हिंदी
Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 9

Simultaneous Linear Equations

Advertisements

Topics

Estimated time: 14 minutes
Maharashtra State Board: Class 10

Definition: Simultaneous Linear Equations

A system of simultaneous linear equations is a pair of linear equations in the same two variables that must be satisfied together.

The standard form is:

\[ a_{1}x + b_{1}y + c_{1} = 0 \]

\[ a_{2}x + b_{2}y + c_{2} = 0 \]

A solution is an ordered pair (x,y) that satisfies both equations.

Maharashtra State Board: Class 10

Method I: Substitution Method

Use this method when one variable can be easily expressed in terms of the other.

Maharashtra State Board: Class 10

Example

Given Equations:

\[ 5x - 3y = 8 \ldots (I) \] 

\[ 3x + y = 2 \ldots (II) \]

Step-by-Step Solution:

Step 1: Pick the easy equation

Equation (II) \[ (3x + y = 2) \] is easier because the variable \[ y \] has a coefficient of \[ 1. \]

Step 2: Express one variable in terms of the other

\[ y = 2 - 3x \ldots (III) \]

Step 3: Substitute into the second equation

\[ 5x - 3(2 - 3x) = 8 \]

Step 4: Solve for \[ x \]

Expand the brackets: \[ 5x - 6 + 9x = 8 \]

Combine like terms: \[ 14x - 6 = 8 \]

Rearrange and solve:

\[ 14x = 8 + 6 \]

\[ 14x = 14 \]

\[ x = 1 \]

Step 5: Substitute back to find \[ y \]

Substitute \[ x = 1 \] into Equation (III):

\[ y = 2 - 3(1) \]

\[ y = 2 - 3 \]

\[ y = -1 \]

Final Answer:

\[ x = 1, y = -1 \] is the solution to the given pair of linear equations.

Maharashtra State Board: Class 10

Method II: Elimination Method

Use this method when coefficients of one variable can be made equal or opposite easily.

Maharashtra State Board: Class 10

Example

Given Equations:

\[ 3x + 2y = 29 \ldots (I) \]

\[ 5x - y = 18 \ldots (II) \]

Step 1: Multiply equation (II) by 2 to align \[ y \] coefficients

\[ 5x \times \mathbf{2} - y \times \mathbf{2} = 18 \times \mathbf{2} \]

\[ \therefore 10x - 2y = \mathbf{36} \ldots (III) \]

Step 2: Add equations (I) and (III)

\[ \begin{array}{r} 3x + 2y = 29 \\ \mathbf{+10x - 2y = 36} \\ \hline \mathbf{13x = 65} \implies \therefore x = \mathbf{5} \end{array} \]

Step 3: Substitute \[ x = 5 \] into equation (I)

\[ 3 \times \mathbf{5} + 2y = 29 \]

\[ \therefore \mathbf{15} + 2y = 29 \]

\[ \therefore 2y = 29 - \mathbf{15} \]

\[ \therefore 2y = \mathbf{14} \implies \therefore y = \mathbf{7} \]

Final Answer:

\[ (x, y) = (5, 7) \] is the solution.

Maharashtra State Board: Class 10

Key Points: Simultaneous Linear Equations

  • Simultaneous equations must be solved together.

  • A solution must satisfy both equations.

  • Substitution replaces one variable using an expression from the other equation.

  • Elimination removes one variable by adding or subtracting suitably modified equations.

Test Yourself

Advertisements
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×