Equations with the unknown on both sides
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Equations with the unknown on both sides. Lesson Objective: An equation is like a set of scales. To keep it balanced, whatever you do to one side you must do to the other. Use this idea to solve equations like: 3x + 1 = x + 7 2 (3x + 1) = 3 (x – 2). Solving equations:.

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Equations with the unknown on both sides

Equations with the unknown on both sides.

Lesson Objective:

An equation is like a set of scales.

To keep it balanced, whatever you

do to one side you must do to the other.

Use this idea to solve equations like:

3x + 1 = x + 7

2 (3x + 1) = 3 (x – 2)


Solving equations
Solving equations:

Only want ‘x’ on one side

2x + 1 = x + 5

Subtract x from each side

x + 1 = 5

Subtract 1 from each side

x = 4

Check your answer. Does the equation balance?

2x4 + 1 = 4 + 5 P


Solving equations1
Solving equations:

Only want ‘x’ on one side

5x - 2 = 2x + 4

Subtract 2x from each side

3x - 2 = 4

Add 2 to each side

3x = 6

Divide each side by 3

x = 2

Check your answer. Does the equation balance?

5x2 - 2 = 2x2 + 4 P


On whiteboards solve each equation

2x + 2 = x + 9

3x + 1 = x + 5

6x – 8 = 4x

5x + 1 = x - 11

x = 7

x = 2

x = 4

x = -3

On whiteboards: Solve each equation


In your books write each equation and solve it to find x

2x – 1 = x + 3

3x + 4 = x + 10

5x – 6 = 2x

4x + 1 = x - 8

2x + 3 = x + 10

4x – 1 = 3x + 7

Extension:

2x - 6 = - 3x + 9

x = 4

x = 3

x = 2

x = -3

x = 7

x = 8

x = 3

In your books: Write each equation and solve it to find x.


Solving equations with brackets
Solving equations with brackets:

2 (x + 3) = x + 11

Multiply out the bracket

2x + 6 = x + 11

Subtract x from each side

x + 6 = 11

Subtract 6 from each side

x = 5


Solving equations with brackets on both sides
Solving equations with brackets on both sides:

2 (3x – 1 ) = 3 (x + 2)

Multiply out the brackets

6x - 2 = 3x + 6

Subtract 3x from each side

3x -2 = + 6

Add 2 to each side

3x = 8

Divide each side by 3

x = 8/3 = 2 2/3


In your books write each equation and solve it to find x1

2 (x + 3) = x + 7

5 (2x - 1) = 3x + 9

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

3 (x – 1) = 2 (x + 1)

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

3 (4x – 3) = 2 (2x + 3)

Extensions: 7(x – 2) = 3 (2x – 7)

3(3x - 1) = 5 (x – 7)

x = 1

x = 2

x = -1

x = 5

x = -4/7

x = 15/8 = 1 7/8

x = - 7

x = -8

In your books: Write each equation and solve it to find x.


How could you check each answer

2 (x + 3) = x + 7

5 (2x - 1) = 3x + 9

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

x = 1 means 2 (1 + 3) = 1 + 7

2 x 4 = 8 P

x = 2 means 5 (2x2 -1) = 3x2 + 9

5 x 3 = 6 + 9 P

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

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

2 x -3 = - 6 P

How could you check each answer?