Simple Equations in One Variable

When one expression is equal to another, the equality of these expressions may hold either for all values of the unknown variables involved or for some particular values of the variables involved. In the former case it is called an identity. For example, $(a+b)(a-b)=a^2-b^2$ which is true for all values of $a$ and $b$ . In the latter case it is called an equation. For example, $4x=8$ which is true only when $x=2$ .

Definition

An equation is a statement in which two algebraic expressions are connected by the sign of equality (=). Each of the expressions on either side of the sign of equality is called a side or member of the equation.

For example, if the expressions $4x-3$ and $2x+1$ are equal in value i.e., $4x-3= 2x+1$ , then this algebraic statement is called an equation where, $4x-3$ and $2x+1$ are the members of the equation. To solve the equation $4x-3 = 2x+1$ means to find the value of the letter $x$ . This letter $x$ is called the variable or the unknown quantity or the root of the equation. Variables are usually represented by alphabets, for example, $a, b, x, y, z$ . The equation in which the variable is of the first order, i.e., an equation in which the highest power of the involved variables is 1 is called a simple or a linear equation.

Solving Linear Equation:

Solving a linear equation is governed by the following rules:

If the same quantity, i.e., equal quantities be added to the two expressions on either side of the equality sign, then the sums are equal.
If the same quantity, i.e., equal quantities be taken away from the two expressions on either side of the equality sign, then the differences are equal.
If the two expressions on either side of the equality sign be multiplied by the same quantity, then the products are equal.
If the two expressions on either side of the equality sign be divided by the same quantity, then the quotients are equal.

Corollary 1:

From rules 1 and 2 we can deduce an important principle, i.e., any term may be transposed from one side of the equality sign to the other by simply changing its sign.

For example, let $x-a=b+c$

Adding $a$ to both sides, we get,

$x-a+a=b+c+a$ [Rule 1]

$x=b+c+a$

Again, subtracting $c$ from both sides, we get,

$x-a-c=b+c-c=b$ [Rule 2]

Thus, we see that $(-a)$ removed from the left side appears as $(+a)$ on the right side. Again, $(-c)$ removed from the right side appears as $(+c)$ on the left side.

Hence, if $x-a=b-c+d$ , we get, $x-a-b+c-d=0$

This is called Transposition.

Corollary 2:

The sign of every term of an equation may be changed without destroying the equality.

For example, let $x-a=b+c$

$\Rightarrow (x-a) \times (-1)=(b+c) \times (-1)$ [Rule 3]

$-x+a=-b-c$

Steps to solve a simple equation:

Simplify all brackets, fractions, etc if required.
Bring all the terms containing the variables on one side and all the constant terms on the other side.
Solve the equation, obtained in Step 2 to get the value of its variable.

Different Forms of Simple Equations

Simple equations, usually, are of three types:

The unknown quantity or variable with any coefficient is equal to a known quantity (i.e., constant). For example, $4x=28$ . The general form of this type of equation is $ax=b$ . The root of this type of equation is obtained by dividing the known quantity by the coefficient of the unknown quantity and is $\dfrac{b}{a}$ .
The sum of the unknown quantity with any coefficient and a known quantity is equal to a known quantity. For example, $2x+3=5$ . The general form of this type of equation is $ax+b=c$ In solving the equation, $b$ is to be transposed to the right-hand side and the equation stands as $ax=b-c$ . The root is found out by dividing the algebraic difference of the known quantities by the coefficient of the unknown quantity and is $\dfrac{c-b}{a}$ .
In this type of equations, there are known and unknown quantities on both sides. For example, $3x+3=5x-6$ . The general form of this type of equation is $ax+b=cx+d$ To solve a simple equation of this type, unknown quantities are to be grouped on one side and the known quantities are to be grouped on the other side. The equation then stands as $ax-cx=d-b$ or, $(a-c)x=d-b$ and the root is found out by dividing the algebraic difference of the known quantities by the algebraic difference of the coefficients of the unknown quantity. Here the root is $\dfrac{d-b}{a-c}$ .

Note: It is very clear that all simple equations are reducible to Type 1.

Example: Solve $\dfrac{1}{2}x+ \dfrac{1}{3}x= x-3$

Solution:

$6 \times \dfrac{1}{2}x + 6 \times \dfrac{1}{3}x = (6 \times x) - (6 \times 3)$

$or, 3x+2x= 6x-18$

$or, -x =-18 \: or \: x=18$

Example: Solve $\dfrac{3x+1}{x+5} = \dfrac{6x-7}{2x+1}$

Solution:

$\dfrac{3x+1}{2x+1} = \dfrac{6x-7}{x+5}$

$or, 6x^2 + 3x +2x +1= 6x^2 +30x -7x -35$

$or, x=2$

Example: If $\dfrac{2x}{a} + \dfrac{y}{2b} =4$ and $3x=2y=a =6$ ; find the value of $b$ .

Solution:

Given: $3x =2y= a =6$

$or, x=2, y=3$ and $a=6$

$\therefore$ $\dfrac{2x}{a} + \dfrac{y}{2b} =4$

$or, \dfrac{2.2}{6} + \dfrac{3}{2b}=4$

$or, \dfrac{3}{2b}=4-\dfrac{2}{3}=\dfrac{10}{3}$

$or, b= \dfrac{9}{20}$

Example: Solve $(x+2)(3x+4)-6x=10+(3x+2)(x+1)$

The left side $=3x^2+10x+8-6x =3x^2+4x+8$

The right side $=10+3x^2+5x+2=3x^2+5x+12$

Hence, $3x^2+4x+8=3x^2+5x+12$

Removing $3x^2$ from both sides we have,

$4x+8=5x+2$

Hence, by transposition,

$4x-5x=12-8$

or, $-x=4$

$\therefore$ $x=-4$

Thus, the required value of $x$ is -4.

Example: Solve: $\dfrac{x}{6}+5=\dfrac{x}{3}+\dfrac{x}{4}$

Since, $\dfrac{x}{6}+5=\dfrac{x}{3}+\dfrac{x}{4}$ ,

multiplying both sides by 12, which is the L.C.M of the denominators, we have,

$12(\dfrac{x}{6}+5)=12(\dfrac{x}{3}+\dfrac{x}{4})$

or, $2x+60=4x+3x=7x$

Hence, by transposition,

$2x-7x=-60$

or, $-5x=-60$

or, $x=12$ (dividing both sides by -5)

Thus the required root of thye equation is 12.

Note: While solving equations, when the root is found out, i.e., the value of the variable is found out, it may be verified by putting this value of the variable in the equation. If it is found that equality of both sides is maintained when we put this value of the variable in the equation, then we may conclude that the root is correct.

Exercise

Solve the following equations:

$3(3x-4) - 2(4x-5) = 6$
$\dfrac{5y-4}{8} - \dfrac{y-3}{5} = \dfrac{y+6}{4}$
$\dfrac{2}{5}(x-1)= 1- \dfrac{3}{5}(3x-5)$
Find the value of $x$ , which makes the two expressions $(x+1)(x+2)$ and $x(x+7)- 6$ equal to each other.
Solve for :
1. $\sqrt{x-3}=2$
2. $\sqrt[3]x-4=0$
3. $2 + \dfrac{2x-3}{2x+3} = \dfrac{3x+4}{x+2}$
4. $2+ \dfrac{2x-3}{2x+3} = \dfrac{3x+4}{x+2}$
5. $(7x - 1) -(x- \dfrac{1-x}{2}) = 5x + \dfrac{1}{2}$
6. $\dfrac{x}{2}+5=\dfrac{x}{3}+7$

Definition

Solving Linear Equation:

Corollary 1:

Corollary 2:

Different Forms of Simple Equations

Exercise

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