: 40.000 units of R and 10.000 of PC = $17.000

c) <u>Slack variables</u>: S3=1000, is the unattended demand of P, the others are 0, that means the restrictions are at the limit.

d) <u>Binding Constaints</u>:

1. 0.3 R+0.6 P ≤ 18.000

2. R+P ≤ 50.000

3. P ≤ 20.000

4. R ≥ 0

5. P ≥ 0

Step-by-step explanation:

I will solve it using the graphic method:

First, we have to define the variables:

R : Regular Gasoline

P: Premium Gasoline

We also call:

PC: Profit contributions

A: Grade A crude oil

• R--> PC: $0,3 --> 0,3 A

• P--> PC: $0,5 --> 0,6 A

So the ecuation to maximize is:

MAX--> PC (R,P) = 0,3R+ 0,5P

The restrictions would be:

1. 18.000 A availabe (R=0,3 A ; P 0,6 A)

2. 50.000 capacity

3. Demand of P: No more than 20.000

4. Both P and R 0 or more.

Translated to formulas:

Answer d)

1. 0.3 R+0.6 P ≤ 18.000

2. R+P ≤ 50.000

3. P ≤ 20.000

4. R ≥ 0

5. P ≥ 0

To know the optimal solution it is better to graph all the restrictions, once you have the graphic, the theory says that the solution is on one of the vertices.

So we define the vertices: (you can see on the graphic, or calculate them with the intersection of the ecuations)

V:(R;P)

• V1: (0;0)

• V2: (0; 20.000)

• V3: (20.000;20.000)

• V4: (40.000; 10.000)

• V5:(50.000;0)

We check each one in the profit ecuation:

MAX--> PC (R,P) = 0,3R+ 0,5P

• V1: 0

• V2: 10.000

• V3: 16.000

• V4: 17.000

• V5: 15.000

As we can see, the optimal solution is  

V4: 40.000 units of regular and 10.000 of premium.

To have the slack variables you have to check in each restriction how much you have to add (or substract) to get to de exact (=) result.  

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I can’t see the photo could you please make it bigger so I can help I
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