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Systems of Linear Equations

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					 Systems of Linear
     Equations
 How to: solve by graphing, substitution, linear
combinations, and special types of linear systems
                 By: Sarah R.
              Algebra 1; E block
     What is a Linear System,
            Anyways?
• A linear system includes two, or more,
  equations, and each includes two or more
  variables.
• When two equations are used to model a
  problem, it is called a linear system.
  Before You Begin…Important
         Terms to know
• Linear system: two equations that form one
  equation
• Solution: the answer to a system of linear
  equation; must satisfy both equations
      ***: a solution is written as an ordered pair:
  (x,y)
• Leading Coefficient: any given number that is
  before any given variable (for example, the
  leading coefficient in 3x is 3.)
• Isolate: to get alone
      Solving Linear Systems by
             Substitution
• Basic steps:
• 1. Solve one equation for    See next
  one of its variables
• 2. Substitute that            page for a
  expression into the other
  equation and solve for the
  other variable
                                step by step
• 3. Substitute that value
  into first equation; solve
                                example!
• 4. Check the solution
     Example: The Substitution
             Method
• Here’s the problem:
     Equation one       • Try this on your
     -x+y=1               own…but if you need
     Equation two         help or a few
     2x+y=-2              pointers…see the next
                          page!
     First, solve equation one for y
     Y=x+1
     Next, substitute the above expression in for “y” in equation two, and solve for x
     Here’s how:
                Equation two
                2x+y=-2
                Substitute “x+1” for y
                2x+ (x+1)=-2
                simplify the above expression
                3x+1=-2
                Subtract one from both sides (because your goal is to solve for x)
                3x=-3
                Solve for x ( divide both sides by 3; since x is being multiplied by three, and you need it alone,
so do the inverse operation: divide by 3)
                X=-1
     Congratulations! You now know x has a value of –1…but you still need to find “y”.
     To do so…
         First, write down equation one
                Y=x+1
                Substitute –1 for x, since you just found that x=-1
                Y= (-1)+1
                Solve the equation for y by adding –1 +1
                Y=0
     So, now what?
                You’re done; simply write out the solution as (-1,0)
     ***Did you remember??
            To write a solution, once you’ve found x and y, you must put x first and then y: (x,y)
  Things to Know About the
     Substitution Method
• 1. It doesn’t matter if you choose to solve for y or
  x first; the answer or solution will be the same
  either way.
• 2. You can also choose to solve equation two
  before equation one; simply follow the same steps,
  just using a slightly rearranged order.
• *** You should always decide whether to solve x
  or y first, or equation one or two first, depending
  on which way is more efficient (See next page!)
   Deciding the Order in Which to Solve
• Here is an instance where it is easier to solve equation two
  first (for x)
        Equation One: 3x-2y=1
        Equation Two: x+4y=3
        By solving equation two first, you are lessening your
  work, because there is no leading coefficient before the x
  in equation two, so you don’t have to worry about dividing
  to isolate the “x”
• Here is an instance where you help yourself by solving
  equation one for y
        Equation One: 2x+y=5
        Equation Two: 3x-2y=11
        You should solve for y in the first equation. Again,
  you lessen your work because there is no leading
  coefficient before the y in equation one, while there are
  leading coefficients with all the other variables.
Solving Linear Systems by Linear
         Combinations
    Solving Systems by means of
        Linear Combinations
• Basic steps:
       1. Arrange the equations with like terms in columns
       2. After looking at the coefficients of x and y, you
  need to multiply one or both equations by a number that
  will give you new coefficients for x or y that are opposites.
       3. Add the equations and solve for the unknown
  variable
       4. Substitute the value gotten in step 3 into either of
  the original equations; solve for other variable
       5. Check the solution in both original equations
  Example: Solving Systems by
     Linear Combinations
• Here’s an example…try it out, but if you
  have any problems, see the next page for a
  guided, step by step explanation
• Solve this linear equation:
     Equation One: 3x+5y=6
     Equation Two: -4x+2y=5
Here’s the original problem:
Solve the linear system
Equation 1: 3x+5y=6
Equation 2: -4x+2y=5
Do you remember the first step?
…put the equations into columns
3x+5y=6
-4x+2y=5
Now, you need to multiply each equation by a number that will cause your leading
coefficients of either x or y to become opposites. In this case, try to get opposite
coefficients for x. to do this, multiply the first equation by four and the second by three.
***You must multiply all terms by 3 or 4
3x+5y=6, when all terms are multiplied by four, this equation will be: 12x+20y=24
-4x+2y=5, when all terms are multiplied by three, this equation will be: -12x +6y=15
Your next step is to add the two revised equations:
     12x+ 20y=24
+ (-12x) + 6y= 15
           26y=39 (sum of equations)
To get the “y” alone, you must divide each side by 26, (you divide since the y is being
multiplied by 26, and to isolate the y you do the inverse operation)
So, you have found “Y”, but you aren't done yet!
What’s left, you may be thinking…well, you have only found “y”…what
about x?
To find x, you have to place “y” into equation 2.
Equation 2: -4x+2y=5
Substitute the value you just found for “y” : 3
                                              2
-4x+2(3)=5
       2
simplify by multiplying 2 by three-halves
-4x+3=5
subtract 3 from both sides because you are working to isolate x
-4x=2
solve for x by dividing both sides by –4 (inverse operation)
x=-1
   2
The solution to the example system is (-1, 3)
                                         2 2
A Final way to Solve Systems:
      Graph and Check
Here’s a method called graph and
             check
• Basic steps:
      1. Put each equation into slope intercept
  form (y=Mx+B)
      2. Graph the two lines (M is your slope;
  B is your Y-intercept)
      3. Find the point that the lines appear to
  intersect at, and then put that solution into
  EACH equation and solve to check for
  accuracy.
  An Example of the Graph and
        Check Method
• Here’s the problem:   • Try this problem
     Equation one         out…but a step by step
     x+y=(-2)             process follows!
     Equation two
     2x-3y=(-9)
The first step is to put the equations into slope-intercept form
Equation one: originally, it was: x+y=(-2) but after putting it into slope intercept, it
reads: y=(-x)-2
Equation two: originally, it was: 2x-3y=(-9), but once in slope intercept, it reads:
                                       y= 2x+3
                                           3
From the above equations, you can make the following conclusions:
Equation one has a slope of –1 and a y intercept of –2
Equation two has a slope of 2 and a y intercept of 3
                               3
***Remember that in the slope intercept form (y=mx+b), m is the slope; b is the y
intercept
now, you will be able to graph the two equations as lines. Once done this, you can
conclude that the lines seem to intercept at (-3,1).
To check this assumption, put (-3) in for x and 1 in for y in BOTH EQUATIONS, and
solve both:
Equation one: (-3)+(-1)=-2
Equation two: 2(-3)-3(1)=-6-3=-9
Since both equations, once solved, equaled what they should have, you know that the
solution to this linear system is (-3,1)
Don’t Let These Fool You…
  Special types of Linear Systems
Linear Systems with NO Solution
• Here’s the problem:
       Equation one: 2x+y=3
       Equation two: 4x+2y=8
• After trying the graph method, you’ll find that the
  lines are parallel( don’t intersect) and therefore
  have no solution
• After trying either of the substitution or linear
  combination methods, you will have an equation
  that cannot be dealt with. You will know that this
  is the case because it will make no senses
  whatsoever. Therefore, you have no solution to
  the system.
    Linear System with MANY
             Solutions
• If you use the graph method, you will see that the
  equations are the same line, and any point on the
  line is a solution.
• If you use linear combinations or substitution, you
  will have a number =number, but both numbers
  will be the same. For example, 7=7 or 1=1. This
  indicates that the systems has many solutions.
Solving Systems of Linear
       Inequalities
     Graphing Systems of Linear
            Inequalities
• Here are some pointers and things to know:
  1. The boundary line on the graph will be dashed if the
  inequality is < or >.
  2. The boundary line will be solid if the inequality is <or >.
  3. You will also notice that graphs of linear inequalities are
  shaded in certain areas. To decide where to shade, pick a
  point that is CLEARLY above the line, and a point that is
  CLEARLY below the line. Put the first point into the
  inequality; solve; then do the same for the other point.
  Whichever point works, you shade that side.
An Example of Graphing Linear
        Inequalities
• Y<4
• Y>1
• Try this one out! Remember the steps; you
  can always go back a page if necessary…or
  go forward one page to get step by step
  guidance.
      So, You Needed Help…
• Here’s the original problem: y<4, y>1
• First, make a few basic conclusions:
      * The line for both boundaries will be
  dotted or dashed because it is < or >.
      *both will be horizontal lines because
  there is no x whatsoever in either equation
Now, you can graph the equation (next page)
             Graphing Errors

• If your graph looked like the previous slide, you
  can congratulate yourself on getting the lines
  drawn correctly. However you forgot to:
• Label the axis
• Label the lines
• Pick points and follow the previously described
  process to find where to shade (between y=1 and
  y=4)
• Write the equation on the line
   In Simpler Terms: Graphing
  Systems of Linear Inequalities
• 1. Sketch the lines of each inequality (remember
  to use dashed lines for < or > and solid lines for <
  or >)
• 2. LIGHTLY SHADE the area that is found by
  choosing points and placing them into the
  equation
• 3. The final result, or answer, is the area that is
  where the shaded planes intersect, for example, in
  the previous problem, the answer is anywhere
  between the boundary lines of y=4 and y=1.
 To Make it Somewhat Easier…
• Basic guidelines for linear systems:
• 1. Use the graphing method to get an approximate
  answer, to check a solution, or to give a visual
  idea of the system
• 2. Using substitution or linear combinations will
  allow you to get an exact and more accurate
  answer
• 3. Substitution helps a lot when used in systems
  that have coefficients of 1 or –1.
• 4. When there isn't a 1 or –1 as coefficients, the
  linear combinations method is efficient.
Fun, Fun: Examples to do on Your Own
      (Answers are on Last Page)
• 1. Solve the following Linear System by graphing
       Equation one: -2x+3y=6
       Equation two: 2x+y=10
• 2. Solve the following Linear system by means of
  substitution
       Equation one: x-6y=-19
       Equation two: 3x-2y=-9
• 3. Solve the following Linear system by means of
  substitution
       Equation one: x+3y=7
       Equation two: 4x-7y=-10
See next page for more; answers on last page
      A Little More Fun: More
              Examples
• 4. Use linear combinations to solve this
  system
      Equation one: -2x-3y=4
      Equation two: 2x-4y=3
• 5. Use linear combinations to solve this
  system
      Equation one: 3x-5y=-4
      Equation two: -9x+7y=8
      Answers to the Examples
• 1. Your graph should show a point of
  intersection, which is your solution, of
  (3,4).
• 2. (-1,3)
• 3. (1,2)
• 4. (-1.5,9)
• 5. (-.5, .5)

				
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posted:12/14/2011
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