Expert: Scott A Wilson - 6/30/2010

Question
I'm just curious, for beginning/intermediate algebra which areas do they  seem to have the most difficulty?

Answer
Equalities
  When x is 4, that is the same as saying x = 4.

Multiplying
  When you see 4x, that means 4*x.  So if x is 2, 4x is 8; if x=11, 4x=44.

One Vaiable
  The major thing to know is how to set up an equation for x and then to solve it.
  Given two constants A and B, most problems can be set us as Ax + B = 0.
  The solution to this is x = -B/A.

Lines
  Also, the equation of a line is y = mx + b where x is the variable, m = slope of the line,
  b = value where the line crosses the y axis, and y is the value of the line.
  That is in the standard from where y is solved in terms of x.

Story Problems
   From the questions that I remember:

 If the problem says, "the difference between X and Y is ...", that means X-Y.

 If the problem says, "the product of X and Y is ...", that means X*Y.

 If it says, "the product of the A minus 2 and B plus 4...". that means (A-2)*(B+4).

 If it says AB, that means A*B, so 5x+7y means 5*x + 7*y.

 If it says, "the quotient of A divided by B...", that means A/B.

 If (5-x) is subtracted from as expression, then add -5+x to that expression.
 Basically, subtracting a negative is the same as adding the positive.


Solving (which might seem hard right now, but gets easier with repetition):

 To solve an equation, seperate the values of x from the constants,
 then divide by what is in front of x.

 For example, if it was 10x+7 = 3(2x+5), first multiply out the right side.
 The 3(2x+5) goes to 3*2x + 3*5, which is 6x + 15.

 The equation is then 10x + 7 = 6x + 15.
 Subtracting 6x from both sides gives 10x-6x + 7 = 6x-6x+15.
 This goes to 4x + 7 = 15.

 Subtracting 7 from both sides gives 4x + 7 - 7 = 15 - 7, which reduces to 4x = 8.

 Once the x's are on one side and the constant is on the other,
 divide by what is in front of the x, giving 4x/4 = 8/4, which reduces to x = 2.


Quadratics (most likely in Algebra II)
   I'm not sure quadratic equations are in algebra 1, but if you want to know what they will teach you about quadratics, they are of the form ax² + bx + c = 0.  The values of a, b, and c are constants and x is the variable.  The value of x² means x*x and is also written x^2.

   The solution to this will be well known by the time it has been used a bunch in the course,
and the heads up are to say that it is x = (-b±√(b²-4ac))(2a).  Again, evaluate b², which is b*b, evaluate 4*a*c, which is 4 times the first constant times the last constant, take the difference, compute the squareroot, add this and subtract this from -b, and divide the two numbers by twice the first.

   As an example, solve 2x² + 10x + 12.  The quadratic says ... well, -b is -10, b² is 100,
4ac is 4*a*c = 4*2*12 = 4*24 = 96, so b²-4ac is 4, and 2a is 4.  We then would get (-10±√4)/4, which is (-10±2)/4, which is -12/4 and -8/4, which reduces to -3 and -2.  What this means is that if we have y = 2x² + 10x + 12, y = 0 at x = -2 and x = -3.

   If we put x = -2 in the equation, we get 2(-2)² + 10(-2) + 12 = 8 - 20 + 12 = 0 and
at x = -3, we get 2(-3)² +10(-3) + 12 = 18 - 30 + 12 = 0.