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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...

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    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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