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x + 1 = 10

conditional, x = 9

y 4 = 7

conditional, y = 11

5 a = 25

conditional, a = 5

x 4 = 9

conditional, x = 36

18 b = 6

conditional, b = 3

y 2 = y 2

identity

x + 4 = x 3

contradiction

x + x + x = 3 x

identity

8 x = 0

conditional, x = 0

m 7 = 5

conditional, m = 2

Literal equations

Literal equations

Some equations involve more than one variable. Such equations are called literal equations .

An equation is solved for a particular variable if that variable alone equals an expression that does not contain that particular variable.

    The following equations are examples of literal equations.

  1. y = 2 x + 7 . It is solved for y .
  2. d = r t . It is solved for d .
  3. I = p r t . It is solved for I .
  4. z = x u s . It is solved for z .
  5. y + 1 = x + 4 . This equation is not solved for any particular variable since no variable is isolated.

Solving equation of the form x + a = b and x a = b

Recall that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side.

This is the this number same as number x = 6 x + 2 = 8 x 1 = 5

    This suggests the following procedures:

  1. We can obtain an equivalent equation (an equation having the same solutions as the original equation) by adding the same number to both sides of the equation.
  2. We can obtain an equivalent equation by subtracting the same number from both sides of the equation.

We can use these results to isolate x , thus solving for x .

Solving x + a = b For x

x + a = b The a is associated with x by addition . Undo the association x + a a = b a by subtracting a from b o t h sides . x + 0 = b a a a = 0 and 0 is the additive identity . x + 0 = x . x = b a This equation is equivalent to the first equation, and it is solved for x .

Solving x a = b For x

x a = b The a is associated with x by subtraction . Undo the association x a + a = b + a by adding a to b o t h sides . x + 0 = b + a a + a = 0 and 0 is the additive identity . x + 0 = x . x = b + a This equation is equivalent to the first equation, and it is solved for x .

Method for solving x + a = b And x a = b For x

To solve the equation x + a = b for x , subtract a from both sides of the equation.
To solve the equation x a = b for x , add a to both sides of the equation.

Sample set b

Solve x + 7 = 10 for x .

x + 7 = 10 7 is associated with x by addition . Undo the association x + 7 7 = 10 7 by subtracting 7 from b o t h sides . x + 0 = 3 7 7 = 0 and 0 is the additive identity . x + 0 = x . x = 3 x is isolated, and the equation x = 3 is equivalent to the original equation x + 7 = 10. Therefore, these two equation have the same solution . The solution to x = 3 is clearly 3. Thus, the solution to x + 7 = 10 is also 3.

Check : Substitute 3 for x in the original equation. x + 7 = 10 3 + 7 = 10 Is this correct? 10 = 10 Yes, this is correct .

Solve m 2 = 9 for m .

m 2 = 9 2 is associated with m by subtraction . Undo the association m 2 + 2 = 9 + 2 by adding 2 from b o t h sides . m + 0 = 7 2 + 2 = 0 and 0 is the additive identity . m + 0 = m . m = 7

Check : Substitute 7 for m in the original equation. m 2 = 9 7 2 = 9 Is this correct? 9 = 9 Yes, this is correct .

Use a calculator to solve this equation. Solve y 2.181 = 16.915 for y .

y 2.181 = 16.915 y 2.181 + 2.181 = 16.915 + 2.181 y = 14.734

On the Calculator
Type 16.915 Press + / Press + Type 2.181 Press = Display reads: 14.734

Solve y + m = s for y .

y + m = s m is associated with y by addition . Undo the association y + m m = s m by subtracting m from b o t h sides . y + 0 = s m m m = 0 and 0 is the additive identity . y + 0 = y . y = s m

Check : Substitute s m for y in the original equation. y + m = s s m + m = s Is this correct? s = s True Yes, this is correct .

Solve k 3 h = 8 h + 5 for k .

k 3 h = 8 h + 5 3 h is associated with k by subtraction . Undo the association k 3 h + 3 h = 8 h + 5 + 3 h by adding 3 h to b o t h sides . k + 0 = 5 h + 5 3 h + 3 h = 0 and 0 is the additive identity . k + 0 = k . k = 5 h + 5

Practice set b

Solve y 3 = 8 for y .

y = 11

Solve x + 9 = 4 for x .

x = 13

Solve m + 6 = 0 for m .

m = 6

Solve g 7.2 = 1.3 for g .

g = 8.5

solve f + 2 d = 5 d for f .

f = 3 d

Solve x + 8 y = 2 y 1 for x .

x = 6 y 1

Solve y + 4 x 1 = 5 x + 8 for y .

y = x + 9

Exercises

For the following problems, classify each of the equations as an identity, contradiction, or conditional equation.

m + 6 = 15

conditional

y 8 = 12

x + 1 = x + 1

identity

k 2 = k 3

g + g + g + g = 4 g

identity

x + 1 = 0

For the following problems, determine which of the literal equations have been solved for a variable. Write "solved" or "not solved."

y = 3 x + 7

solved

m = 2 k + n 1

4 a = y 6

not solved

h k = 2 k + h

2 a = a + 1

not solved

5 m = 2 m 7

m = m

not solved

For the following problems, solve each of the conditional equations.

h 8 = 14

k + 10 = 1

k = 9

m 2 = 5

y + 6 = 11

y = 17

y 8 = 1

x + 14 = 0

x = 14

m 12 = 0

g + 164 = 123

g = 287

h 265 = 547

x + 17 = 426

x = 443

h 4.82 = 3.56

y + 17.003 = 1.056

y = 18.059

k + 1.0135 = - 6.0032

Solve n + m = 4 for n .

n = 4 m

Solve P + 3 Q 8 = 0 for P .

Solve a + b 3 c = d 2 f for b .

b = a + 3 c + d 2 f

Solve x 3 y + 5 z + 1 = 2 y 7 z + 8 for x .

Solve 4 a 2 b + c + 11 = 6 a 5 b for c .

c = 2 a 3 b 11

Exercises for review

( [link] ) Simplify ( 4 x 5 y 2 ) 3 .

( [link] ) Write 20 x 3 y 7 5 x 5 y 3 so that only positive exponents appear.

4 y 4 x 2

( [link] ) Write the number of terms that appear in the expression 5 x 2 + 2 x 6 + ( a + b ) , and then list them.

( [link] ) Find the product. ( 3 x 1 ) 2 .

9 x 2 6 x + 1

( [link] ) Specify the domain of the equation y = 5 x 2 .

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Source:  OpenStax, Algebra i for the community college. OpenStax CNX. Dec 19, 2014 Download for free at http://legacy.cnx.org/content/col11598/1.3
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