Pair of Linear Equations

A linear equation in two variables is an equation that can be written in the form \(ax + by + c = 0\), where \(a, b, c\) are real numbers, and \(a \neq 0\) or \(b \neq 0\). The graph of a linear equation in two variables is always a straight line.

When we have two such equations, they form a pair of linear equations in two variables:

• \(a_1x + b_1y + c_1 = 0\)
• \(a_2x + b_2y + c_2 = 0\)

Here, \(a_1, b_1, c_1, a_2, b_2, c_2\) are real numbers, and \(a_1^2 + b_1^2 \neq 0\) and \(a_2^2 + b_2^2 \neq 0\).

Representing Situations Algebraically

To represent a real-life situation as a pair of linear equations:

1. Identify the unknowns: Assign variables (e.g., \(x\) and \(y\)) to the quantities you need to find.
2. Formulate equations: Translate the given conditions into two separate linear equations based on the relationships between the variables.

Example: The cost of 1kg potatoes and 2kg tomatoes was ₹30. After two days, the cost of 2kg potatoes and 4kg tomatoes was ₹66.

• Let the cost of 1kg potatoes = ₹\(x\).
• Let the cost of 1kg tomatoes = ₹\(y\).
• Equation 1: \(x + 2y = 30\)
• Equation 2: \(2x + 4y = 66\)

The graphical method involves plotting the two linear equations on a coordinate plane. The solution to the pair of equations is the point of intersection of the two lines.

Steps for Graphical Solution:

1. Find at least two solutions for each equation: For each equation, choose arbitrary values for \(x\) (or \(y\)) and calculate the corresponding \(y\) (or \(x\)) values. These form coordinate pairs \((x, y)\).

• Example: For \(x + 2y = 30\):
• If \(x=0\), \(2y=30 \Rightarrow y=15\) (Point: \((0, 15)\))
• If \(y=0\), \(x=30\) (Point: \((30, 0)\))
• If \(x=10\), \(2y=20 \Rightarrow y=10\) (Point: \((10, 10)\))

1. Plot the points: Mark these coordinate pairs on a graph paper.
2. Draw the lines: Draw a straight line passing through the points for each equation. Extend the lines.
3. Identify the intersection: Observe the point where the two lines intersect. The coordinates of this point \((x, y)\) represent the unique solution to the pair of linear equations.
4. Verify the solution: Substitute the \(x\) and \(y\) values of the intersection point into both original equations to ensure they satisfy both.

Graphical Interpretation of Solutions

The nature of the lines plotted reveals the type of solution:

• Intersecting Lines: If the lines intersect at a single point, the system has a unique solution. This is a consistent system.
• Condition: \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
• Coincident Lines: If the lines overlap completely (one line lies exactly on top of the other), the system has infinitely many solutions. This is a consistent and dependent system.
• Condition: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
• Parallel Lines: If the lines are parallel and do not intersect, the system has no solution. This is an inconsistent system.
• Condition: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

The substitution method is an algebraic technique to solve a pair of linear equations by expressing one variable in terms of the other from one equation and substituting this expression into the second equation.

Steps for Substitution Method:

1. Choose an equation and isolate a variable: Select one of the two equations and express one variable (say, \(y\)) in terms of the other variable (\(x\)). This is often easiest if a variable has a coefficient of 1 or -1.

• Example: Given \(7x - 15y = 2\) (1) and \(x + 2y = 3\) (2).

From (2), we can write \(x = 3 - 2y\) (3).

1. Substitute the expression: Substitute this expression for the chosen variable into the other equation. This will result in a linear equation in only one variable.

• Substitute \(x = 3 - 2y\) into (1):

\(7(3 - 2y) - 15y = 2\)

1. Solve the single-variable equation: Solve the resulting equation to find the value of the remaining variable.

• \(21 - 14y - 15y = 2\)
• \(21 - 29y = 2\)
• \(-29y = 2 - 21\)
• \(-29y = -19\)
• \(y = \frac{-19}{-29} = \frac{19}{29}\)

1. Substitute back to find the other variable: Substitute the value found in step 3 back into the expression from step 1 (or any original equation) to find the value of the first variable.

• Substitute \(y = \frac{19}{29}\) into (3):

\(x = 3 - 2\left(\frac{19}{29}\right)\) \(x = 3 - \frac{38}{29}\) \(x = \frac{3 \times 29 - 38}{29} = \frac{87 - 38}{29} = \frac{49}{29}\)

1. Verify the solution: Check if the values of \(x\) and \(y\) satisfy both original equations.

Solution: \(x = \frac{49}{29}, y = \frac{19}{29}\)

The elimination method involves manipulating the equations (multiplying by constants) so that the coefficients of one variable become numerically equal (or additive inverses). Then, by adding or subtracting the equations, that variable is eliminated, leaving a single-variable equation.

Steps for Elimination Method:

1. Make coefficients equal: Multiply one or both equations by a suitable non-zero constant so that the coefficients of one variable (either \(x\) or \(y\)) are numerically equal.

• Example: Given \(2x + 3y = 9\) (1) and \(3x + 4y = 5\) (2).

To eliminate \(x\), multiply (1) by 3 and (2) by 2:

• \(3 \times (2x + 3y = 9) \Rightarrow 6x + 9y = 27\) (3)
• \(2 \times (3x + 4y = 5) \Rightarrow 6x + 8y = 10\) (4)

1. Add or subtract the equations: If the equal coefficients have the same sign, subtract one equation from the other. If they have opposite signs, add the equations. This eliminates one variable.

• Subtract (4) from (3):

\((6x + 9y) - (6x + 8y) = 27 - 10\) \(6x + 9y - 6x - 8y = 17\) \(y = 17\)

1. Solve the single-variable equation: Solve the resulting equation for the remaining variable.

• We found \(y = 17\).

1. Substitute back to find the other variable: Substitute the value found in step 3 into any of the original equations to find the value of the eliminated variable.

• Substitute \(y = 17\) into (1):

\(2x + 3(17) = 9\) \(2x + 51 = 9\) \(2x = 9 - 51\) \(2x = -42\) \(x = -21\)

1. Verify the solution: Check if the values of \(x\) and \(y\) satisfy both original equations.

Solution: \(x = -21, y = 17\)

Many real-life situations can be modeled and solved using pairs of linear equations. The process involves translating the problem into mathematical form and then applying one of the algebraic methods (substitution or elimination) to find the solution.

General Steps for Word Problems:

1. Read and Understand: Read the problem carefully to identify what is given and what needs to be found.
2. Define Variables: Assign variables (e.g., \(x, y\)) to the unknown quantities. Clearly state what each variable represents.
3. Formulate Equations: Translate the conditions given in the problem into two linear equations based on the relationships between the variables.
4. Choose a Method: Decide whether the substitution method or the elimination method would be more efficient for solving the system of equations. Sometimes, one method is clearly simpler than the other.
5. Solve the Equations: Apply the chosen algebraic method to find the values of the variables.
6. Check and Interpret: Verify your solution by substituting the values back into the original word problem conditions. Ensure the answer makes sense in the context of the problem.

Common Types of Word Problems:

• Age Problems: Involve relationships between ages at different points in time.
• Example: The age of a father is equal to the sum of the ages of his 6 children. After 15 years, twice the age of the father will be the sum of ages of his children. Find the age of the father.
• Cost/Price Problems: Involve costs of multiple items.
• Example: 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and one pen.
• Number Problems: Involve relationships between digits of a number or properties of numbers.
• Geometric Problems: Involve dimensions or properties of geometric figures (e.g., perimeter, area).
• Example: Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
• Distance/Speed/Time Problems: Less common in basic linear equations, but can appear.