SOLUTIONS TO LINEAR EQUATIONS
 

Objective:  To learn how to solve linear equations that do not appear to be   linear.

Recall 1: Watch how quickly the fractional equation

3 / (x + 2) = 2 / (3x - 1)

reduces to a linear equation.  Multiply both sides by (x + 2)(3x - 1) to  get

(x + 2)(3x - 1)[3 / (x + 2)] = (x + 2)(3x - 1)[2 / (3x - 1)].

which simplifies to

3(3x - 1) = 2(x + 2).

Notice that this is a linear equation whose solution is x = 1.

Recall 2: Some rational equations that convert to linear ones have no solutions.  Here is an example.  In attempting to solve the equation

 x / (x - 2)= 2 + 2 / (x - 2) ,

 we see that the common denominator is x - 2 and that by multiplying the  2 by  (x - 2) / (x - 2) we have
 
x / (x - 2)= 2(x - 2) / (x - 2) + 2 / (x - 2)

 or

x = 2(x - 2) + 2

 and so x = 2 is a candidate for a solution.  However, if we check on this  solution, we find that it cannot satisfy the equation, since it would  require us to divide by zero.  We must conclude that this equation has  no solutions.

Recall 3: There are equations that contain powers of x greater than  1, but which quickly convert to linear equations.  Take, for example,  the equation

x^2 + 2x + 3 = x^2 - x - 3.

Simply by subtracting x^2 from both sides, the equation becomes linear.

Recall 4: Occasionally, the disguise is more subtle.  For example,  the equation

2x^2 + 7 = (x - 3)(2x + 5)

is immediately reduced to a linear equation once the product on the  right-hand side is multiplied out and 2x^2 is subtracted from both sides.

Examples
 
Example 1. Show that the equation

6 / (x^2 - 4) = 5 / (x + 2) + 3 / (x - 2)

 can be reduced to a linear equation.

Solution: The common denominator of the right-hand side is

(x + 2)(x - 2),

which is the same as x^2 - 4.  The equation therefore reduces to

 6 / (x^2 - 4) = [5(x - 2) + 3(x + 2)] / (x^2 - 4)

and hence to

6 = 5(x - 2) + 3(x + 2).

Using the distributive law, we see that this is equivalent to

6 = 5x - 10 + 3x + 6,

which is clearly linear.

(x - 3)(x + 4) = (x + 5)(x + 2)
 is linear.

Solution: Multiply out each side to get

x^2 + x - 12 = x^2 + 7x + 10.

Subtract x^2 from both sides to get

x - 12 = 7x + 10,

which is clearly linear.

 sqrt (4 - 3x) = 5

 is linear.

Solution: You should convince yourself that squaring both sides of an  equation gives you a valid equation.  In symbols, this means:

if A = B, then A^2 = B^2.

 By squaring both sides of the equation we have 4 - 3x = 25.

Solution: This poses an interesting problem.  We may add 4(x - 2)^2 to both sides to get

(x - 1)^2 = 4(x - 2)^2.

 But now what?  Can we simply take the square root of both sides?  Does  A^2 = B^2, tell us that A = B?
The answer is:

A^2 = B^2, implies that A = ±B.

 In the case of our problem it tells us that

(x - 1) = ±2(x - 2).

 This means that we have reduced our equation to two different linear  equations, namely

(x - 1) = 2(x - 2)
 and
(x - 1) = -2(x - 2).

The solution to the first is x = 3; the solution to the second is x = 5 / 3

Taking square roots of both sides of an equation may introduce extra solutions to an equation, since A^2 = B^2, implies that A = ±B.

 a / (x^2 - 4) = b / (x + 2) + c / (x - 2)

 solve for x in terms of a, b and c.

Solution: Treat the a, b and c as if they are numbers and proceed as  illustrated in the solution to Example 1.  The common denominator of  the right-hand side is x^2 - 4.  Reduce the right hand side to a single  fraction so the equation looks like

 a / (x^2 - 4) = [b(x - 2) + c(x + 2)] / (x^2 - 4)

 Here is a special fact that comes in handy at times like this.

If  A / B = C / B and B different from 0, then A = C.

 In other words, whenever two fractions are equal, their numerators are  equal--as long as their denominators are non-zero.  Applying this  special fact to our equation, we see that

a = b(x - 2) + c(x + 2).

 The distributive law then tells us that a = bx - 2b + cx + 2c.  Collect  the terms involving x on the left side and the terms not involving x on  the right side to get bx + cx = a + 2b - 2c.  Then use the distributive  law to factor out the x from the left-hand side.  You get

x(b + c) =  a + 2b - 2c.

As long as b is different from -c, we have a solution; x = (a + 2b - 2c) / (b+c).

When b = -c we have two cases:

i) a + 2b - 2c = 0 (that is, a = 4c), then our equation reduces to 0 = 0, so any real number, except for 2 and -2, is a solution.

ii) a is different from 4c, then the equation has no solutions.

Solve for the variable in the following problems:
 
1. x^2 + 3x - 5 = x^2 + 5x + 3

 
2. 2x^2 - 3x + 7 = (2x + 3)(x - 2)
 

 
3. (s + 4)(2s - 1) = (2s + 3)(s - 2)

 
4. 2/ (x + 5)  = 3 / (x - 2)

 
5. 1 / (x - 1) + 3 / (x - 1)  = 2

 
6.   1/ (x - 2) + 5 / (2 - x) = 1

 
7. 2 / (2y - 1) = 7 / (x + 4)   [Hint:  (2y - 1)*y + 4) is a common divisor.]

 
8.  3x / (2x - 6)= 3 + 3 / (x - 3) [Hint:  Factor out the 2 from the denominator of the left-hand side.]

 
9.  3 / (t - 3)+ 2 / (t - 4) =  5/ [(t - 3)(t - 4)] [Hint:  (t - 3)(t - 4) is a  common divisor.]

 
10.  7 / [(2x + 3)(x - 2)] = 4 / (2x + 3) + 2 / (x - 2)  [Hint:  (2x + 3)(x - 2) is a common divisor.]

 
11.  (4 - 3x)/ (6x - 8) = 5 / (3x - 4)  [Hint:  Factor out a 2 from the denominator of the first term.]

 
12.  2 / (x - 2) + 7 / (x+2) = 8 / (x^2 - 4)   [Hint: x^2 - 4 = (x - 2)(x + 2).]

 
13. sqrt (7 - 3y) = 6

 
14.  

 
15.  Solve the following equation for x in terms of a and b, where a different from b:
   (x + a) / (x - b)= a / b .

 
16. 2 + 2 / (2 + 2/x) = 5 / (x + 1)    [Hint: Simplify the compound fraction on the left-hand side.]

ANSWERS
 
1.  x = -4

 
2.  x = 13 / 2

 
3.  s = -1 / 4

 
4.  x = -19

 
5.  x = 3

 
6.  x = -2

 
7.  y = 15 / 12

 
8.  x = 4
 

10. x = 9/ 8
 

11.  x = -2
 

12. x = 2
 

13.  y = -29 / 3
 

14. x = 1/ 5
 

15.  x = 2ab / (a - b)
 

 16. x = 1.