In case algebraic equation, you can solve for the variable because two algebraic statements make an equation and hence it can be true for a certain value of variable or for a set of values of the variable. But an algebraic expression is just one statement. To solve an algebraic expression by simplification means pruning of the
given expression. If the given expression can be reduced to the lowest
possible level by any algebraic methods then we can ‘solve’ the
algebraic expression by simplification. There are a number of method to
solve an algebraic expression. Some of the methods are, grouping terms
and performing the addition/subtraction, use of identities, factoring,
rationalization etc.

## How to Solve an Algebraic Expression

How to solve an algebraic expression is a trick question to answer. Actually here the word ‘solve’ is to be considered with the given directions. Thus ‘solve algebraic expression’ could mean simplifying the algebraic expression, translating the algebraic expression, evaluating the algebraic expression etc. as the case may be. Hence with the question ‘how to solve a algebraic expression’ the direction to the question must be given. That is, ‘solve an algebraic expression’ make sense only if the question specifies whether to simplify, translate, evaluate for certain values of the variable etc.

##
Solved Examples

**Question 1: **Solve for x, x

^{2} – 5x + 6.

** Solution: **
**Step 1:**

Given quadratic expression x^{2} – 5x + 6

There is no any common factor.

Step 2:

Split the middle term -5x as, -3x - 2x. That is,

=> x^{2} – 5x + 6 = x^{2} - 3x - 2x + 6

Step 3:

Now find ‘x’ is a common factor for the first two terms and -1 as the common factor for the last two terms.

By partial factoring

=> x^{2} – 5x + 6 = x(x - 3) - 2(x - 3)

Now we see the expression (x – 3) is common factor and using that as a factor,

=> x^{2} – 5x + 6 = (x – 3)(x – 2)

Thus, we ultimately find that x^{2} – 5x + 6 can be factored and simplified as

(x – 3)(x – 2).

=> x^{2} – 5x + 6 = (x – 3)(x – 2). **answer**

**Question 2: **If the husband of a family earns 20 dollars per day and if the total
income per day of the family with his wife’s earning is 35 dollars, what
is the earning of his wife per day?

** Solution: **
**Step 1:**

Given:

Husband of a family earns 20 dollars per day

Let his wife earns x dollars per day

Total income per day = 35 dollars

Step 2:

The approach is,

20 + x = 35 . . . . Translation from word model to algebraic model

x = 35 – 20 = 15 . . . . Algebraic operation.

x = 15,

=> His wife earns 15 dollars per day. **answer**

**Question 3: **If log(

$\frac{x + y}{2}$) =

$\frac{1}{2}$(log x + log y), prove that x = y.

** Solution: **
Given log($\frac{x + y}{2}$ ) = $\frac{1}{2}$(log x + log y)

log($\frac{x + y}{2}$ ) = $(log x + log y)^{\frac{1}{2}}$

=> $\frac{x + y}{2}$ = $\sqrt{(xy}$

Squaring both sides, we have

$\frac{x^2 + y^2 + 2xy}{4}$ = xy

or $x^2 + y^2 + 2xy$ = 4xy

or $x^2 + y^2 + 2xy$ - 4xy = 0

or $x^2 + y^2 - 2xy$ = 0

or $(x - y)^2$ = 0

=> x = y. Proved