Basic Concepts of Algebra

5. EQUATIONS 1

5.    EQUATIONS 1

5.1  DEFINITION:
                              An equation is a statement that two quantities are equal. The equation, x+2=9
      tells us that a certain number added to 2 is equal to 9. When x is replaced by 7, the statement is   
      true so we say that 7 is the solution to the equation. In this chapter, our objective will be to find a 
      replacement number for the variable that will make the equation a true statement.

5.2  TYPES OF EQUATIONS:
                              There are basically two types of equations, conditional equations and identities.
                              A conditional equation is one that is true for only certain values of a variable.  ...          For example, x+2=9 is true for x=7. Therefore, x+2=9 is a  conditional equation.
                              An identity is a statement that is true for all admissible values.  Thus, x+5=(2x+10)/2 is an identity.
        Consider:
                          4(x-1)+5+x=3(x+2)+2x-5
        SOLUTION:
                            4x-4+5+x=3x+6+2x-5
                            4x+x+5-4=3x+2x+6-5
                            5x+1        =5x+1
                              5x-5x    =1-1
                                      0  =  0
When an equation is solved and 0=0, the solution is the entire set of real numbers. Therefore the             
        equation is an identity. In other words, x can equal any number.
        For example, if x=2, then
                            4(2-1)+5+(2)=3(2+2)+2(2)-5
                                      4    +5+2=12            +4-5
                                                    11  =11
        Or, if x=3, then

                            4(3-1)+5+(3)=3(3+2)+2(3)-5
                                    8      +5+3=15            +6      -5
                                                      16  =16

        Try any other value for x. It will satisfy the equation.

5.3  SOVLVING EQUATIONS    (SINGLE-STEP)
                In order to find the solution (or root) of an equation we find the number(s) which, when used 
          as the value of the variable(s), makes the statement true.
               
                Think about the following equations.
1.    x+4=11
2.    x-5=10
3.    2x =14
4.    x÷3=5
               



                These statements would read:
1.    A number plus 4 is equal to 11.
2.    A number minus 5 is equal to 10.
3.    Two times a number is equal to 14.
4.    A number divided by 3 is equal to 5.
      Think about each equation as a balance scale (beam balance). If one side is changed in some way,
        the  other side must be changed in exactly the same way to keep both sides balanced (in    equilibrium).
      Thus, from the examples,
1.    x+4        =11              ……To balance
      x+4-4    =  11-4……Subtrat 4 from both sides
      ∴x          =7                   

2.  x-5        =10                ……To balance
      x-5+5=10+5……Add 5 to both sides
      ∴x            =15

3.      2x    =      14                    ……To balance
   
        2x/2  = 14/2                      ……Divide both sides by 2.
              ∴x=7

4.      x/3          =5        ……To balance
   
        x/3  x 3  =5 x3  ……Multiply both sides by 3
      ∴    x        =15
           
      Remember:
                            To solve an equation, we isolate the variable on one side of the equal sign by using       
                            either addition, subtraction, multiplication or division. Thus, if the term is linked to
                            the variable by:
                        - Addition, then we subtract (as in 1)
                        - Subtraction, then we add (as in 2)
                        - Multiplication, then we divide (as in 3)
                        - Division, then we multiply (as in 4)

∙    You may find it helpful to work the problems quickly by using the following steps.
    If x+4=11
    Then x=11-4    Subtract 4 from both sides
      So x=7                 

2.  If x-5=10
    Then x=10+5    Add 5 to both sides
    So x=15               


3.  If 2x  =14
    Then x=14/2            Divide both sides by 2
      So  x = 7                             
4. If x/3=5                    Multiply both sides by 3
    Then x=5 x 3     
    So    x=15
∙  You may solve the equations without writing out the steps taken.
    Thus, x+4=11
              x=11-4
              x=7
   




    EXERCISE 5.3
    Solve the equations:

1.  x+6=6

2.  5x=30

3.  1/2 x=5

4.  x-8=25

5.  7x=49

6.  12x=0

7.  1/4 x=-1

8.  9+x=0

9.  x-4=44

10. x-11=22






5.4 SOVLING EQUATIONS (MULTI-STEP)
  ∙  When solving multi-step equation, use the reverse order of operations and isolate the variable on
      one side of the equation.
      Example 1.    5x-15=50
      Step1. Perform the inverse operations for addition and subtraction first 5x-15+15=50+15
                    Add 15 to both sides.    5x=65
      Step2. Perform the inverse operations or multiplication and division 5x/5=65/5 divide both sides by
                    5.    x=13
      Step3. Check. Substitute the value for x
                                                    5(13)-15=50
                                                          65-15=50
                                                                    50=50         
                    The solution is x=13





∙    Some equations may have variable terms on both sides. If so, you need to group all the variable
    terms on one side of the equation.
    Example 2.        8x+7=5x+1
    Step1. Group the variable 8x-5x+7=5x-5x+1 Subtract 5x from both sides 3x+7=1
    Step2. Subtract 7 from both sides 3x+7-7=1-7        3x=-6
    Step3. Divide both sides by 3.        3x/3=(-6)/3  ∴x^ =-2
    Step4. Check
                            8(-2)+7=5(-2)+1
                            -16+7=-10+1
                                        -9=-9
                The solution is x=-2

∙    Some equations contain parentheses. Remove the parentheses by multiplying each term within the
    parentheses by the factor.
    Example 3.      7(x+1)=35
    Step1. Multiply each term in the parentheses by 7.            7x+7=35
    Step2. Subtract 7 from both sides 7x+7-7=35-7          7x=28
    Step3. Divide both sides by 7.          7x/7=28/7              x=4
    Step4. Check.
                              7(4+1)=35
                                          35=35
                                                           
    The solution is x=4
∙    To solve equations with like terms, first combine the like terms, then simplify each side of the
    equation, then solve.
    Example 4.      5x+7-8x=15+4
    Step1. Simplify by combining like terms    5x-8x=-3x;  15+4=19
                5x+7-8x=15+4
                -3x+7=19
    Step2. Isolate the variable -3x+7-7=19-7  subtract 7 from both sides    -3x=12
    Step3. Divide both sides by -3, so that you solve for x, not –x      (-3x)/(-3)=12/(-3)        x=-4
    Step4. Check
                            5(-4)+7-8(-4)=15+4
                            -20+7+32=19
                                                19=19
    The solution is x=-4
   





  EXERCISE 5.4
    Solve the equations.

1. 7x-12=9

2. 9x+7=34

3. 4x+5x=45

4. 3x+2-6x=-19

5. 10x+7=6-x+12

6. 7x+3=38

7. 5x-17=23

8. 11x-12=9x+2

9. 10(x-2)=50

10. 15(2+x)=7(x+10)







5.5    SUMMARY EXERCISE for Ch5.
          Solve the following equations:

1.  x-10=19

2.  n-7=14

3.  a+12=23

4.  x/4=12

5.  4x=28

6.  -7+5x=10x-2

7.  6x-2+2x=14

8.  5y+12-3y=-2y-13-y

9.  2n-5n+11=38

10. 5t-13t+2t=-70+t

11. 5x+3=8(x-3)

12. 2(x+2x)-6=30

13. 3(y-9)-2=-35

14. x+11+3x=20+7x

15. 6(x-2)=5x

16. 2(x+5)=4-x

17. 4(2x-7)=3(4x-9)

18. 9-2x=1-4x

19. 9x-4(2x-3)=5(x+2)

20. 7+2(3x-1)=8x-3(2x+5)