Solving Linear Equations 
                                                                             
                                                                             
                                                                           
                          I.       Linear Equations 
                                   a.  Definition:  A linear equation in one unknown is an equation in 
                                       which the only exponent on the unknown is 1. 
                                    
                                   b.  The General Form of a basic linear equation is:axbc. 
                       
                                   c.  To Solve: the goal is to write the equation in the form variable = 
                                       constant. 
                       
                                   d.  The solution to an equation is the set of all values that check in the 
                                       equation. 
                       
                          II.      A STEP BY STEP PROCEDURE FOR SOLVING LINEAR 
                                   EQUATIONS: 
                                              1.      Remove any parentheses or grouping symbol. 
                                              2.      Multiply every term on both sides of the equation by the 
                                                      L.C.D. of all fractions appearing in the equation. This 
                                                      will get rid of all fractions. 
                                              3.      Combine similar terms on each side of the equation. 
                                              4.      Add or subtract terms on both sides of the equation to 
                                                      get the unknowns on one side and the number on the 
                                                      other side. 
                                              5.      Divide both sides of the equation by the coefficient of 
                                                      the unknown. 
                                              6.      Simplify the answer if necessary. 
                                              7.      Check your answer!  
                       
                       
                       
                          III.     Examples: 
                                   1. 3x12(x5) 
                                        3x 1 2x10                                           (Step 1) 
                                        2x3x12x102x                                       
                                      x110                                                   
                                        1     1                                              (Step 4) 
                                      x  9                                                   (Step 4) 
                                    
                                     Answer                                                    Check 
                                      x  9                                                   3(9)12(95) 
                                                                                               2712(14) 
                                                                                               2828 (It checks!) 
                                    
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                                                  2.  x  2 x 1  1(x4)                                                                 L.C.D. = 30 
                                                        5     3         2     3
                                                       x  2 x 1  x  4                                                                 (Step 1) 
                                                      5      3        2      3      3
                                                      (30) x (30) 2 x(30) 1 (30) x (30) 4                                             (Step 2) 
                                                              5             3                2             3             3
                                                      6x20x1510x40 
                                                      14x1510x40                                                                      (Step 3) 
                                                      10x            10x                                                                (Step 4) 
                                                      24x1540 
                                                                 15     15                                                              (Step 4) 
                                                      24x55 
                                                      x  55                                                                             (Step 5) 
                                                             24
                                                      x  55                                                                              (Step 6) 
                                                             24
                                                  Answer:  x  55  
                                                                           24
                                                   
                                                   
                                                  3.  1 (x3) 1(1 x x)                                                                 L.C.D. = 4 
                                                        2                       4         2
                                                       
                                                      1 x 3 11 x x                                                                    (Step 1) 
                                                      2         2           4        2
                                                      (4) 1 x(4) 3 (4)1(4) 1 x(4) x                                                   (Step 2) 
                                                            2              2                      4              2
                                                      2x64x2x                                                                          
                                                      2x                2x                                                              (Step 4) 
                                                      64x 
                                                   4  4                                                                                 (Step 4) 
                                                      2x 
                                                  Answer:  x2 
                                                   
                                                   
                                                  4. 3(x2)6(x1) 
                                                                            2
                                                      3x63x6                                                                           (Step 1) 
                                                      3x     3x                                                                         (Step 4) 
                                                      66 
                                                      This is a true statement which implies x can be any real number we 
                                                      want it to be. 
                                                  Answer:  x = Every real number 
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                                  5. 2x10 1(4x12) 
                                                2
                                    2x102x6                                              (Step 1) 
                                    2x      2x                                            (Step 4) 
                                    106 
                                    This is an untrue statement which implies that no value of x will 
                                    satisfy the equation. 
                                  Answer:  There is no solution. 
                                   
                                   
                                   
                                  6.  2x63        5    1                                L.C.D. = 3(2x1) 
                                      2x1         2x1 3
                                    3(2x1)2x63(2x1)33(2x1)                 5    3(2x1)1 
                                              2x1                             2x1               3
                                     
                                    3(2x1)2x63(2x1)33(2x1)                 5    3(2x1)1 
                                              2x1                             2x1               3
                                    3(2x6)9(2x1)3(5)(2x1) 
                                    6x1818x9152x1 
                                    12x9142x 
                                    10x914 
                                    10x5 
                                    x    5  
                                         10
                                    x   1  
                                           2
                                     
                                    Check 
                                    2(1)6                                      It is crucial to check your 
                                         2      3        5     1              answer when x appears in the 
                                     2(1)1          2(1)1 3                  denominator of a fraction! 
                                         2                 2
                                    16351 
                                       0          0 3
                                     
                                               
                                         undefined 
                                    Answer: Since x  1  does not check the answer is: “There is no 
                                                             2
                                    solution.”                                               
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                     Practice Problems: 
                      
                     1.  x7 11                                          17. 2(m5)m3 
                      
                     2.                                                   18.  s  3s1     s3 
                        n310                                                2 5  10
                      
                                                                              8k3 7k1           1
                     3.  y 84                                           19.    6    4 2 
                      
                     4. 8y  48                                           20. 3(w5)2(w2) w1 
                                                                                  4           6
                      
                     5. 1 x                                              21. Solve for r:      D  rt 
                            4
                      
                     6. 15  a                                            22. Solve for W:  P 2L2W  
                              3
                      
                     7. 10y 40                                          23. Solve for x:    3ax14 
                      
                     8.  4 a  3                                          24. Solve for x:    3ax4a 73ax 
                        9      2
                      
                     9.  3 y   5                                       25. Solve for x:    3ax4ax 73a 
                          4       11
                      
                     10. 3s71 
                      
                     11. 7  11z7 
                      
                     12.  d 1 7 
                          3
                      
                     13. 3d20d 
                      
                     14. 5y34y2 
                      
                     15. 5b8b13 
                      
                     16. 2(w3)2 
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