System of Equations

Algebra is distinct from Arithmetic because of the presence of variables. Thus while 65 means just that, 12x can take any value depending on the value that x, the variable, takes. A Variable, can assume different values, depending on conditions given, if any. Thus if x belongs to set of Real Numbers (usually written as x∈R), x can assume any real value where as if x belongs to set of Integers (usually written as x ∈ Z), x can assume only integral values.

Expressions:

Variables and constants combine to form Terms. Thus 9x or –6xy or x² are terms. The constant is called the coefficient of the corresponding variable. One or more terms together form an Algebraic Expression. Thus   x + y or 3x – xy + 5y or x² – 8x + 15 are all examples of expressions. Remember expressions do not have solutions i.e. x² – 8x + 15 does not boil down to the solution that x = 3 or 5.

Degree of an algebraic expression: 

The degree of an expression is same as the maximum value of the sum of the powers of the variables in any term. The degree of the expression has nothing to do with the number of variables in the algebraic expression. Thus all of the following x³ – 3x + 5; x + y + z + xyz; a²b + b²a has a degree of 3.

Expressions of degree 1 are called Linear expressions, usually called them a straight lines of degree 2 are called Quadratic expressions and of degree 3 are called Cubic expressions

An equation in one variable and with degree n can have at maximum n real roots. The number of real roots can be less than n also because few of them may be Imaginary.

Functions:

Functions are just a language of Algebra. We can refer to the expression x

2

${\mathrm{x}}^2$ – 8x + 15 as y i.e.

 y = x

2

${\mathrm{x}}^2$ – 8x + 15. Now what will be the numerical value of y? Obviously the value of y also keeps changing as the value of x changes i.e. the numerical value of y depends on the value assumed by x. If x = –2, y = 35, if x = 0, y = 15 and if x = 2, y = 3. This dependency is expressed as y = f(x) and is read as y is a function of x. In y = f(x) = x

2

${\mathrm{x}}^2$ – 8x + 15, the right most side explicitly states how y is related to x.

Thus functions is nothing new and even the quadratic expression that we are so used to can be expressed in the language of functions. In this example, if we need to find the value of y when x = 5, we simply mean to evaluate f(5). To evaluate f(5), we just substitute the value of x as 5 in the expression x

2

${\mathrm{x}}^2$ – 8x + 15 and find that f(5) = 0

Equations:

Only when an expression is equated to another expression (even a constant is an expression), do we get an equation. Thus x

2

${\mathrm{x}}^2$ – 8x + 15 is not an equation but x

2

${\mathrm{x}}^2$ – 8x + 15 = 0 is.

Roots or solution set to an equation are those values of the variables which satisfy the equation.

Thus, as seen from the table and graph of the expression x

2

${\mathrm{x}}^2$ – 8x + 15, the expression takes a value equal to 0 only when x = 3 or when x = 5.

Thus 3, 5 is the solution set or the roots of the equation.

Arithmetic operations on Algebraic expressions:

Addition & Subtraction:

The addition or subtraction of algebraic expression follows the same rules as the arithmetic addition/subtraction. We can add the coefficients of the terms with the same degree and in the same variables only.

For example,

(2X + 3Y) + (5X + 4Y) = 7X + 7Y

 = 7(X + Y)

For example,

(3X

2

${\mathrm{X}}^2$ + 4XY + 7Z

2

${\mathrm{Z}}^2$) – (5XY – 4Z

2

${\mathrm{Z}}^2$)

= 3X

2

${\mathrm{X}}^2$ + 4XY + 7Z

2

${\mathrm{Z}}^2$ – 5XY + 4Z

2

${\mathrm{Z}}^2$

= 3X

2

${\mathrm{X}}^2$ – XY + 11Z

2

${\mathrm{Z}}^2$

Multiplication: 

The multiplication of any 2 algebraic expressions follows the distributive property of multiplication and the index rules.

For example,

(2X + 3Y) (5X + 4Y)

= (2X) (5X + 4Y) + (3Y) (5X + 4Y)

= 10X

2

${\mathrm{X}}^2$ + 8XY + 15XY + 12Y

2

${\mathrm{Y}}^2$

= 10X

2

${\mathrm{X}}^2$ + 23XY + 12Y

2

${\mathrm{Y}}^{\mathrm{2<a\; name="more"></a>}}$

Division: 

Just as in the normal arithmetic, division and multiplication of algebraic expressions follow similar rules.

For example, divide the expression

f(x) = 3x

2

${\mathrm{x}}^2$ + 4x + 3 by (x – 2).

Solution:

Just as in normal arithmetic division, the quotient of the division process is 3x +  5 and the remainder is 23.

Factorisation: 

The expression x

2

${\mathrm{x}}^2$ – 5x + 6 can be written as product of (x – 2) and (x – 3). This process of writing an expression as a product of different expressions is called as Factorisation. The expressions (x – 2) and

(x – 3) are called factors of x

2

${\mathrm{x}}^2$ – 5x + 6. Just as in Arithmetic, in Algebra also a factor of an expression can completely divide the expression i.e. the remainder is zero. The process of factorization is useful to find the roots of any equation.

Thus x

2

${\mathrm{x}}^2$ – 5x + 6 = 0 ⇒$\Rightarrow$ (x – 2) (x – 3) = 0. The product of two numbers can be 0 only if atleast one of them is zero. Thus the LHS of the above equation will become 0 if x = 2 or 3 and the equation will be satisfied.

Remainder Theorem:

To identify whether a given expression is a factor of another expression, we can take help of Remainder Theorem.

According to the remainder theorem, when any expression f(x) is divided by (x – a), the remainder is f(a). (a is any constant in this example).

Thus when the expression, x

3

${\mathrm{x}}^3$ + x

2

${\mathrm{x}}^2$ + 4 is divided by x + 1, the remainder is (−1)

3

${\left(-1\right)}^3$ + (−1)

2

${\left(-1\right)}^2$ + 4 i.e. 4.

Factor Theorem:

We can also use the remainder theorem or more specifically the Factor theorem to identify if an expression is a factor of another expression. As already seen, a expression is said to be a factor of another expression only when the remainder is 0 when the latter is divided by the former.

Thus (x – a) is a factor of f(x) if and only if f(a) = 0.

Factor theorem also helps us in factorising higher degree equations. Consider the equation

f(x) = x

3

${\mathrm{x}}^3$ + 6x

2

${\mathrm{x}}^2$ – 19x – 24 = 0.

By hit and trial (basically substituting values of x as –2, –1, 1 or 2), we see that f(–1) = (−1)

3

${\left(-1\right)}^3$ + 6 (−1)

2

${\left(-1\right)}^2$ – 19 (–1) – 24 = –1 + 6 + 19 – 24 = 0.

Thus, we can deduce that (x + 1) is a factor of f(x).  i.e.

x

3

${\mathrm{x}}^3$ + 6x

2

${\mathrm{x}}^2$ – 19x – 24 = (x + 1) ´ g(x),  where g(x) is another algebraic expression in variable x.

Using common sense we can gather that g(x) is a quadratic expression. Why? Only (x + 1) (ax

2

${\mathrm{x}}^2$ + bx + c) will have a term in x

3

${\mathrm{x}}^3$, x

2

${\mathrm{x}}^2$, x and a constant.

Thus, x

3

${\mathrm{x}}^3$ + 6x

2

${\mathrm{x}}^2$ – 19x – 24 = (x + 1)(ax

2

${\mathrm{x}}^2$ + bx + c).

By visual check we can ascertain that a = 1 and c = –24. How?

Equating coefficient of x

3

${\mathrm{x}}^3$ on RHS and LHS, we get a = 1 (on the RHS only x × ax

2

${\mathrm{x}}^2$ will result in term in  x

3

${\mathrm{x}}^3$ ).

Equating the constant terms of RHS and LHS, we get c = –24 (only 1 × c will result in a constant on RHS).

Thus now we have x

3

${\mathrm{x}}^3$ + 6x

2

${\mathrm{x}}^2$ – 19x – 24

= (x + 1)(x

2

${\mathrm{x}}^2$ + bx – 24).

To find b, equate the coefficient of either x

2

${\mathrm{x}}^2$ or of x of the two sides of the equation. Equating the coefficient of x, we have –19 = b – 24 giving us b = 5 (when the LHS is expanded, the only terms in x are bx – 24x).

Thus finally, x

3

${\mathrm{x}}^3$ + 6x

2

${\mathrm{x}}^2$ – 19x – 24

= (x + 1)(x

2

${\mathrm{x}}^2$ + 5x – 24).

Consistency of Equations:

When a system of equations has at least one solution, we say that the system is consistent. When it has no solution we say that the system is inconsistent. Let the system of equation be

a

1

${\mathrm{a}}_1$ x + b

1

${\mathrm{b}}_1$ y + c

1

${\mathrm{c}}_1$ = 0

a

2

${\mathrm{a}}_2$ x + b

2

${\mathrm{b}}_2$ y + c

2

${\mathrm{c}}_2$ = 0

⇒xb

1

c

2

−b

2

c

1

=yc

1

a

2

−a

1

c

2

=1a

1

b

2

−a

2

b

1

$\Rightarrow {\displaystyle \frac{\mathrm{x}}{{\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1}=\frac{\mathrm{y}}{{\mathrm{c}}_1{\mathrm{a}}_2-{\mathrm{a}}_1{\mathrm{c}}_2}=\frac{1}{{\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1}}$

    and a

1

b

2

≠a

2

b

1

${\mathrm{a}}_1{b}_2\ne a_2{b}_1$

(i)      If a

1

a

2

≠b

1

b

2

,${\displaystyle \frac{a_1}{a_2}\ne \frac{b_1}{b_2},}$ the equations are consistent with unique solution.

(ii)     If a

1

a

2

=b

1

b

2

=c

1

c

2

,${\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2},}$ the equations are consistent with infinite solutions.

(iii)     a

1

a

2

=b

1

b

2

≠c

1

c

2

,${\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2},}$ the equations are inconsistent i.e, no solution.