# [SOLVED] Help! Many algebra problems :(

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• Nov 27th 2006, 12:20 PM
Elite00
[SOLVED] Help! Many algebra problems :(
Hey, I'm a high school student and recently came back to school after a long period of absence (appendicitis :() and my teacher gave me a 2-page packet with a bunch of problems from the chapters I missed, and he gave me all the notes to learn how to do them on my own. I am struggling very much and these problems are due tomorrow! I would greatly appreciate if anyone could help me out, even if you just solve 1 or 2.

1. Find the slope of the line between the points (-10, -4) and (-3, -3)

2. Write the equation in slope-intercept form for the line that contains the point (-2,7) and is parallel to y = 3x – 4

3. Write an equation in slope-intercept form for the line that contains the point (8,-1) and is perpendicular to the graph y=4x-3

4. Solve for x in the proportion 6x-3/9 = 3x/8

5. Use the table provided to (a) Write the linear regression equation and (b) tell how many students were enrolled in AP statistics for the 4th year.
Year 1 2 3 5 6 7
Enrollment 33 30 43 47 58 65

6. Solve for x in the following equation 2/5x + 6/5 = x-3

7. Solve and graph the solution(s) for the following inequalities 2x – 3 > 1 or 3x+7 < 1

8. Solve and graph the following |3x+2| > 4

9. Simplify the expressions completely by leaving only positive exponents:
(3x^3 y^2 / y^-1)^3 (x^-2 / 2y^-1) ^-3(3y/x^-4)^-2

10. Tell whether or not the following equation represents a function. Then give the domain, range, and zeros. f(x) = 9 – x^2

11. Let f(x) = 4x^2 and g(x) = 2x-3. Find each of the following:

a. (f – g)(x) b. (f x g)(x) c. g ( f (x) )

12. Find the inverse of the following function. Then use the composition of functions to prove that your equations are inverses of each other. f ( x ) = x-1 / 3

13. Graph the following piecewise function and give the domain, range, and zeros.
f (x ) = {2x + 3, x<4
{x -1, 4 < x < 9

14. Solve the following systems of equations:
a. {s = 9 + 5t b. {3x + 2y =12
{2s = 30 – 2t {5x-3y = -18

15. A lab technician is trying to make an 8% solution by combining a 2% and 12% solution. If he is trying to make a 150mL solution, how many mL of each solution should be mixed?

16. Graph, shade the feasible region and label the corner points for the following system of inequalities: {y < -2x -2
{y > 1/2x
{x > -3

17. Write the pair of parametric equations as a single equation of and y
{x ( t ) = 3t -1
{y ( t ) = t + 4

18. Let A = [ 8 4 ] and b= [6 -2] Show each of the following: A – B and BA
[4 1 ] [0 4]

19. Let C = [2 4] Find the inverse of C.
[1 0]

20. Solve the following system of equations: { -3x + 6y – 4z = 8
{x + 2z – 4y = -3
{8y=z

21. Solve 4(x+5)^2 = 20. express the answer exactly and if necessary, rounded to the nearest tenth.

22. Factor 2x^2 – 10x – 48 =0 and solve for x

23. Find the exact values of x in the following quadratic equation. 2x^2 – 5x -12 = 0

24. Simplify the following complex number: 3 + i / 2 – 3i

25.Solve the Inequality 4x – 1 > 8 – x^2 and graph it on a number line.
• Nov 27th 2006, 12:25 PM
topsquark
Quote:

Originally Posted by Elite00
1. Find the slope of the line between the points (-10, -4) and (-3, -3)

The slope between two distinct points $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2, y_2)$ is defined as:
$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$.

So
$\displaystyle m = \frac{-3 - -4}{-3 - -10} = \frac{-3 + 4}{-3 + 10} = \frac{1}{7}$

-Dan
• Nov 27th 2006, 12:28 PM
topsquark
Quote:

Originally Posted by Elite00
2. Write the equation in slope-intercept form for the line that contains the point (-2,7) and is parallel to y = 3x – 4

The slope-intercept form of a line is $\displaystyle y = mx + b$ where m is the slope and the point (0, b) is the y-intercept: the point where the line crosses the y-axis.

So the slope of the line parallel to $\displaystyle y = 3x - 4$ is m = 3. (Parallel lines have the same slope.) We know this line passes through the point (-2, 7), so applying the slope-intercept form for the line:
$\displaystyle 7 = 3 \cdot -2 + b$

$\displaystyle 7 = -6 + b$

$\displaystyle b = 13$

So the line will be $\displaystyle y = 3x + 13$.

-Dan
• Nov 27th 2006, 12:31 PM
topsquark
Quote:

Originally Posted by Elite00
3. Write an equation in slope-intercept form for the line that contains the point (8,-1) and is perpendicular to the graph y=4x-3

This is done in the same way as problem 2, except that the relationship between the slopes of perpendicular lines is
$\displaystyle m_2 = \frac{-1}{m_1}$

So the slope of the line you are interested in is $\displaystyle m = -\frac{1}{4}$.

-Dan
• Nov 27th 2006, 12:36 PM
Quick
Quote:

Originally Posted by topsquark
This is done in the same way as problem 2, except that the relationship between the slopes of parallel lines is
$\displaystyle m_2 = \frac{-1}{m_1}$

So the slope of the line you are interested in is $\displaystyle m = -\frac{1}{4}$.

-Dan

just to clarify, he means perpendicular ;)
• Nov 27th 2006, 01:12 PM
topsquark
Quote:

Originally Posted by Quick
just to clarify, he means perpendicular ;)

I've fixed it in the original post. Thanks for the spot, Quick. :)

-Dan
• Nov 27th 2006, 01:17 PM
topsquark
Quote:

Originally Posted by Elite00
4. Solve for x in the proportion 6x-3/9 = 3x/8

I presume this is
$\displaystyle \frac{6x - 3}{9} = \frac{3x}{8}$

$\displaystyle \frac{3(2x - 1)}{9} = \frac{3x}{8}$

$\displaystyle \frac{2x - 1}{3} = \frac{3x}{8}$

Now, I'm going to "clear the fractions." The least common multiple of 3 and 8 is 24:

$\displaystyle 24 \cdot \frac{2x - 1}{3} = 24 \cdot \frac{3x}{8}$

$\displaystyle 8 (2x - 1) = 3 (3x)$

$\displaystyle 16x - 8 = 9x$

$\displaystyle 16x - 9x = 8$

$\displaystyle 7x = 8$

$\displaystyle x = \frac{8}{7}$

-Dan
• Nov 27th 2006, 01:25 PM
topsquark
Quote:

Originally Posted by Elite00
6. Solve for x in the following equation 2/5x + 6/5 = x-3

I presume this is:
$\displaystyle \frac{2}{5x} + \frac{6}{5} = x - 3$

I'm going to add the fractions: The least common multiple of 5x and 5 is 5x.

So
$\displaystyle \frac{2}{5x} + \frac{6}{5} \cdot \frac{x}{x} = x - 3$

$\displaystyle \frac{2 + 6x}{5x} = x - 3$

$\displaystyle 5x \cdot \frac{2 + 6x}{5x} = 5x (x - 3)$

$\displaystyle 2 + 6x = 5x^2 - 15x$

$\displaystyle 5x^2 - 21x - 2 = 0$

As this doesn't factor, use the quadratic formula:
If $\displaystyle ax^2 + bx + c = 0$ then $\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

So:
$\displaystyle x = \frac{21 \pm \sqrt{21^2 - 4 \cdot 5 \cdot (-2)}}{2 \cdot 5}$

$\displaystyle x = \frac{21 \pm \sqrt{481}}{10}$

-Dan
• Nov 27th 2006, 04:58 PM
topsquark
Quote:

Originally Posted by Elite00
7. Solve and graph the solution(s) for the following inequalities 2x – 3 > 1 or 3x+7 < 1

First:
$\displaystyle 2x - 3 > 1$

$\displaystyle 2x > 4$

$\displaystyle x > 2$

The solution set here is $\displaystyle (2, \infty)$.

Second:
$\displaystyle 3x+7 < 1$

$\displaystyle 3x < -6$

$\displaystyle x < -2$

The solution set is $\displaystyle (-\infty, -2)$

-Dan
• Nov 27th 2006, 05:03 PM
topsquark
Quote:

Originally Posted by Elite00
8. Solve and graph the following |3x+2| > 4

This is actually two inequalities:
$\displaystyle 3x + 2 > 4$
and
$\displaystyle 3x + 2 < -4$

The first one gives:
$\displaystyle 3x + 2 > 4$

$\displaystyle 3x > 2$

$\displaystyle x > \frac{2}{3}$

and the second one gives:
$\displaystyle 3x + 2 < -4$

$\displaystyle 3x < -6$

$\displaystyle x < -2$

Combining both of these gives a solution set of:
$\displaystyle (-\infty, -2) \cup (\frac{2}{3}, \infty)$

-Dan
• Nov 27th 2006, 05:23 PM
topsquark
Quote:

Originally Posted by Elite00
9. Simplify the expressions completely by leaving only positive exponents:
(3x^3 y^2 / y^-1)^3 (x^-2 / 2y^-1) ^-3(3y/x^-4)^-2

$\displaystyle \left ( \frac{3x^3y^2}{y^{-1}} \right )^3 \left ( \frac{x^{-2}}{2y^{-1}} \right )^{-3} \left ( \frac{3y}{x^{-4}} \right )^{-2}$

The rules of exponents are as follows: (Let's keep it simple and assume for now that a, b, x, y are positive integers.)
$\displaystyle a^x a^y = a^{x+y}$
$\displaystyle \frac{a^x}{a^y} = a^{x-y}$
$\displaystyle (ab)^x = a^x b^x$
$\displaystyle a^0 = 1$
$\displaystyle a^{-x} = \frac{1}{a^x}$
$\displaystyle (a^x)^y = a^{xy}$

The first thing to do is get rid of those pesky negative exponents:
$\displaystyle (3x^3y^2y)^3 \left ( \frac{y}{2x^2} \right )^{-3} (3x^4y)^{-2}$

$\displaystyle (3x^3y^3)^3 \left ( \frac{2x^2}{y} \right )^{3} \left ( \frac{1}{3x^4y} \right )^{2}$

$\displaystyle (3^3x^{3 \cdot 3}y^{3 \cdot 3}) \left ( \frac{2^3x^{2 \cdot 3}}{y^3} \right ) \left ( \frac{1}{3^2x^{4 \cdot 2}y^2} \right )$

$\displaystyle (27x^9y^9) \left ( \frac{8x^6}{y^3} \right ) \left ( \frac{1}{9x^8y^2} \right )$

$\displaystyle \frac{(27x^9y^9)(8x^6)}{(y^3)(9x^8y^2)}$

$\displaystyle \frac{27 \cdot 8}{9} \cdot \frac{x^9 \cdot x^6}{x^8} \cdot \frac{y^9}{y^3 \cdot y^2}$

$\displaystyle (3 \cdot 8) \cdot \frac{x^{15}}{x^8} \cdot \frac{y^9}{y^5}$

$\displaystyle 24x^7y^4$

-Dan
• Nov 27th 2006, 05:32 PM
topsquark
Quote:

Originally Posted by Elite00
10. Tell whether or not the following equation represents a function. Then give the domain, range, and zeros. f(x) = 9 – x^2

A function is a relation (a set of ordered pairs (x,f(x)) ) such that there is only one f(x) for each x.

A simple way to test if you have a function is to graph f(x). If you can draw ANY vertical line through the graph that cuts the graph in more than one place then f(x) is not a function.

The graph of f(x) = 9 - x^2 shows that it is indeed a function. (See attachment below.)

There are no values of x such that f(x) is not defined. (Typically such a value of x would make the denominator of the function 0. We don't have any fractions here.) So the domain of the function is all real numbers: $\displaystyle (-\infty, \infty)$.

The range is a bit trickier. Note that the function has a maximum value: f(max) = 9. There is no lower bound to the function, so the range of the function is $\displaystyle (-\infty, 9]$.

Zeros of a function are values of x such that f(x) = 0. So we need to solve
$\displaystyle f(x) = 0 = 9 - x^2$

$\displaystyle x^2 = 9$

$\displaystyle x = \pm 3$

So the zeros of the function are x = -3 and x = +3.

-Dan
• Nov 27th 2006, 05:41 PM
topsquark
Quote:

Originally Posted by Elite00
11. Let f(x) = 4x^2 and g(x) = 2x-3. Find each of the following:

a. (f – g)(x) b. (f x g)(x) c. g ( f (x) )

$\displaystyle f(x) = 4x^2$
$\displaystyle g(x) = 2x - 3$

a) $\displaystyle (f - g)(x) = f(x) - g(x) = (4x^2) - (2x - 3) = 4x^2 - 2x + 3$

b) $\displaystyle (f \times g)(x) = f(x) \times g(x) = (4x^2) \cdot (2x - 3) = 8x^3 - 12x^2$

c) $\displaystyle g(f(x)) = g(4x^2) = 2(4x^2) - 3 = 8x^2 - 3$

-Dan
• Nov 27th 2006, 05:49 PM
topsquark
Quote:

Originally Posted by Elite00
12. Find the inverse of the following function. Then use the composition of functions to prove that your equations are inverses of each other. f ( x ) = x-1 / 3

I'm assuming this is: $\displaystyle f(x) = \frac{x - 1}{3}$

A function and its inverse have a kind of symmetry: calling y = f(x), the graph of f(x) and it's inverse are reflections over the line y = x. This means that, in order to find the inverse of the function y = f(x) what we need to do is switch the x and y in the equation: x = f(y). Once we solve this for y we have our inverse function.

So:
$\displaystyle y = \frac{x - 1}{3}$

Switch x and y:
$\displaystyle x = \frac{y - 1}{3}$

Now solve for y:
$\displaystyle x = \frac{y - 1}{3}$

$\displaystyle 3x = y - 1$

$\displaystyle y = 3x + 1$

So the inverse function is $\displaystyle g(x) = 3x + 1$. (Typically we use the notation $\displaystyle f^{-1}(x)$ for the inverse function.)

Now we need to show that these functions are inverses of each other. If this is true then we should be able to show that
$\displaystyle f(g(x)) = x$
and
$\displaystyle g(f(x)) = x$

I'll do the first one.
$\displaystyle f(g(x)) = f(3x + 1) = \frac{(3x + 1) - 1}{3} = \frac{3x}{3} = x$ (Check!)

-Dan
• Nov 27th 2006, 05:58 PM
topsquark
Quote:

Originally Posted by Elite00
14. Solve the following systems of equations:
a. {s = 9 + 5t b. {3x + 2y =12
{2s = 30 – 2t {5x-3y = -18

a)
$\displaystyle s = 9 + 5t$
$\displaystyle 2s = 30 - 2t$

There are several ways to solve systems of equations. This one is called the "substitution method." It's occasionally more trouble that the other methods, but it has the advantage of being one of the more systematic methods.

What we want to do is solve one of the equations for one of the unknown variables. We then insert this value into the second equation. This provides us with an equation in a single unknown. We can solve this and use this value in either of the original equations to find the value of the other variable.

For example:
$\displaystyle s = 9 + 5t$
$\displaystyle 2s = 30 - 2t$

As it happens, the top equation is already solved for s. I will use this value of s in the second equation:
$\displaystyle 2(9 + 5t) = 30 - 2t$

This is an equation in a single unknown, t, which we can solve:
$\displaystyle 18 + 10t = 30 - 2t$

$\displaystyle 12t = 12$

$\displaystyle t = 1$

We now use this value of t in either of the original equations to get a value for s:
$\displaystyle s = 9 + 5 \cdot 1 = 14$

So the solution to the set of equations is (s, t) = (14, 1).

You can do the same for problem b). I get that (x, y) = (0, 6).

-Dan
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