Force

Linear Equation

Unit: 11
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Class 9: Mathematics

Equation, Linear Equation, Simultaneous Linear Equations, Solution of the Equations, Methods of solving simultaneous equations, Example Questions with...

Equation

The mathematical statement consisting of an equal sign is called an equation.

Example: \(2x + 3y = 0\)

Linear Equation

The equation of degree one is called a linear equation. For example, \(x + y = 3\), the power of the variables \(x\) and \(y\) is one. The graph of the linear equation is a straight line.

 

Simultaneous Linear Equations

Simultaneous linear equations are a set of two or more linear equations involving the same set of variables. The simultaneous linear equations satisfy only specific values of the variables.

Example: \(x + y = 3\) and \(4x - y = 2\) 

 

Solution of the Equations

The values of the variables that satisfy the given equations are called the solutions of the equations.

Example: \(x + y = 3\) and \(4x - y = 2\) for this equations the solution is \(x = 1, y = 2\).

 

Methods of Solving Simultaneous Linear Equations

To find the solution of simultaneous linear equations, you can use methods like substitution, elimination, or graphical methods:

1. Substitution: Solve one equation for one variable, substitute it into the other equation, and solve for the second variable.

2. Elimination: Add or subtract the equations to eliminate one of the variables, making it easier to solve for the remaining variable.

3. Graphical Method: Plot the equations on a graph, and the intersection point(s) represent the solution(s) to the system.

 

Example Questions with Solutions

\( \begin{aligned} & \text{Q1. Solve by substitution method: } x + y = 3, 4x - y = 2 \\ & \text{Solution: Given equations,} \\ & x + y = 3 \quad -------- (i) \\ & 4x – y = 2 \quad -------- (ii) \\ & \text{Now, From equation (i),} \\ & y = 3 – x \quad --------- (iii) \\ & \text{Now, putting the value of y in equation (ii),} \\ & or, 4x – (3 - x) = 2 \\ & or, 5x = 2 + 3 \\ & \therefore x = \frac{5}{5} = 1 \\ & \text{Now, putting the value of x in equation (iii),} \\ & y = 3 – 1 \\ & \therefore y = 2 \\ & \text{Thus, the is the solution is } (x, y) = (1, 2) \end{aligned} \)
\( \begin{aligned} & \text{Q1. Solve by elimination method: } x + y = 3, \; 4x - y = 2 \\ & \text{Solution: Given equations,} \\ & x + y = 3 \quad \text{-------- (i)} \\ & 4x - y = 2 \quad \text{-------- (ii)} \\ & \text{Now, adding equation (i) and equation (ii)}: \\ & \phantom{4}x + y = 3 \\ & + \; 4x - y = 2 \\ & \underline{\phantom{5x + 0y =} \quad \quad } \\ & 5x + 0y = 5 \\ & or, x = \frac{5}{5} \\ & \therefore x = 1 \\ & \text{Now, putting the value of } x \text{ in equation (i):} \\ & or, x + y = 3 \\ & or, 1 + y = 3 \\ & or, y = 3 - 1 \\ & \therefore y = 2 \\ & \text{Thus, the solution is } (x, y) = (1, 2). \end{aligned} \)

 

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