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Module 7: Algebra

Isolation (Solving for a Variable)

Tutorial 30: How to Isolate and Solve for a Variable in an Algebraic Expression

It’s important to keep these points in mind:

  1. Terms, variables, and numbers are moved from one side to the other by performing an opposing operation on something to be moved using itself.
    1. Addition and Subtraction are opposite.
    2. Multiplication and Division are opposite.
    3. Exponents and Roots are opposite.
  2. What is done to one side of the equation has to be done on the other side to keep the value of the equation the same.
  3. Movement and operations occur in reverse BEDMAS order (SAMDEB).

Below are a few examples where we'll solve for x {"version":"1.1","math":"x"}.

Click the boxes to see more:

10 x − 12 x + 4 = 36 {"version":"1.1","math":"10x−12x+4=36 "}

− 2 x + 4 − 4 = 36 − 4 {"version":"1.1","math":"−2x+4−4=36−4 "}

− 2 x = 32 {"version":"1.1","math":"−2x=32 "}

− 2 x ( − 2 ) = 32 ( − 2 ) {"version":"1.1","math":"−\dfrac{2x}{(−2)}=\dfrac{32}{(−2)} "}

x = − 16 {"version":"1.1","math":"x=−16 "}

x − 12 + 4 = 112 {"version":"1.1","math":"\sqrt{x}-12+4=112"}

x − 8 = 112 {"version":"1.1","math":"\sqrt{x}−8=112 "}

x − 8 + 8 = 112 + 8 {"version":"1.1","math":"\sqrt{x}−8+8=112+8 "}

x = 120 {"version":"1.1","math":"\sqrt{x}=120 "}

x 2 = 120 2 {"version":"1.1","math":"\sqrt{x}^2=120^2 "}

x = 14400 {"version":"1.1","math":"x=14400 "}

x 2 ( x + 2 5 = 12 x ( x + 2 ) 24 {"version":"1.1","math":"\dfrac{x^2(x+2}{5}=\dfrac{12x(x+2)}{24} "}

x 2 ( x + 2 ) 5 = x ( x + 2 ) 2 {"version":"1.1","math":"\dfrac{x^2(x+2)}{5}=\dfrac{x(x+2)}{2} "}

2 x 2 ( x + 2 ) = 5 x ( x + 2 ) {"version":"1.1","math":"2x^2(x+2)=5x(x+2) "}

2 x = 5 {"version":"1.1","math":"2x=5 "}

x = 5 2 {"version":"1.1","math":"x=\dfrac{5}{2} "}

Here are some extra tips/notes:

  • If possible, reduction of fractions/equations and combining like terms before doing anything else can make the work easier.
  • To get rid of denominators in fractions, cross-multiply the denominator from one side to the numerator of the other side and work from there.
  • SAMDEB is the order of isolation.

Left Side/Right Side Check

To verify whether an answer is correct, conduct a left side right side check (LS/RS).

In this series, we'll use the same examples we used in the boxes above (Examples 1, 2, and 3).

Example: 10 x − 12 x + 4 = 36 {"version":"1.1","math":"10x−12x+4=36 "}

Left Side

10 x − 12 x + 4 {"version":"1.1","math":"10x−12x+4 "}

= − 2 ( − 16 ) + 4 {"version":"1.1","math":"= −2(−16)+4 "}

= 32 + 4 {"version":"1.1","math":"= 32 + 4 "}

= 36 {"version":"1.1","math":"= 36 "}

Right Side

36 {"version":"1.1","math":"36"}

 

Example: x − 12 + 4 = 112 {"version":"1.1","math":"\sqrt{x}−12+4=112 "}

Left Side

x − 12 + 4 {"version":"1.1","math":"\sqrt{x}-12+4"}

= ( 14400 ) − 8 {"version":"1.1","math":"=\sqrt{(14400)}−8 "}

= 120 − 8 {"version":"1.1","math":"=120−8 "}

= 112 {"version":"1.1","math":"=112 "}

Right Side

112 {"version":"1.1","math":"112"}

Example: x 2 ( x + 2 5 = 12 x ( x + 2 ) 24 {"version":"1.1","math":"\dfrac{x^2(x+2}{5}=\dfrac{12x(x+2)}{24} "}

Left Side

x 2 ( x + 2 ) 5 {"version":"1.1","math":"\dfrac{x^2(x+2)}{5} "}

= 5 2 2 ( 5 2 + 2 ) 5 {"version":"1.1","math":"=\dfrac{\frac{5}{2}^2(\frac{5}{2}+2)}{5} "}

= ( 25 4 ) ( 5 2 + 4 2 ) 5 {"version":"1.1","math":"=\dfrac{(\frac{25}{4})(\frac{5}{2}+\frac{4}{2})}{5} "}

= ( 25 4 ) ( 9 2 ) 5 {"version":"1.1","math":"=\dfrac{(\frac{25}{4})(\frac{9}{2})}{5} "}

= 225 8 5 → 225 8 × 1 5 → 45 8 × 1 1 {"version":"1.1","math":"=\dfrac{\frac{225}{8}}{5} \rightarrow \dfrac{225}{8} \times \dfrac{1}{5} \rightarrow \dfrac{45}{8} \times \dfrac{1}{1} "}

= 45 8 {"version":"1.1","math":"=\dfrac{45}{8} "}

Right Side

12 x ( x + 2 ) 24 {"version":"1.1","math":"\dfrac{12x(x+2)}{24} "}

= 12 ( 5 2 ) ( 5 2 + 2 ) 24 {"version":"1.1","math":"=\dfrac{12(\frac{5}{2})(\frac{5}{2}+2)}{24} "}

= 30 ( 5 2 + 4 2 ) 24 {"version":"1.1","math":"=\dfrac{30(\frac{5}{2}+\frac{4}{2})}{24} "}

= 30 ( 9 2 ) 24 {"version":"1.1","math":"=\dfrac{30(\frac{9}{2})}{24} "}

= 135 24 → 45 8 {"version":"1.1","math":"= \frac{135}{24} \rightarrow \frac{45}{8} "}

Why is it solved this way? The whole point of using opposite operations on both sides of the equation is to make insignificant results relative to the operation. For example, if we have + 4 {"version":"1.1","math":"+4"} on the left side and subtract 4 {"version":"1.1","math":"4"} from both sides, the difference on the left side will be 0 {"version":"1.1","math":"0"} (insignificant number in addition/subtraction). This is similarly explored in multiplication/division operations where the insignificant result is 1 {"version":"1.1","math":"1"}.

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