SOLVING ALGEBRAIC EQUATIONS WITH SPECIAL CASES

When we solve algebraic equation in one variable, we may face three different situations.

One solution :

The final answer will result in the form “x = a” (the variable will equal SOMETHING)

Only one real number can make the equation true.

Infinitely Many Solutions :

The final answer will result in the form “a = a” (the same number will be on both sides of the equal sign).

Any real number can make the equation true.

No Solution :

The final answer will result in the form “a = b” (where a is a different number than b).

There is no possible answer for this equation.

Solve the following equations. Some equations will have single answer, other will have no solution and still other will have infinitely many solutions.

Example 1 :

2x + 2x + 2 = 4x + 2

Solution :

2x + 2x + 2 = 4x + 2

By combining like terms, we get

4x + 2 = 4x + 2

By applying any values for x, we will get the same values on both sides. For infinite values of x, it becomes true. So, the algebraic equation has infinitely many solution.

Example 2 :

3(x - 1) = 2x + 9

Solution :

3(x - 1) = 2x + 9

By distributing 3.

3x - 3 = 2x + 9

Subtracting 2x on both sides.

3x - 2x - 3 = 9

x - 3 = 9

Add 3 on both sides.

x = 9 + 3

x = 12

So, the equation has one solution.

Example 3 :

2x + 8 = 2(x + 4)

Solution :

2x + 8 = 2(x + 4)

Using distributive property in the right side.

2x + 8 = 2x + 8

For all values of x, it is true. So, it has an infinitely many solution.

Example 4 :

2x - x + 7 = x + 3 + 4

Solution :

2x - x + 7 = x + 3 + 4

Combining like terms, we get

x + 7 = x + 7

For all values of x, it is true. So, it has an infinitely many solution.

Example 5 :

-2(x + 1) = 2x + 5

Solution :

-2(x + 1) = 2x + 5

Distributing -2

-2x - 2 = 2x + 5

Subtract 2x

-2x - 2x - 2 = 5

-4x - 2 = 5

Add 2 on both sides.

-4x - 2 + 2 = 5 + 2

-4x = 7

Divide by -4

x = -7/4

So, it has one solution.

Example 6 :

Scott has 12 more 50 cents coins than 20 cent coins, and their total value is $10.20. How many 20 cent coins does Scott have?

Solution :

Let x be number of coins in 20 cents.

Number of 50 cents coins = x + 12.

0.20x + 0.50(x + 12) = 10.20

0.20x + 0.50x + 6 = 10.20

0.70x = 10.20 - 6

0.70x = 4.2

x = 4.2/0.70

x = 6

So, number if 20 cent coins is 6.

Example 7 :

At the supermarket a carton of juice costs $1.50 more than a bottle of milk. If 3 bottles of milk and 2 cartons of juice cost $8, how much does a carton of juice cost?

Solution :

Let x be the cost of a bottle of milk.

x + 1.50 = a carton of juice

3 bottles of milk + 2 cartons of juice = $8

3x + 2(x + 1.50) = 8

3x + 2x + 3 = 8

5x = 8 - 3

5x = 5

x = 1

Carton of juice = 1 + 1.50.

= $2.50

Cost of carton of juice is $2.50.

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