Solving Problems Algebraically and Graphically

Key Questions

  • Answer:

    See explanation


    In algebra you have letters to indicate the existence of an unknown value.

    These letters may be combined in various configurations with known values.

    You have two main groups of configuration
    1: Expression. This is a collection of numbers and letters without an
    equals sign. This is just a statement and is not what is termed

    2.Equation. This has an equal sign that literally means the intrinsic
    value of everything on one side of the equals is 'EXACTLY' the
    same intrinsic value as on the other side.

    The thing is: something can look quite different but have the same value. For example, take the value three ( 3 ). This is 'set in stone'. We all know what it means. However ( 5-2 ) does not look the same but its intrinsic value is.

    So we have some form of configuration on one side of an equals sign and another form of configuration on the other.

    To obtain a solution we move things round in a way that is mathematically correct until we have just #color(red)("one")# of what we wish to find the value of on one side of the equals and everything else on the other.

    Example: Given that #3=x+6# find the value of #x#

    To have #x# on its own we need to remove the 6. The problem is that all on the left #color(red)("must")#be equal in value to all on the right. So if we remove 6 from the right we #color(red)("must")# also remove it from the left. Remove is 'subtract' so we have:

    Subtract 6 from both sides giving:

    #(3)-6=(x+6) -6# Note that I have put brackets round the original parts to show what is going on. They serve no other purpose in this case.

    From this action we now end up with:

    #-3 = x#

    Convention is that what we are looking for is on the left so we write:

    #x= -3#

    There are a large number of correct mathematical manipulations. Knowing them is down to experience.

    If this helps please let me know.

  • Answer:

    From the graph you should be able to determine some #(x,y)# coordinates. Use these point to solve the general equation form for the constants.


    The number of point you need will depend upon the shape of the graph: 2 sets for a linear graph; 3 sets for a parabola and so on.


    1. Linear Graph
    graph{(y-(5x-3))(x^2+(y+3)^2-0.01)((x-1)^2+(y-2)^2-0.01)=0 [-5.15, 7.337, -4.03, 2.213]}
    2 possible points that could be read from this graph:
    #color(white)("XXX")(x,y) in { (0,-3), (1,2)}#

    General form of a linear equation (one version):

    Using the points we read from the graph for #(x,y)#
    Two equations in two variables can be easily solved as :
    #color(white)("XXX")b=-3 and a=5#
    So the equation can be written as

    2. Parabolic Graph
    graph{(y-(3*x^2 -2x +4))((x-1)^2+(y-5)^2-0.01)((x+1)^2+(y-9)^2-0.01)(x^2+(y-4)^2-0.01)=0 [-4.23, 6.87, 3.568, 9.115]}
    3 possible points that could be read from this graph:
    #color(white)("XXX")(x,y) in {(-1,9), (0,4), (1,5)}#

    Using the general form for a parabola of:

    and the points we read from the graph:
    Again (without the details) we have 3 equations in 3 variables that can be solved as
    #color(white)("XXX")a=3, b=(-2), c=4#
    So the equation can be written as

    More complex polynomials will require more points but the process is identical.

  • Answer:

    See explanation.


    To solve the problrm graphically means to plot function graphs in the single coordinate plane and look for points where the graphs meet.


    To graphically solve a system:


    you have to draw two graphs on a single coordinate plane:

    graph{(x+y-7)(x-y-5)=0 [-0.875, 19.125, -4.88, 5.12]}

    The graphs cross at the point #(6,1)# this means that the system has a solution: