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:
\(3x + 4 = 31\)
\(3 \times 9 + 4 = 31\)
\(27 + 4 = 31\)
\(31 = 31\)
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.
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: |
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.
Check Answer: |
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 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: |
Having completed the practice questions, you are now ready for other algebra lessons, like Substitution.
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