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# Free Example of Intermediate Algebra Essay

1. Solve the following system of equations graphically... Solving for values of X and Y;

3x-4y=24;

x: O 12 16 20

y: -6 3 6 9

3x+2y=6;

x: 2 4 6 8

y: 0 -3 -6 -9

The point where the two lines intersect gives the solution to the equation. This is x=4 and y= -3

2. Show that (2, 5) is or is not a solution to the system of equations: 5x-4y=20 2x-3y=-1

If (2,5) is a solution to the equations, then x=2 and y=5

Substituting this values into the equations, 5(2)-4(5)=20

10-20=20, this is not true

Also 2(2)-3(5)=-1

4-15=-1, this is not also true-therefore these value are not a solution to the given equation.

3. Solve the system of equations by substitution. x+3y=5; 4x+5y=13

x+3y=5;

Reverse equation

x=-3y+5

Substituting x in equation 2 then solving for y;

4x+5y=13

4(-3y+5) +5y=13

-12y+20+5y=13

-7y=-7

y=1

Back substituting this y-value into the revised first equation;

x=-3y+5

x=-3(1) +5

x=2

Solution is (2, 1).

4. Solve the system of equations by elimination (addition) 7x+2y=5 2x+3y=16

You obtain the opposite coefficients of y by multiplying the first equation by 3 and the second equation by 2.

3(7x+2y=5) = 21x+6y=15

2(2x+3y=16) = - 4x+6y=32

_________

17x =-17

x=-1

Substitute x in the first equation

7(-1) +2y=5

-7+2y=5

Y=6

Solution is (-1, 6).

5. A candy manufacture sells 60 pounds of a candy mix for$66. It is composed of two candies, one worth $.90 a pound and the other is worth $1.50 a pound. How many pounds of each should be in the 60-pound mixture?

Let one of the candies to be A and the other to be B.

A+B=60

(0.9)A+(1.5)B=66

A=60-B; and 0.9A=66-1.5B or A=(66-1.5B)0.9

Therefore; 60-B=(66-1.5B)0.9

60-B=73.33-1.66B

0.66B=13.33

B=20.197

Since A=60-B

A=60-20.197

=39.803

6. Draw the graphs of systems of equations that are consistent, inconsistent, and dependent.

7. Determine whether or not {(1, 2), (4, 4), (5, 4), (7, 8)} is a function and then list the domain and range

Any given set represent a relation because the x's and y's are being related to each other. In the above set there are two points of y-value which are similar: (4. 4) and (5, 4). Therefore since y=4 gives two possible destinations then this relation is not a function.

Domain=Set of all input values of x= {1, 4, 5, 7}

Range=Set of all input values of y= {2, 4, 4, 8}

8. Find the domain of f(x) = Square root of x+2

The expression defining function f contains a square root hence the domain must positive or zero. The expression under the radical has to satisfy the condition

x +2 >= 0 for the function to take real values.

The domain, in interval notation, is given by

(+infinity, +2)

9. Explain, in words, how to use the vertical line test to determine if a graph is a function.

The vertical line test provides an easy way to check whether the equation that has been graphed is a function. The vertical line test states that a graph is the graph of a function if no vertical line intersects the graph more than once. To use the vertical line test, you look at a graph and determine if there are any places where a vertical line intersects the graph more than once. If you cannot find any such places, then this is the graph of a function. If you can, then this is not the graph of a function.