# Algebra Lesson - Transposition

This may be your first algebra lesson. In case you are rather uncomfortable with algebra, you may want to first go through Basic Algebra - An Introduction - that would give you a good foundation before this lesson.

This Algebra Lesson introduces a technique known as 'transposition'. This is the most common way to solve algebra equations. A quick review here of the basic principles - all equations have two sides: a Left Side (LS) and a Right Side (RS). In the example below 3x + 4 is on the Left Side of the equation and 31 is on the Right Side of the equation:

3x + 4 = 31
LS     RS

The common transposition method is to do the same thing (mathematically) to both sides of the equation, with the aim of bringing like terms together and isolating the variable (or the unknown quantity).

So, to solve this equation, first subtract 4 from both sides of the equation. This will get rid of number 4 from the LS

3x + 4 - 4 = 31 - 4
That will give us:
3x = 27
Now, looking at the LS we have 3x. So we need to divide it by 3 to isolate x, and we need to do the same to the RS.
$$\frac{{3x}}{3} = \frac{{27}}{3}$$
Now that gives us:
$$x = \frac{{27}}{3} = 9$$
Check Our Answer:
Now, we have x = 9. We can substitute this value in the original equation to check if our answer is correct:

3x + 4 = 31

3 × 9 + 4 = 31

27 + 4 = 31

31 = 31

So, our answer x = 9 is correct.

That's the algebra lesson on transposition. Now, you are ready to review some examples to further develop your understanding of transposition.

### Algebra Lesson - Transposition: Some Examples

 Example 1 x + 6 = 14 x + 6 - 6 = 14 - 6 x = 14 - 6 x = 8 Note: Since 6 is with a + sign on the LS, we subtract 6 from both sides. +6 and –6 cancel each other on the LS. Check Answer: x + 6 = 14 8 + 6 = 14 Example 2 x - 3 = 6 x - 3 + 3 = 6 + 3 x = 6 + 3 x = 9 Note: Since 3 is with a – sign on the LS, we add 3 to both sides. –3 and +3 cancel each other on the LS. Check Answer: x - 3 = 6 9 - 3 = 6 Example 3: 4x = 16 $$\frac{{4x}}{4} = \frac{{16}}{4}$$ $$x = \frac{{16}}{4}$$ x = 4 Note: Since 4 is multiplying x on the LS, we divide both sides by 4. On the LS 4x divided by 4 gives leaves x. Check Answer: 4x = 16 4 × 4 = 16 Example 4: $$\frac{x}{5} = 12$$ $$\frac{x}{5} \times 5 = 12 \times 5$$ x = 12 × 5 x = 60 Note: Since 5 is dividing x on the LS, we multiply both sides by 5. So on the LS 5 divided by 5 is cancelled out leaving x. Check Answer: $$\frac{x}{5} = 12$$ $$\frac{{60}}{5} = 12$$ Example 5: 3x + 4 = 34 - 2x 3x + 4 + 2x = 34 - 2x + 2x 5x + 4 = 34 5x + 4 - 4 = 34 - 4 5x = 30 $$\frac{{5x}}{5} = \frac{{30}}{5}$$ $$x = \frac{{30}}{5}$$ x = 6 Note: Since there is a –2x on the RS we add a +2x on both sides. On the RS –2x and +2x cancel each other out. Since there is a +4 on the LS we subtract 4 from both sides. On the LS +4 and –4 cancel each other out. Since 5 multiplies x on the LS we divide both sides with 5. On the LS 5 divided by 5 cancel each other leaving x. Check Answer: 3x + 4 = 34 - 2x 3 × 6 + 4 = 34 - 2 × 6 18 + 4 = 34 - 12 22 = 22 Example 6: $$\frac{{3x + 4}}{2} + \frac{{2x + 1}}{3} = 4$$ $$6 \times \left( {\frac{{3x + 4}}{2} + \frac{{2x + 1}}{3}} \right) = 6 \times 4$$ $$6 \times \left( {\frac{{3x + 4}}{2}} \right) + 6 \times \left( {\frac{{2x + 1}}{3}} \right) = 24$$ 3 × (3x + 4) + 2 × (2x + 1) = 24 9x + 12 + 4x + 2 = 24 13x + 14 = 24 13x + 14 - 14 = 24 - 14 13x = 10 $$\frac{{13x}}{{13}} = \frac{{10}}{{13}}$$ $$x = \frac{{10}}{{13}}$$ Note: This equation involves fractions and we should try to get rid of fractions first. We do this by multiplying each term with 6 (lowest common multiple for 2 and 3 is 6). 2 cancels out 6 by 3 in the first term and 3 cancels out 6 by 2 in the second term leaving 3(3x+4) + 2(2x+1) = 24 Collecting like terms together we get 13x + 14 = 24 Since there is a +14 on the LS, we subtract 14 from both sides. On the LS the +14 cancels out –14. Since 13 multiplies x on the LS we divide both sides by 13. On LS 13 divided by 13 cancels out each other, leaving x equal to 10 divided by 13. Check Answer: You can do the math by substituting $$x = \frac{10}{13}$$ into the original equation $$\frac{{3x + 4}}{2} + \frac{{2x + 1}}{3} = 4$$

Transposition (Rearranging Equations) - Introduction
What is transposition? What is it used for?
Transposition (Rearranging Equations) - 1
How to transpose (or rearrange) equations?
How to solve equations?
1. Remove fractions
2. Remove brackets
3. Move added/subtracted terms
4. Divide by the number next to the letter
Transposition (Rearranging Equations) - 2
Include added and subtracted terms
Transposition (Rearranging Equations) - 3
Add brackets to the set of equations that we've already learned how to transpose
Transposition (Rearranging Equations) - 4
Include all the steps

### Algebra Lesson - Transposition: Practice Questions

Algebra Worksheets and Practice Questions

Having completed the practice questions, you are now ready for other algebra lessons, like Substitution.

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