Page 168 - Maths Class 06

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(c) 5 + 4x – 4x + 2a when x = 3, and a = 5
(d) x²y + x²y² – xy² when x = 1 and y = 2.
(e) 4a – 3b + c when a = 2, b = 3 and c = 5.
(f) a² – 2b² + 3c² when a = 0, b = 1 and c = 1.
m 2
2. Eval u ate if m = 6 and n = 3.
3 n
xy
3. Eval u ate – (x + w), if x = 25, y = 36 and w = 20.
w
4. If x = 7, y = 6 and z = 4 find the value of 3x + (2y × z).

Linear Equation

In the previous concepts we have learnt literal numbers or variables, constant, algebraic expressions, and
how a variable is used in a pattern.
In this chapter, we will learn the linear equations and the solution of linear equations using the concept of
balance or equality on either side of an equation.

An equation in which the highest power of the variables involved is 1 is called a linear equation.

For example, 2x + 3 = 7, x + y = 9 and a b+ = 2.5, etc.
Clearly, the sign of equality in an equation divides it in two sides, namely, the left hand side and the
right-hand side, written as LHS and RHS respectively.

Solving an Equation
To solve an equation means to find a number which when substituted for the variable in the equation,
makes its LHS equal to the RHS. This number which satisfies the equation is called the solution or root of
the equation.
To find the root (solution) of an equation, i e. . to solve and equation, we can follow these methods:

1. Trial and error method 2. Systematic method 3. Transposition method
Trial And Error Method

In this method, we try different values for the variable to make both sides of an equation equal. We stop
this trial as soon as we get a particular value of the varible which makes the LHS equal to the RHS. This
particular value is said to be the root of the equation.
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EXAMPLE 1. Solve 2x - = 5, using the trial and error method.
SOLUTION : We try different values of x to find LHS = RHS.

S.No. Value of x LHS =(2x - )3 RHS =( )5 LHS = RHS

1. 1 2 1 2 3´ = - = - 1 5 No
2. 2 2 2 3´ - = 4 3- = 1 5 No

3. 3 2 3 3´ - = 6 3- = 3 5 No

4. 4 2 4 3´ - = 8 3- = 5 5 Yes

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