If you were asked to solve a formula for a specified variable, how would you go about doing it?
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1. Look for an equals or an inequality sign (=, <, >, ≠, ≤, ≥). If none of these appear there is no equation to solve, so it may be an expression containing at least one appearance of the variable. The question may actually be to simplify the expression or to factorise the expression. Let's assume that there is one equals sign and the expression on the left is the LHS and the expression on the right in the RHS.

2. Note where the variable appears in LHS and RHS. 

3. a) If possible, move all the terms containing the variable to either LHS or RHS and all other terms to the other side of the equals sign. A term preceded by plus (+) changes to minus (-) if moved to the other side; and a term preceded by minus if moved to the other side. (A term is a group of one or more variables (represented by letters) and/or a constant (a regular number) with a plus (the first term in an expression has an "invisible" plus in front of it) or minus sign in front of the group. An expression is  a single term or a group of terms separated by plus or minus signs.)

b) It may not be possible to simply move all terms to either LHS or RHS because of parentheses (brackets ( ) { } [ ]). If the parentheses do not contain the required variable there is no need to remove them; but if the variable is in parentheses, then they must be removed, using the distributive property (whatever is immediately outside the parentheses has to be multiplied by each of the terms inside the parentheses.

4. At this stage don't attempt to resolve any fractions (indicated by division / or ➗) but just treat them as any other term, whether they contain the variable or not, and move them to LHS or RHS.

5. At this stage all the terms containing the variable should be on LHS or RHS and all other terms on the other side.

6. Look at the degree of the variable. The degree is its exponent (power). If none of the occurrences of the variable have an exponent then the degree is 1. The equation will be more difficult to solve if the degree is greater than 2, or is a fraction, or if there are fractions in which the variable is in the denominator. If the variable is under a square root sign (√) the equation will be more difficult to solve.

7. Solving equations involves many techniques, too many to discuss here, so we'll now assume degree-1, or linear equations, involving one variable to be solved in terms of other variables and constants. All the terms containing the variable should now be on either LHS or RHS, with everything else on the other side. To collect all the terms of the variable on one side of the equals sign is the aim in solving any equation. The next step is to simplify LHS and RHS terms as much as possible. For example, the separate terms 4x and 3x can be combined as 7x. The same rule applies to any constants or algebraic quantities.

8. After simplification (which isn't always necessary), if there is more than one term containing, the subject variable can be factorised by writing x outside parentheses and each term (with x removed) placed inside the parentheses. The parentheses are only necessary if there was more than one term containing the variable. For example, if the terms were 3x+xy+6xz, then this becomes x(3+y+6z). If there was only one term then this step is unnecessary. Treat the parentheses and everything in it as a single unit, like a parcel. 

9. The variable can now be isolated by transferring the parenthesised "parcel" to the other side of the equation as a divisor of everything on that side. If there is no parcel but simply a number or an algebraic quantity, then the whole of the other side of the equation is just divided by the quantity or number.

10. The fraction on the other side of the equation may need to be reduced to its simplest form by dividing numerator and denominator by a common factor.

An example may help to illustrate the method.

p is the variable to be found (the subject):

10q + 5p - 6 = 20a + 2p + b, separate p terms from the rest:

5p-2p=20a+b-10q+6, simplify:

3p=20a+b-10q+6,

p=(20a+b-10q+6)/3. final solution.

(If the variable had been on RHS with the expression on LHS, just swap places without changing any signs.)

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