Two methods are covered here:
Method 1: using inverse operationsA quick review of the basic principles - all equations have two sides: a Left Side (LS) and a Right Side (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 isolate the variable (or the unknown quantity).
For example:
5x + 8 = 3x – 6
We want to get rid of the number 8 from the left side.
So we subtract 8 from both sides of the equation.
5x + 8 | = | 3x – 6 | original equation |
– 8 | = | – 8 | subtract 8 from both sides |
5x | = | 3x – 14 | resulting equation |
Next, we want to get rid of 3x from the right side.
So, we subtract 3x from both sides of the equation.
5x | = | 3x – 14 | result from above |
– 3x | = | – 3x | subtract 3x from both sides |
2x | = | – 14 | resulting equation |
Now, we want to get rid of the coefficient 2.
So, we divide 2 from both sides of the equation.
2x | = | – 14 | result from above |
÷2 | = | ÷2 | divide both sides by 2 |
x | = | – 7 | resulting solution |
Now check your answer x = –7 by plugging it back into the original equation.
5x + 8 = 3x – 6
LS: 5 × (– 7) + 8 = – 35 + 8 = – 27
RS: 3 × (– 7) – 6 = – 21 – 6 = – 27
LS = RS, answer is correct.
In this method we isolate the variable by moving like terms to one side of the equation. To maintain the equality of the equation, when removing a term from one side of the equation we perform the opposite operation to the other side.
For example:
5x + 8 = 3x – 6
To remove + 8 from the LS, we subtract 8 from the RS
To remove + 3x from the RS, we subtract 3x from the LS
To remove the coefficient 2, we divide 2 on the RS
When you have grasped Method 2, it is faster because it allows you to perform several steps at the same time to isolate the variable.
For example:
5x + 8 = 3x – 6
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