TECHNIQUES IN GETTING THE LEAST COMMON DENOMINATOR IN COLLEGE ALGEBRA

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*techniques in getting the least common denominator in college algebra*, here's the result:

REVIEW/TUTORIAL FOR THE COLLEGE ALGEBRA PLACEMENT TEST | ||||

1. | Evaluate x^{2} 3x + 9 for x = 1 | |||

2. |
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3. | Simplify and factor: 5x 2(y x) x | |||

4. |
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5. |
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6. |
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7. |
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8. | Multiply: (8x^{3} y)(3x^{2}y^{7}) | |||

9. | Find the slope of the line: 2x 3y + 10 = 0 | |||

10. |
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11. |
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12. |
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13. |
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14. | Find the solution set for x: 2x + 1 > 3x + 7 | |||

15. | Solve for x: 2x^{2} + 3x 2 = 0 | |||

16. | Solve for x: 2x^{2} 3x 4 = 0 | |||

17. | Solve for x: 2x^{2} 6x + 5 = 0 | |||

18. |
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19. |
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20. |
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21. |
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22. | Factor completely: x^{4 } 81. | |||

23. |
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24. | The graphs of the system of equation consists of lines that | |||

a) intersect b) meet in one point c) are parallel | ||||

25. |
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26. | Find the solution set for x: x^{2} + 6 > 5x | |||

27. | If f(x) = x^{3} + 1, find f(3) | |||

28. | Graph: y = 1 | |||

29. | Graph: x = 3 | |||

30. | Graph: 2x + 3y = 2 |

ANSWERS FOR THE REVIEW/TUTORIAL | ||||||||||||||||

1. | Answer | 5 | ||||||||||||||

Evaluate x^{2} 3x + 9 for x = 1 | ||||||||||||||||

When substituting x = 1 in x^{2} | ||||||||||||||||

be sure to do the exponent before the multiplication by 1 | ||||||||||||||||

to get (1)^{2 = } 1. So 1 3 + 9 = 5 | ||||||||||||||||

2. | Answer | |||||||||||||||

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3. | Answer | 2(3x - y) | ||||||||||||||

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) | ||||||||||||||||

4. | Answer | |||||||||||||||

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5. | Answer | |||||||||||||||

Write the numerator as a single fraction with denominator of 4, | ||||||||||||||||

and write the denominator as a single fraction with denominator of 6, | ||||||||||||||||

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Be sure to invert and multiply to get | ||||||||||||||||

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

6. | Answer | |||||||||||||||

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7. | Answer | |||||||||||||||

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8. | Answer | 24x^{5}y^{8} | ||||||||||||||

When multiplying exponent expressions with the same base, keep the | ||||||||||||||||

base and add the exponents thus, x^{3}(x^{2}) = x^{5} and y(y^{7}) = y^{8} | ||||||||||||||||

9. | Answer | |||||||||||||||

Write the equation of the line in the form y = mx + b or where m is the slope. | ||||||||||||||||

3y = 2x 10 | ||||||||||||||||

10. | Answer | |||||||||||||||

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11. | Answer | |||||||||||||||

Multiply both sides of the equation by the denominator x 3 | ||||||||||||||||

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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 | ||||||||||||||||

13. | Answer | |||||||||||||||

We factor x^{2} 3x = x(x 3) and 9 x^{2} = (3 x)(3 + x) | ||||||||||||||||

Remember the difference of squares factorization a^{2} b^{2} = (a b)(a + b). | ||||||||||||||||

Thus we get | ||||||||||||||||

Use the fact that x – 3 = – (3 – x), so that | ||||||||||||||||

Putting this together we get | ||||||||||||||||

14. | Answer | x < 6 | ||||||||||||||

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. | ||||||||||||||||

15. | Answer |
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Consider the standard quadratic equation ax^{2} + bx + c = 0, | ||||||||||||||||

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For the solution of 2x^{2} + 3x 2 = 0, a = 2, b = 3, and c = 2. | ||||||||||||||||

Since the discriminant b^{2} 4ac = 3^{2} 4(2)( 2) = 9 + 16 = 25 is a | ||||||||||||||||

perfect square, we can factor directly: 2x^{2} + 3x 2 = 0, | ||||||||||||||||

(2x 1)(x + 2) = 0 | ||||||||||||||||

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16. | Answer | |||||||||||||||

For the solution of 2x^{2} 3x 4 = 0, we have a = 2, b = 3, c = 4. | ||||||||||||||||

The discriminant b^{2} 4ac = ( 3)^{2 } 4(2)( 4) = 9 + 32 = 41 is not a | ||||||||||||||||

perfect square so that we must use the quadratic formula and get | ||||||||||||||||

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17. | Answer | |||||||||||||||

For the solution of 2x^{2} 6x + 5 = 0, a = 2, b = 6, c = 5 | ||||||||||||||||

The discriminant b^{2} 4ac = ( 6)^{2 } 4(2)( 5) = 36 40 = 4. | ||||||||||||||||

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

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We can reduce the answer by factoring 2 in the numerator | ||||||||||||||||

18. | Answer | |||||||||||||||

The negative exponent means we have to take the reciprocal of what is in | ||||||||||||||||

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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 | ||||||||||||||||

Add the equations to get: | ||||||||||||||||

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: | ||||||||||||||||

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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. | ||||||||||||||||

20. | Answer | Answer: 3 < x < 7 | ||||||||||||||

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 | ||||||||||||||||

21. | Answer | |||||||||||||||

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Since the denominators of the two fractions are now the same, | ||||||||||||||||

we can add the numerators getting | ||||||||||||||||

22. | Answer | (x 3)(x +3)(x^{2 }+ 9) | ||||||||||||||

Recall the factorization of the difference of squares a^{2} b^{2} = (a b)(a + b) | ||||||||||||||||

x^{4} 81 is a difference of squares namely (x^{2} )^{2} and 9^{2}. | ||||||||||||||||

x^{4} 81 = (x^{2} 9)( x^{2} + 9). The first factor is again a difference of squares: | ||||||||||||||||

(x^{2} 9) = (x 3)(x +3) | ||||||||||||||||

x^{4} 81 = (x^{2} 9)( x^{2} + 9) = (x 3)(x +3)(x^{2 }+ 9) | ||||||||||||||||

x^{2 }+ 9, the sum of squares, does not factor | ||||||||||||||||

23. | Answer | |||||||||||||||

The least common denominator is 24y. | ||||||||||||||||

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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. | ||||||||||||||||

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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. | ||||||||||||||||

25. | Answer | 2 | ||||||||||||||

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26. | Answer | x < 2 or x > 3 | ||||||||||||||

x^{2} 5x + 6 > 0 | ||||||||||||||||

To find the solution set for x set the inequality equal to zero and factor and solve for x: | ||||||||||||||||

x^{2} 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: | ||||||||||||||||

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

For ¥ < x < 2, say x = 0. | ||||||||||||||||

Substituting x = 0 into x^{2} 5x + 6 gives us (0)^{2} 5(0) + 6 = 6 | ||||||||||||||||

which is greater than zero. | ||||||||||||||||

Thus ¥ < x < 2 is part of the solution set. | ||||||||||||||||

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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 | ||||||||||||||||

4^{2} 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, ¥). | ||||||||||||||||

27. | Answer | 28 | ||||||||||||||

If f(x) = x^{3 }+ 1, f(3) = 3^{3 }+ 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: | ||||||||||||||||

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and if we let y =0 the x intercept is (1,0) | ||||||||||||||||

Slope-Intercept method: | ||||||||||||||||

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to find a second point by moving, 2 units to the left (0 2 = 2) | ||||||||||||||||

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