TECHNIQUES IN GETTING THE LEAST COMMON DENOMINATOR IN COLLEGE ALGEBRA
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REVIEW/TUTORIAL FOR THE COLLEGE ALGEBRA PLACEMENT TEST

1. Evaluate x2 3x + 9 for x = 1
2.
 Multiply and simplify:
3. Simplify and factor: 5x 2(y x) x
4.
 Simplify:
5.
 Simplify and write as a single fraction:
6.
 Simplify as much as possible:
7.
 Simplify by rationalizing the denominator:

8. Multiply: (8x3 y)(3x2y7)

9. Find the slope of the line: 2x 3y + 10 = 0
10.
 Write as a single fraction in terms of
11.
 Solve for all values of x:
12.
 Solve for all values of x: = 5
13.
 Multiply and simplify:
14. Find the solution set for x: 2x + 1 > 3x + 7
15. Solve for x: 2x2 + 3x 2 = 0
16. Solve for x: 2x2 3x 4 = 0
17. Solve for x: 2x2 6x + 5 = 0
18.
 Evaluate:
19.
 Solve for x and y:
20.
 Find the solution set for x:
21.
 Write as a single fraction.
22. Factor completely: x4 81.
23.
24. The graphs of the system of equation consists of lines that
a) intersect b) meet in one point c) are parallel
25.
 Simplify:
26. Find the solution set for x: x2 + 6 > 5x
27. If f(x) = x3 + 1, find f(3)
28. Graph: y = 1
29. Graph: x = 3
30. Graph: 2x + 3y = 2

Evaluate x2 3x + 9  for x = 1
When substituting x = 1 in   x2
be sure to do the exponent before the multiplication by   1
to get    (1)2 = 1.   So     1 3 + 9 = 5

 When multiplying so that

 becomes

Using the distribute law -2(y - x) = -2y + 2x we get:
5x 2(y x) x
= 5x 2y + 2x x
= 6x -2y
= 2(3x - y)

 To divide by 3, we invert and multiply

Write the numerator as a single fraction with denominator of 4,
and write the denominator as a single fraction with denominator of 6,

 Now divide by
Be sure to invert and multiply to get

Here we divided the denominator 4 and the numerator 6 by 2.

,

 since x4 is positive. Similarly

 Since we write

When multiplying exponent expressions with the same base, keep the
base and add the exponents thus, x3(x2) = x5 and y(y7) = y8

Write the equation of the line in the form y = mx + b or where m is the slope.
3y = 2x 10

 Write and rationalize Now we are adding

Multiply both sides of the equation by the denominator x 3

 Use the fact that
1 3(x 3) = x
1 3x + 9 = x
10 = 4x

12. Answer x = 5,   x = 5
There are two values of x whose absolute value is 5
|-5| = 5   and   |5| = 5

We factor x2 3x = x(x 3)   and   9 x2 = (3 x)(3 + x)
Remember the difference of squares factorization
a2 b2 = (a b)(a + b).
Thus we get
Use the fact that x – 3 = – (3 – x),   so that

Putting this together we get

2x + 1 > 3x + 7
Subtract 3x  from both sides x + 1 > 7
Subtract 1 from both sides x > 6
Multiply by 1, we get x < 6
Remember multiplying an inequality by a minus changes the sense of the arrow.

 and x = 2
Consider the standard quadratic equation ax2 + bx + c = 0,
 whose solution is
For the solution of 2x2 + 3x 2 = 0, a = 2, b = 3, and c = 2.
Since the discriminant b2 4ac = 32 4(2)( 2) = 9 + 16 = 25 is a
perfect square, we can factor directly: 2x2 + 3x 2 = 0,
(2x 1)(x + 2) = 0
 2x 1 = 0 x + 2 = 0
 x = 2

For the solution of 2x2 3x 4 = 0, we have a = 2, b = 3, c = 4.
The discriminant b2 4ac = ( 3)2 4(2)( 4) = 9 + 32 = 41 is not a
perfect square so that we must use the quadratic formula and get
 two real roots: Thus

For the solution of 2x2 6x + 5 = 0, a = 2, b = 6, c = 5

The discriminant b2 4ac = ( 6)2 4(2)( 5) = 36 40 = 4.

Since the discriminant is negative, the two roots are imaginary.

 Thus, here we use

We can reduce the answer by factoring 2 in the numerator

The negative exponent means we have to take the reciprocal of what is in
 the parentheses and then square.
Remember when raising a fraction to a power, both the denominator and
numerator are raised to that power

19. Answer x = 3, y = 2 Solve for x and y there are two methods that are often used.
Addition method: Multiply the second equation by 5

5x 3y = 21
5x 25y = 35
28y = 56

Substitute back into the second equation to get:
x + 5( 2) = 7
x 10 = 7
x = 7 + 10 = 3
Thus x = 3, y = 2.

Substitution method: Solve for x in the second equation to get:
x = 7 5y
Substitute for x in the first equation to get:

 5(–7 – 5y) – 3y = 21 –35 – 25y – 3y = 21 –35 – 28y = 21 Add 35 – 28y = 21 + 35 – 28y = 56 Divide by – 28

Substitute in equation two to get x. Substitute the value y = 2
in x = 7 5y to get x = 7 5(2) = 7 + 10 = 3
Thus x = 3, y = 2.

We can do the problem algebraically or geometrically.
Algebraically:
|x - 2| < 5 means 5 < x 2 < 5
So by adding to 2 all three parts of the inequality we get
5 < x 2 < 5
5 + 2 < x 2 + 2 < 5 + 2
3 < x < 7

Geometrically:
|x - 2| < 5 means that the distance (in both directions) from x to 2 is less than 5.
So if we move 5 units to the right (up the axis) from 2 we get 7 and 5 units
to the left (down the axis) from 2 we get 3

 The least common denominator is xy. Thus
Since the denominators of the two fractions are now the same,
we can add the numerators getting

22. Answer (x 3)(x +3)(x2 + 9)
Recall the factorization of the difference of squares a2 b2 = (a b)(a + b)
x4 81 is a difference of squares namely (x2 )2 and 92.
x4 81 = (x2 9)( x2 + 9). The first factor is again a difference of squares:
(x2 9) = (x 3)(x +3)
x4 81 = (x2 9)( x2 + 9) = (x 3)(x +3)(x2 + 9)
x2 + 9, the sum of squares, does not factor

The least common denominator is 24y.
 Thus
Here since the denominators are the same we add the numerator.

24. Answer The lines are parallel.
If you solve for y in each equation to get the form y = mx + b,
you can examine the slopes m and the y intercepts b.
 Equation 1 becomes: Equation 2 becomes:
which reduces to

Since the slopes are equal and the y intercepts are not, the lines are
parallel. In the case that both the slopes and the y intercepts were equal,
the lines would be the same.
In the case that the slopes are unequal, the lines intersect in one point.

 Think of as as

 is the cube root of 8 often written as

 So that

 the meaning of a negative exponent.

 is the square root of 4

 often written Thus

26. Answer x < 2  or  x > 3
x2 5x + 6 > 0
To find the solution set for x set the inequality equal to zero and factor and solve for x:
x2 5x + 6 = 0
(x 3)(x 2) = 0
x = 3, x = 2
Since the inequality is strictly greater than zero, neither of these are in the solution set.
These two numbers divide the x axis into three sections:

 ¥ < x < 2, 2 < x < 3, and 3 < x < ¥

Pick any "test" number for each section and substitute into the inequality:

For ¥ < x < 2, say x = 0.

Substituting x = 0 into x2 5x + 6 gives us (0)2 5(0) + 6 = 6
which is greater than zero.

Thus ¥ < x < 2 is part of the solution set.
 For 2 < x < 3, we can test with or

 Substituting into x2 5x + 6  gives us
which is less than zero.

Thus 2 < x < 3, is NOT in our solution set.
For 3 < x < ¥, we can substitute x = 4   to get
42 5(4) +6 = 16 20 + 6 = 12 > 0.
Thus 3 < x < ¥ is part of the solution set.
Thus the complete solution set is {x| ¥ < x < 2 or 3 < x < ¥}
which can also be written in interval notation as ( ¥, 2) (3, ¥).

If f(x) = x3 + 1, f(3) = 33 + 1 = 28

28. Answer The horizontal line one unit above the x axis.
29. Answer The vertical line 3 units to the left of the y axis.

30. Answer The graph of  2x + 3y = 2   is
To graph 2x + 3y = 2, we need to find any two points which lie on the line
and connect them with the straight edge. There are several ways to do this.
We outline two common methods.
Intercept method:

 If we let x = 0, the y intercept is 3y = 2 or
and if we let y =0 the x intercept is (1,0)

Slope-Intercept method:

 Solving for y in the form y = mx + b we get:
 From this form we see that the y intercept is
 From this point we can use the slope, which is
to find a second point by moving, 2 units to the left (0 2 = 2)

 and 3 units up to get the second point

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