OML Search

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:

\(3x + 4 = 31\)

\(3 \times 9 + 4 = 31\)

\(27 + 4 = 31\)

\(31 = 31\)

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:
\(4x = 16\)
\(4 \times 4 = 16\)

Example 4:

\(\frac{x}{5} = 12\)

\(\frac{x}{5} \times 5 = 12 \times 5\)

\(x = 12 \times 5\)

\(x = 60\)

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
\(3 \times 6 + 4 = 34 - 2 \times 6\)
\(18 + 4 = 34 - 12\)
\(22 = 22\)

Example 6:

\(\frac{{3x + 4}}{2} + \frac{{2x + 1}}{3} = 4\)

\(6 \times \left( {\frac{{3x + 4}}{2} + \frac{{2x + 1}}{3}} \right) = 6 \times 4\)

\(6 \times \left( {\frac{{3x + 4}}{2}} \right) + 6 \times \left( {\frac{{2x + 1}}{3}} \right) = 24\)

\(3 \times (3x + 4) + 2 \times (2x + 1) = 24\)

\(9x + 12 + 4x + 2 = 24\)

\(12x + 14 = 24\)

\(12x + 14 - 14 = 24 - 14\)


\(12x = 10\)

\(\frac{{12x}}{{12}} = \frac{{10}}{{12}}\)

\(x = \frac{{10}}{{12}} = \frac{5}{6}\)




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 12x + 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 12 multiplies x on the LS we divide both sides by 12. On LS 12 divided by 12 cancels out each other, leaving x equal to 10 divided by 12.

Check Answer:
You can do the math by substituting \(x = \frac{5}{6}\) into the original equation
\(\frac{{3x + 4}}{2} + \frac{{2x + 1}}{3} = 4\)




 

Algebra Lesson - Transposition: Practice Questions

We shall be adding Practice Questions in the near future.

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


Useful Links:
More Algebra Lessons at Math.com
More Algebra Lessons on A Math Refresher

You can use the Mathway widget below to practice Algebra or other math topics. Try the given examples, or type in your own problem. Then click "Answer" to check your answer.

OML Search

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

[?] Subscribe To This Site

XML RSS
follow us in feedly
Add to My Yahoo!
Add to My MSN
Subscribe with Bloglines



Math TutorsMath Tutors