LINEAR EQUATIONS
 

Objectives:  To recall how to solve linear equations in one variable.

Recall 1:  A linear equation in one variable x is an equation that  contains x, but not x^2 or x^3 or x to any other exponent.
For example,  the following are linear equations in one variable:

(-5)(x - 4)/3 + x = 2 - 3x,

6(x - 5) = 3x - 6,

4.23x + 1.3 = 2.23 - 3.3(1 - 2x).

Recall 2:  The equality sign of an equation divides the equation into  two expressions--a left-hand side and a right-hand side.
A solution to  an equation is a number that can be substituted for x making the value  of the left-hand side equal to the value of the right hand side.
For  example, x = -2, x = 8 and x = 1 are the respective solutions to the three  equations above.

Recall 3:  Two equations are considered to be mathematically the  same or equivalent if they have the same solution.  So,

 (-5)(x - 4)/3 + x = 2 - 3x,
 
and

(-5)(x - 4) + 3x = 6 - 9x,

 are equivalent, since they both have the solution x = -2.

Recall 4:  Linear equations in one variable are occasionally disguised  in unnecessarily complicated form.  For example

 (-5)(x - 4)/3 + x = 2 - 3x
 
and

7x + 14 = 0

are equivalent linear equations, since they both have the solution x = -2.

Recall 5: Any linear equation is mathematically equivalent to an  equation of the form
 

ax + b = 0

 where a and b are fixed real numbers.

We call this the standard form  of a linear equation of one variable x.

For example, the first  equation of Recall 4 is equivalent to the second; and the second is in  standard form, where a = 7 and b = 14.

Recall 6:  Aside from the usual associative, commutative and  distributive laws of arithmetic the following three laws may be used to  obtain equivalent equations.
 

 If a, b and c are any real numbers with a=b, then 

  1. a + c = b + c,
  2. ac = bc,
  3. if c different from 0, then a / c = b / c .
 

It is generally easy to find the solution to a linear equation in one variable.  Simply use the rules of arithmetic, together with the three rules above, to gather all terms involving the unknown x to one side and all the known numbers to the other side.

Examples
 
Example 1.      Find a solution to 4x + 5 = 6 - 7x.

Solution:     Write a sequence of equivalent equations, using the rules of arithmetic and the rules of Recall 6.

  7x + 4x + 5 = 7x + 6 - 7x,             rule 1

  11x + 5 = 6,                                collecting x's

  11x + 5 + -5 = 6 + -5,                  rule 1

  11x = 1,                                      arithmetic

   11 / 11 x = 1 / 11 ,                      rule 3

   x = 1 / 11                                  arithmetic.


 
Example 2.       Find a solution to 3(x - 7) = 2 - 10x.
 
 Solution:      First write an equivalent equation that does not contain parentheses.
 
   3x - 21 = 2 - 10x,          distributive law

  10x + 3x - 21 = 10x + 2 - 10x,       rule 1

  13x - 21 = 2,                  collecting x's

  13x - 21 + 21 = 2 + 21,  rule 1

  13x = 23,                      arithmetic

  13x / 13= 23 / 13 ,         rule 3

   x =23 / 13 ,                  arithmetic.


 
Warning 1:   Always check your answers. In algebra, it is usually simple  to check an answer.  Substitute the solution for the unknown variable  and check that the equation is satisfied.  In the Example 2, the solution is x = 23 / 13.  If you substitute this value for x in the equation

3(x - 7) = 2 - 10x,

you find that the left-hand side has the value - 204 /13 , and the right-hand side has the same value, since 2 - 10( 23 / 13) = - 204 / 13 .


 
Example 3.      Find a solution to  (-5)(x - 4)/3 + x = 2 - 3x.

Solution:         We shall use the three rules of Recall 6, together with the rules of arithmetic to find the standard form of the equation.  The idea is to gather all terms involving the unknown x to one side and all the known numbers to the other side.

     (-5)(x - 4)/3 + x + (-x) = 2 - 3x + (-x),     from rule 1

    (-5)(x - 4)/3 = 2 - 4x,                             collecting x's

   3(-5)(x - 4)/3 = 3(2 - 4*x),                      from rule 2

  -5(x - 4) = 6 - 12x,                               distributive law and simplification

  -5x + 20 = 6 - 12x,                              distributive law

  12x - 5x + 20 = 12x + 6 - 12x,              from rule 1

  7x + 20 = 6,                                          col lecting x's

  7x + 20 + -20 = 6 + -20,                         from rule 1

  7x = -14,                                           ;    arithmetic

  7x / 7 = -14 / 7 ,                                    from rule 3

  x = -2,                                                   arithmetic.


                                                                                                         
Note

The examples that you encountered had unique solutions.  Not every equation that you write will have a unique solution.  We must distinguish between equations and -- what we generally call -- identities.  For example,

3x + 5 = 2x + x + 5

is an identity.  That means that the left-hand side always has the same value as the right-hand side, no matter what value you give x.  If you attempt to find a solution, you will eventually come to the equation x = x.  Any equation that is equivalent to x = x is called an identity.  Any equation that is not equivalent to x = x has a unique solution, or no solution.  For example, the equation 2(x+2) = 3(x+2) has no solution because you can rewrite it as 3=4, which is never true.


 
Example 4.   Find the solution to [5(x + 3) - x +15] / 2 = 3(x + 5) - x.

 Solution:   Multiply both sides by 2 to get

 5(x + 3) - x +15 = 2(3(x + 5) - x).

Use the distributive law to get

 5x + 15 - x +15= 6x + 30 - 2x.

Collect the x's and use arithmetic to get

4x + 30 = 4x + 30.

You can now see that this means that x = x.  So the original equation was an identity and there is no unique solution -- any value of x will be valid in this equation.

Warning 2:  Your biggest frustrations will occur when you work out an  exercise with haste and carelessness.  It does not take many mistakes to  shake your self confidence; that is a human condition.  Studies show that  most students find math difficult because of a lack of self confidence  rather than a lack of ability.  So take your time and check your results.


 
Note

In the examples above, we were dealing with linear equations in the variable x.  A linear equation in one variable could involve any single letter.  For example,

3t - 2 = 5,

(2 - 4y) / 5 = y,

and
 
3b / 5 - 6b = 1,

are linear equations in t, y and b respectively.


 
EXERCISES

Solve the following equations:
 
1.  x + 5 = 9
 

 
2. 4 = b - 12
 

 
3.  x - 3.45 = 2.67
 

 
4.  4y - 8 = 10
 

 
5.  -3x - 5 / 6  = x / 3
 

 
6. 2z / 3  -5 = 3z
 

 
7. (x + 2.3)/ 3.2 = 4.5
 

 
8.  4x - 8 = 7x + 3

 
9.  3(y - 3) + (y - 3) = (y - 3)
 

10. 3(t - 7) = 2 - 10t

 
11.  2(4x - 5) + 3x = 5(x - 1)
 

 
12.  7(3x - 5) + 3(x - 1) = 0
 

 
13.  (x + 3) / 4  + x / 6 = 5

 
14. (2R - 1) / 6  - (3R - 4) /  3  = 3/4

 
15. (2x + 4) / 3 - (x - 1)/ 2 = 1 - x / 6
 

 
16.   [7(3x - 5) + 3(x - 1)] / 3= 1 - x / 6

 
17. (x + 4)/ 3 + x / 2 = 3(3 - 2x)

 
18 [2(4x - 5) + 3x] / 4 - (4x - 5) / 3 + 3x = 5(x - 1)

ANSWERS
 
1. x = 4

 
2. b = 16

 
3.  x = 6.12

 
4.  y =  9 / 2

 
5. x = -1 / 4

 
6.  x = -15 / 7

 
7.  x = 12.1

 
8.  x =  -11 / 3

 
9.  y = 3
 

10.  t = 23 / 13
 

11. x = 5 / 6
 

12.  x = 38 / 24   or 19 / 12
 

13.  x = 102 / 10
 

14. R = 15 / 24  or 5 / 8
 

15,. x = -5 / 2
 

16.  x = 82 / 49
 
 

17.  x = 46 / 41
 

18. x = -25