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Help so confused in my math class.

  1. Sep 24, 2008 #1
    It's so annoying and boring, I'm doing algebra and triginometry for college students.

    I feel like I did all this already in high school but it was a coupla years ago and I've forgotten it and this teacher is just so boring, but I try to adjust the best I can. That's why I just needed some help. She doesn't really explain anything and she goes so quickly...

    Right now we're on like inequalities and factoring binomials. Atleast they're better than some of the stuff we were doing.

    Here's something I didn't know what the hell the professor was talking about if you wanna help me out on it and get me up to date on it:

    Actually well these are all the topics I've done so far:

    Integral Exponents & Scientific Notation
    Functions, Domain, Functional Notation, Vertical Line Test
    Graphing Linear Functions, Slope, Equations of Lines, Introduction to Graphing Calculator
    Solving Systems of Equations: Graphically and Algebraically
    Linear & Compound Inequalities & Absolute Value Equations
    Graphing Linear Inequalities(by hand and with a calculator)

    I was confused on the solving systems of equations and absolute value equations the most, how do you do them? Here are some examples if anyone wants to help me out:

    |4y + 3| = |4y + 5| How would I find the solution set of that?

    |4x - 9| = |2x + 1|

    |2x/3 - 2| = |x/3 + 3| How would I solve that?

    And lastly: Graph each inequality:

    y < -1/3x

    Graph the solution set of each system of inequalities:

    x - y <= 4
    y >= 2x - 4

    For that one I would change the sign of the first one to an equal sign and draw a solid line through 4 on both axis right? But how would I do the 2nd one?
  2. jcsd
  3. Sep 24, 2008 #2


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    marik1234, this is still early in the semester, so the newer material (assuming you're a PreCalculus student) has not come in the progression yet.

    The absolute value equations need to be handled with relatively simple logic. You have usually two conditions for each expression in the absolute value function. The expression may be positive, or it may be negative, or it may be zero (actually that was three conditions, not two, but a long time has passed since I last worked with these).
    The absolute value itself is never negative, but you need to still account for the separate conditions. Drawing a number line and graphing is always helpful in solving absolute value equations.
  4. Sep 24, 2008 #3


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    Hi marik1234. I'm only in year 11 but these questions you've offered are quite simple for our class, so I will do my best to explain them to you.
    [tex]|4y + 3| = |4y + 5|[/tex]

    Firstly, if you have an absolute value, say:
    [tex]|x+1| = y[/tex]

    Then this means that [tex]x+1 = \pm y[/tex]

    Now if you have an absolute value on both ends of the equals sign such as your question it can be deduced as:

    [tex]\pm(4y+3) = \pm(4y+5)[/tex]

    However, out of the 4 possible outcomes, 2 would give the same results. For e.g. if you take the positive of both ends and multiple by -1, you end up with the negative of both ends so these are equal.

    Therefore, all you need is to find: [tex]4y+3 = \pm(4y+5)[/tex]

    For the second and third question, it is the same idea. Just solve:
    [tex]4x-9 = \pm(2x+1)[/tex] and [tex]2\frac{x}{3}-2 = \pm(\frac{x}{3}+3)[/tex] respectively.

    As for the graphing, for equations in the form y=mx+b (where m=gradient, b=y-intercept) they are straight lines.
    The inequality: [tex]y<\frac{-1}{3}x[/tex] just treat it as an equality, graph it and then find the section of the inequality later. The [tex]\frac{-1}{3}[/tex] is the gradient and b=0 therefore it passes through the origin (0,0). Using this and the fact that the inequality is not equal to (which mean use a dotted line in graphing), all you need is to use a point on either end of the graphed line, and test it by substituting the point into the inequality. If the inequality is satisfied, all of the graph on that end of the line is shaded in, if it doesn't satisfy then shade in the other end of the line.

    The last 2 inequalities are of the same process, just graph the line and then test a point on either end of the line to see if it is satisfied by the inequality.

    [tex]y\geq2x-4[/tex] this is in the form of [tex]y=mx+b[/tex] so it is quite obvious that the gradient (m) = 2 and the y-intercept (b) = -4

    If you are unsure how to graph the lines, a good way to find how it is drawn is by substituting x=0 into the function (this will give you the y-intercept) because the entire y-axis is where x=0. Then sub y=0 into the function to find the x-intercept and simply join these two points.
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