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EXAMPLES & SOLUTIONS
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System of equations combine two equations that assist in solving problem.

Two equations are often needed to solve a system of equations.
The two equations form some type of relationship to discover a solution.
There can be one solution, no solution or infinite solutions to the problem.

WORD PROBLEM:  ONE SOLUTION SYSTEM OF EQUATION

Indentify the components of the equations

A farm wants to build a total of 7 storage areas in the form of stables and barns. The goal is to place 20 cows in each barn and 5 horses in each stable. How many total barns and total stables are needed to hold 80 cows and horses?

The equation is solved by knowing how many of the 80 cows and horses can be placed in each barn and stable.

x  = number of barns (unknown variable)
y  = number of stables (unknown variable)

Gathering Information for Equation 1:

Variable 1 : unknown number of barns (x) 
Variable 2: unknown number of stables (y) 
Constanttotal number of barns and stables (7)

Operator: "total" is a keyword for add, '+'
Terms: x, y, 7
Coefficient: 1x  (unknown number of barns), and 1y is multiplied by itself (unknown number of stables) each time a barn or stable is added.

Expression:
x + y

Equation:
x + y = 7

Gathering Information for Equation 2:

Variable 1 :  unknown number of barns (x) 
Variable 2: unknown number of stables (y) 
Constant: total number of cows and horses (80)
Operator: "total" is a keyword for add, '+'
Terms: 20x , 5y, 80
Coefficient: 20 cowsx (# of barns) is multiplied by 20 each ttime a (20 cows per barn), and y multiplied by 5 (5 horses per stable) each time a barn or stable is added

Expression:
20x + 5y

Equation:
20x + 5y = 80

SOLVE WITH A GRAPH

Graph the system of equation

Equation #1

x + y = 7

change to slope-intercept form
y = -x + 7

therefore,

if x = 0, then
y = 0 + 7
y = 7 
--y-intercept (0,7)

if y = 0, then

0 = -x + 7
     -7 = -x + 7 - 7
-7  = -x 
x = 7
7 = x 
--x-intercept (7,0)

Equation #2

20x + 5y =80

change to slope-intercept form
 5y -20x + 80


therefore,

if x = 0, then
y = 0 + 16
y = 16 
--y-intercept (0,16)

if y = 0, then

0 = -4x + 16
-16 = -4x + 16 -16
-16 = -4x 


x = 4
4 = x 
--x-intercept (4,0)
y = -4x + 16
 5            5          5
 -4       -4

Place x- & y- interecept on graph

(0, 16) 

14

16

10

12

Equation #2

20x + 5y = 80
y = -4x + 16

m =  4
            1

 (0,  7)

Equation #1

x + y = 7
y = -x + 7

m =  -1
             1

(4,0)

 (3, 4) = Point of intersection

4

6

2

-5

-4

-3

-2

-1

-2

1

2

3

4

5

6

7

(7, 0) 

Equation #1
slope = -1
y-intercept = (0,7)
Equation #2
slope = 3
y-intercept = (0, 16)
Graphically these linear lines  INTERSECT
at
one pont
thus, there is 
one solution (3,4) 

SOLVE: ELMINATION OR SUBSTITUTION

Discover the solution with the steps of elimination OR substitution

ELMINATION 

         x + y = 7
20x + 5y = 80


choose -5 to eliminate y
(-5)(x) +  (-5)(y) = (-5)(7) --multiply by -5 to solve for x


   -5x  -  5y    =   -35  --updated equation
   20x  +  5y    =     80
    15x  +   0      =    45
                  15x   =    45


                        x   =  3  


substitute x=3 into the equation to solve for y
       x + y = 7
       3 + y = 7
3 -3  + y = 7 - 3
       0 + y = 4
               y = 4

x = 3, y = 4

Solution (3, 4)

  15             15

SUBSTITUTION

         x + y = 7
20x + 5y = 80


change one equation to slope-intercept form
x + y = 7
x - x + y = -x + 7
  --subtract x from both sides
       0  + y = -x + 7 
                y = -x + 7
 
--updated equation

sustitute the y equation into the standard form
 20x + 5 (-x + 7)  =  80
       20x -5x + 35  =  80
                 15x  + 35  =  80
        15x + 35 - 35 = 80 - 35
                     15x + 0  = 45

                              15x =  45

                           x = 3


sustitute x in equation to solve for y
        x + y  =  7
        3 + y  =   7
3 -3  + y  =  7 - 3
         0 + y  =  4
                 y  =  4

x = 3, y = 4

Solution (3, 4)

 15        15

Solution is at the intersecting point (3, 4).
The farm needs 3 barns and 4 stables.
This means that each barn can hold 20 cows and each stable can hold 5 horses.

Check  your answer:
                           
 20x + 5y

if x = 3 and y = 4, then
     = 20(3) + 5(4)
     = 60 + 20
     = 80 (cows and horses)

WORD PROBLEM:  NO SOLUTION SYSTEM OF EQUATION

ARRIVING SOON!!!

WORD PROBLEM:   MANY SOLUTION SYSTEM OF EQUATION

ARRIVING SOON!!!

WORD PROBLEM:  PAROBLAS, NO SOLUTION, ONE SOLUTION AND INFINITE SOLUTIIONS

ARRIVING SOON!!!
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