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Solving linear equations:  Section 2

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Canceling

The unknown on both sides

Simple fractional equations

Canceling

If equal terms appear on both sides of an equation,
then we may "cancel" them.

x + b + d  = c + d.

d appears on both sides.  Therefore, we may cancel them.

x + b  = c.

Theoretically, we can say that we subtracted d from both sides.

Finally, on solving for x:

x  = cb.

Problem 17.   Solve for x :

x² + x − 5  =  x² − 3

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Cancel the x²'s:

x − 5  =  −3
 
x  =  −3 + 5 = 2.

Problem 18.   Solve for x :

xa + b  =  a + b + c

Cancel the b's but not the a's.  On the left is −a, but on the right is +a.  They are not equal.

Transpose −a:

x  =  a + a + c
 
x  =  2a + c.

The unknown on both sides

Example 3.   Solve for x :

4x − 3 = 2x − 11.
 
      1.  Transpose the x's to the left and the numbers to the right:
 
4x − 2x = −11 + 3.
 
      2.  Collect like terms, and solve:
2x = −8
x = −4.

This is another example of doing algebra with your eyes.  In the first line, you should see that 2x goes to the left as −2x, and that −3 goes to the right as +3.

As a general rule for solving any linear equation, we can now state the following:

Transpose all the terms that involve the unknown to the left, and add them;
transpose the remaining terms to the right;
make 1 the final coefficient of the unknown, by dividing or multiplying.

Problem 19.   Solve for x :

15 + x = 7 + 5x
 
x − 5x = 7 − 15
 
−4x = −8
 
x = 2.

Problem 20.   Solve for x :

1.25x − 6 = x  
 
1.25xx = 6  
 
(1.25 − 1)x = 6   On combining the like terms
on the left.
.25 x = 6  
 
x = 24.   On multiplying both sides
by 4.

Problem 21.   Remove parentheses, add like terms, and solve for x :

(8x − 2) + (3 − 5x) = (2x − 1) − (x − 3)
 
8x − 2 + 3 − 5x = 2x − 1 − x + 3
    on removing the parentheses
 
3x + 1 = x + 2
 
3xx = 2 − 1
 
2x = 1
 
x = 1
2

Simple fractional equations

   Example 4.        x
2
  =  4.

Since 2 divides on the left, it will multiply on the right:

x = 2· 4    
 
  = 8.

Problem 22.   Solve for x :

x
 5 
= 3
 
x = 15
 
x = −15.
  Problem 23.   x
4
= 1
2
 
    x = 4·  1
2
= 2.

Example 5.   Solve for x:

4
x
  =  5.

Solution.   In the standard form of a simple fractional equation, x is in the numerator.  But we can easily make that standard form by taking the reciprocal of both sides.

  x
4
  =   1
5
 
This implies
  x   =   4 ·   1
5
  =   4
5
.
  Problem 24.   2
x
= 1
3
 
    x
2
= 3
 
    x = 6.

Example 6.  Fractional coefficient.

3x
4  
= y
 
     Since 4 divides on the left, it will multiply on the right:
 
3x = 4y.
 
     And since 3 multiplies on the left, it will divide on the right:
 
x = 4y
 3
In other words,  3
4
 goes to the other side as its reciprocal 4
3
.
Note that  3
4
 is the coefficient of x :
3x
4
3
4
x .

Coefficients go to the other side as their reciprocals!

Problem 25.   2x
= a
 
    x = 3a
 2
Problem 26.   5
8
x = ab
 
      x = 8
5
(ab)
  Problem 27.   4
5
x  +  6 = 14
 
      4
5
x = 14 − 6 = 8
 
        x = 5
4
· 8 = 10.
  Problem 28.   A = ½xB
 
         Exchange sides:  
 
  ½xB = A
 
  x = 2A
 B

The reciprocal of ½ is 2.

Problem 29.   The Celsius temperature C is related to the Fahrenheit temperature F  by this formula,

F   =   9
5
C + 32

a)   What is the Fahrenheit temperature when the Celsius temperature
a)    is 10°?

F = 9
5
· 10 + 32
 
  = 18 + 32
 
  = 50°

b)  Solve the formula for C.

Exchange sides:

9
5
C 32 = F
 
  9
5
C = F − 32
 
    C = 5
9
(F − 32)

c)  What is the Celsius temperature when the Fahrenheit temperature
c)   is 68°?

20°. This is Problem 13 of Lesson 1.


Next Lesson:  Word problems

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