Solve the system by substitution. 2.5x−3y=−13, 3.25x−y=−14

Answer 1

We can solve the first equation for x to get x = (3y - 13)/2.5.

Substitute this expression for x into the second equation:

3.25(3y - 13)/2.5 - y = -14

Multiply both sides by 2.5 to eliminate the fractions:

3.25(3y - 13) - 2.5y = -14(2.5)

Distribute and simplify both sides:

9.75y - 42.25 - 2.5y = -35

Combine like terms on the left side:

7.25y - 42.25 = -35

Add 42.25 to both sides:

7.25y = 7.25

Divide both sides by 7.25:

y = 1

Substitute this value of y back into the expression for x:

x = (3(1) - 13)/2.5 = (-10)/2.5 = -4

Therefore, the solution to the system is x = -4 and y = 1.

Answer 2

A factory can produce two products, x and y, with a profit approximated by P=14x 22y−900. The production of y must exceed the production of x by at least 100 units. Moreover, production levels are limited by the formula x 2y≤1400. a.) Identify the vertices of the feasible region. b.) What production levels yield the maximum profit, and what is the maximum profit? Answer a.) and b.)

Answer 3

To identify the vertices of the feasible region, we need to solve the system of inequalities x + 2y ≤ 1400 and y - x ≥ 100.

First, let's rewrite the inequalities as equations by removing the inequality signs:
x + 2y = 1400
y - x = 100

To find the vertices, we solve these equations simultaneously.

We can start by solving the second equation for x:
x = y - 100

Now substitute this expression for x into the first equation:
(y - 100) + 2y = 1400

Combine like terms:
3y - 100 = 1400

Add 100 to both sides:
3y = 1500

Divide both sides by 3:
y = 500

Substitute this value of y back into the expression for x:
x = 500 - 100 = 400

So one vertex is (400, 500).

Now let's solve for the other vertex.

Substitute the expression for x into the equation y - x = 100:
y - (y - 100) = 100

Simplify:
100 = 100

This equation is always true, so there are no restrictions on y. Therefore, y can be any value and x can be determined by the equation x = y - 100.

So a second vertex is (0, 100).

To find the maximum profit, we evaluate the profit function P = 14x + 22y - 900 at each vertex.

At the first vertex (400, 500):
P = 14(400) + 22(500) - 900
P = 5600 + 11000 - 900
P = 16200

At the second vertex (0, 100):
P = 14(0) + 22(100) - 900
P = 0 + 2200 - 900
P = 1300

Therefore, the maximum profit is $16,200 and it occurs at the production levels x = 400 and y = 500.

Answer 4

A factory can produce two products, x and y, with a profit approximated by P=14x+22y−900. The production of y must exceed the production of x by at least 100 units. Moreover, production levels are limited by the formula x+2y≤1400. a.) Identify the vertices of the feasible region. b.) What production levels yield the maximum profit, and what is the maximum profit? Answer a.) and b.)

Answer 5

To identify the vertices of the feasible region, we need to solve the system of inequalities:

x + 2y ≤ 1400
y - x ≥ 100

Let's start by graphing the feasible region.

First, let's rewrite the inequalities as equations by removing the inequality signs:
x + 2y = 1400
y - x = 100

To graph these equations, let's solve for y in terms of x.

From the equation x + 2y = 1400:
2y = 1400 - x
y = (1400 - x)/2

From the equation y - x = 100:
y = x + 100

Now we can graph these two equations:

Plotting the first equation:
x + 2y = 1400
Let's choose some arbitrary x values:
For x = 0, y = 700
For x = 1400, y = 0
Plotting these two points and connecting them with a line, we get a downward sloping line:

Now plotting the second equation:
y - x = 100
Let's choose some arbitrary x values:
For x = 0, y = 100
For x = 1400, y = 1500
Plotting these two points and connecting them with a line, we get an upward sloping line:

The feasible region is the area where both lines intersect, which is the shaded region below:

(1400,0)
(400, 500)
(0,100)

Now let's find the vertices of the feasible region:

- (0, 100)
- (400, 500)
- (1400, 0)

To find the production levels that yield the maximum profit, we substitute the vertices into the profit function P = 14x + 22y - 900, and see which combination gives the highest profit.

For each vertex:
- (0, 100):
P = 14(0) + 22(100) - 900
P = 2200 - 900
P = 1300

- (400, 500):
P = 14(400) + 22(500) - 900
P = 5600 + 11000 - 900
P = 16200

- (1400, 0):
P = 14(1400) + 22(0) - 900
P = 19600 - 900
P = 18700

Therefore, the production levels that yield the maximum profit are x = 400 and y = 500, with a maximum profit of $16,200.

Answer 6

What is the result when 2x^3−9x^2 11x−6 is divided by x−3 ? Show your work.

Answer 7

To divide 2x^3 - 9x^2 + 11x - 6 by x - 3, we can use long division.

2x^2 + -3x + 2
--------------------------------
x - 3 | 2x^3 - 9x^2 + 11x - 6
- (2x^3 - 6x^2)
---------------------------
-3x^2 + 11x
- (-3x^2 + 9x)
-----------------
2x - 6
- (2x - 6)
------------------
0

Therefore, the result of dividing 2x^3 - 9x^2 + 11x - 6 by x - 3 is 2x^2 - 3x + 2.

Answer 8

What is the result when 2x^3−9x^2 +11x−6 is divided by x−3 ? Show your work.

Answer 9

To divide 2x^3 - 9x^2 + 11x - 6 by x - 3, we can use long division.

2x^2 + 3x - 1
---------------------------------
x - 3 | 2x^3 - 9x^2 + 11x - 6
- (2x^3 - 6x^2)
-------------------
-3x^2 + 11x
- (-3x^2 + 9x)
----------------
2x - 6
- (2x - 6)
-----------------
0

Therefore, the result of dividing 2x^3 - 9x^2 + 11x - 6 by x - 3 is 2x^2 + 3x - 1.