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TECHNIQUES IN GETTING THE LEAST COMMON DENOMINATOR IN COLLEGE ALGEBRA
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You landed on this page because you entered a search term similar to this: 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. 
 
3.  Simplify and factor: 5x 2(y x) x  
4. 
 
5. 
 
6. 
 
7. 
 
8.  Multiply: (8x^{3} y)(3x^{2}y^{7})  
9.  Find the slope of the line: 2x 3y + 10 = 0  
10. 
 
11. 
 
12. 
 
13. 
 
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. 
 
19. 
 
20. 
 
21. 
 
22.  Factor completely: x^{4 } 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. 
 
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  
 
 
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  
 
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,  
 
Be sure to invert and multiply to get  
Here we divided the denominator 4 and the numerator 6 by 2.  
6.  Answer  
,  
 
7.  Answer  
 
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  
 
11.  Answer  
Multiply both sides of the equation by the denominator x 3  
 
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 
 
Consider the standard quadratic equation ax^{2} + bx + c = 0,  
 
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  
 
 
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  
 
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.  
 
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  
 
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:  
 
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  
 
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.  
 
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.  
 
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  
 
 
 
 
 
 
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:  
 
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.  
 
 
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:  
 
and if we let y =0 the x intercept is (1,0)  
SlopeIntercept method:  
 
 
 
to find a second point by moving, 2 units to the left (0 2 = 2)  
 
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