Pair of Linear Equations in Two Variables Notes

3

Pair of Linear Equations
in Two Variables

Detailed Theory • Very Easy English • Board Exam Focus

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Chapter Roadmap

1. Linear Equation in Two Variables 2. Pair of Linear Equations 3. Graphical Method 4. Consistency & Number of Solutions 5. Substitution Method 6. Elimination Method 7. Cross-Multiplication Method 8. Word Problems 9. 10 Extra Board Questions 10. Final Revision

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1. Linear Equation in Two Variables

Meaning

A linear equation in two variables is an equation that can be written in the form

\[ \boxed{ax+by+c=0} \] where \(a,b,c\) are real numbers and \(a,b\) are not both zero.

Here \(x\) and \(y\) are the two variables. The highest power of each variable is \(1\).

Examples

• \(2x+3y=7\)
• \(x-y+5=0\)
• \(4x+7y-12=0\)

Not linear in two variables: \(x^2+y=5\), \(xy=6\), \(x+\frac1y=3\), because they do not have the required linear form.

Solution of a Linear Equation

A pair \((x,y)\) is called a solution of the equation if it makes the equation true.

Three Detailed Solved Questions

Question 1. Check whether \((2,1)\) is a solution of \(3x+2y=8\).

Solution:

1. Given equation: \[ 3x+2y=8 \]
2. Given pair is \((2,1)\), so \(x=2\) and \(y=1\).
3. Substitute \(x=2,y=1\): \[ 3(2)+2(1)=8 \]
4. Calculate: \[ 6+2=8 \]
5. Thus the left-hand side equals the right-hand side.

Answer: Yes, \((2,1)\) is a solution.

Question 2. Find two solutions of \(x+2y=6\).

Solution:

1. Given: \[ x+2y=6 \]
2. Choose \(y=0\).
3. Then: \[ x+2(0)=6 \]
4. So \(x=6\).
5. First solution is \((6,0)\).
6. Now choose \(y=1\).
7. Then: \[ x+2(1)=6 \]
8. \[ x+2=6 \]
9. \[ x=4 \]
10. Second solution is \((4,1)\).

Answer: Two solutions are \(\boxed{(6,0),(4,1)}\).

Question 3. Find the value of \(y\) when \(x=3\) in \(2x+y=11\).

Solution:

1. Given: \[ 2x+y=11 \]
2. Given \(x=3\).
3. Substitute: \[ 2(3)+y=11 \]
4. \[ 6+y=11 \]
5. Subtract \(6\) from both sides: \[ y=11-6 \]
6. \[ \boxed{y=5} \]

2. Pair of Linear Equations in Two Variables

Standard Form

A pair of linear equations in two variables can be written as

\[ \boxed{a_1x+b_1y+c_1=0} \] \[ \boxed{a_2x+b_2y+c_2=0} \]

Both equations together form a pair of linear equations in two variables.

Meaning of Common Solution

The common solution is the ordered pair \((x,y)\) that satisfies both equations at the same time.

For example:

\[ x+y=5,\qquad x-y=1 \]

The pair \((3,2)\) satisfies both equations, so it is their common solution.

Three Detailed Solved Questions

Question 1. Check whether \((3,2)\) is the solution of \(x+y=5\) and \(x-y=1\).

1. First equation: \[ x+y=5 \]
2. Substitute \(x=3,y=2\): \[ 3+2=5 \]
3. This is true.
4. Second equation: \[ x-y=1 \]
5. Substitute: \[ 3-2=1 \]
6. This is also true.
7. Therefore the pair satisfies both equations.

Answer: \((3,2)\) is the common solution.

Question 2. Write the standard-form coefficients \(a_1,b_1,c_1,a_2,b_2,c_2\) for \(2x+3y-7=0\) and \(4x-y+5=0\).

1. Compare the first equation with \(a_1x+b_1y+c_1=0\).
2. First equation: \[ 2x+3y-7=0 \]
3. Therefore: \[ a_1=2,\quad b_1=3,\quad c_1=-7 \]
4. Second equation: \[ 4x-y+5=0 \]
5. Therefore: \[ a_2=4,\quad b_2=-1,\quad c_2=5 \]

Answer: \(a_1=2,b_1=3,c_1=-7,a_2=4,b_2=-1,c_2=5\).

Question 3. Determine whether \(3x+2y=7\) and \(6x+4y=14\) represent the same equation.

1. Write the second equation: \[ 6x+4y=14 \]
2. Divide every term by \(2\): \[ \frac{6x}{2}+\frac{4y}{2}=\frac{14}{2} \]
3. Therefore: \[ 3x+2y=7 \]
4. This is exactly the first equation.

Answer: Yes, both equations represent the same line.

3. Graphical Method of Solving a Pair

Basic Idea

Each linear equation in two variables represents a straight line on the coordinate plane.

To solve a pair graphically:

1. Find at least two points for the first equation.
2. Plot those points and draw the first straight line.
3. Find at least two points for the second equation.
4. Plot those points and draw the second straight line.
5. The point where the two lines intersect is the common solution.

Three Possible Graphical Cases

Lines	Point of Intersection	Number of Solutions
Intersecting	One point	One unique solution
Parallel	No common point	No solution
Coincident	Every point common	Infinitely many solutions

Three Detailed Solved Questions

Question 1. Solve graphically: \(x+y=5\) and \(x-y=1\).

1. First equation: \[ x+y=5 \]
2. Choose \(x=0\): \[ 0+y=5\Rightarrow y=5 \] Point: \((0,5)\).
3. Choose \(x=5\): \[ 5+y=5\Rightarrow y=0 \] Point: \((5,0)\).
4. Second equation: \[ x-y=1 \]
5. Choose \(x=1\): \[ 1-y=1\Rightarrow y=0 \] Point: \((1,0)\).
6. Choose \(x=3\): \[ 3-y=1\Rightarrow y=2 \] Point: \((3,2)\).
7. Plot the two pairs of points and draw both straight lines.
8. The lines intersect at \((3,2)\).

Answer: Graphical solution is \(\boxed{(3,2)}\).

Question 2. What is the graphical nature of \(2x+3y=6\) and \(4x+6y=12\)?

1. Write the equations: \[ 2x+3y=6 \] \[ 4x+6y=12 \]
2. Divide the second equation by \(2\): \[ 2x+3y=6 \]
3. Both equations are identical.
4. Therefore, both equations represent the same straight line.
5. Every point on this line satisfies both equations.

Answer: The lines are coincident and there are infinitely many solutions.

Question 3. What happens graphically when two lines are parallel?

1. Parallel lines never meet.
2. Therefore, there is no point common to both lines.
3. A common solution would require a common point.
4. Since no common point exists, no ordered pair satisfies both equations.

Answer: Parallel lines give no solution.

4. Consistency and Number of Solutions

Meaning of Consistent and Inconsistent

A pair of linear equations is called consistent if it has at least one solution.

It is called inconsistent if it has no solution.

Condition for Unique Solution

\[ \boxed{\frac{a_1}{a_2}\neq\frac{b_1}{b_2}} \]

Then the pair has a unique solution.

Condition for No Solution

\[ \boxed{\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}} \]

Then the pair has no solution.

Condition for Infinitely Many Solutions

\[ \boxed{\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}} \]

Then the pair has infinitely many solutions.

Three Detailed Solved Questions

Question 1. Determine the number of solutions of \(2x+3y=7\) and \(4x-y=5\).

1. Write in standard form: \[ 2x+3y-7=0 \] \[ 4x-y-5=0 \]
2. Therefore: \[ a_1=2,\;b_1=3,\;a_2=4,\;b_2=-1 \]
3. Calculate: \[ \frac{a_1}{a_2}=\frac24=\frac12 \]
4. Calculate: \[ \frac{b_1}{b_2}=\frac3{-1}=-3 \]
5. Since: \[ \frac12\neq-3 \]

Answer: The pair has a unique solution.

Question 2. Determine the number of solutions of \(3x+2y=5\) and \(6x+4y=12\).

1. Write standard form: \[ 3x+2y-5=0 \] \[ 6x+4y-12=0 \]
2. Thus: \[ a_1=3,b_1=2,c_1=-5 \] \[ a_2=6,b_2=4,c_2=-12 \]
3. Calculate: \[ \frac{a_1}{a_2}=\frac36=\frac12 \]
4. Calculate: \[ \frac{b_1}{b_2}=\frac24=\frac12 \]
5. Calculate: \[ \frac{c_1}{c_2}=\frac{-5}{-12}=\frac5{12} \]
6. Thus: \[ \frac12=\frac12\neq\frac5{12} \]

Answer: No solution; the pair is inconsistent.

Question 3. Determine the number of solutions of \(2x-3y+4=0\) and \(4x-6y+8=0\).

1. Here: \[ a_1=2,b_1=-3,c_1=4 \]
2. And: \[ a_2=4,b_2=-6,c_2=8 \]
3. Calculate: \[ \frac{a_1}{a_2}=\frac24=\frac12 \]
4. \[ \frac{b_1}{b_2}=\frac{-3}{-6}=\frac12 \]
5. \[ \frac{c_1}{c_2}=\frac48=\frac12 \]
6. All three ratios are equal.

Answer: Infinitely many solutions; the pair is consistent and dependent.

5. Substitution Method

Basic Idea

In the substitution method, we first express one variable in terms of the other variable. Then we substitute this expression into the second equation.

Step-by-Step Method

1. Take either equation.
2. Make one variable the subject.
3. Substitute that expression in the other equation.
4. Solve the resulting one-variable equation.
5. Put the value back into either original equation.
6. Find the second variable.
7. Write the ordered pair as the final answer.

Three Detailed Solved Questions

Question 1. Solve by substitution: \(x+y=7,\;x-y=1\).

1. First equation: \[ x+y=7 \]
2. Make \(x\) the subject: \[ x=7-y \]
3. Second equation: \[ x-y=1 \]
4. Substitute \(x=7-y\): \[ (7-y)-y=1 \]
5. Remove brackets: \[ 7-y-y=1 \]
6. Combine like terms: \[ 7-2y=1 \]
7. Subtract \(7\): \[ -2y=1-7 \]
8. \[ -2y=-6 \]
9. Divide by \(-2\): \[ y=3 \]
10. Put \(y=3\) in \(x=7-y\): \[ x=7-3=4 \]

Answer: \(\boxed{x=4,\;y=3}\), so solution is \(\boxed{(4,3)}\).

Question 2. Solve by substitution: \(2x+3y=13,\;x+y=5\).

1. From the second equation: \[ x+y=5 \]
2. Make \(x\) the subject: \[ x=5-y \]
3. Substitute into the first equation: \[ 2(5-y)+3y=13 \]
4. Multiply: \[ 10-2y+3y=13 \]
5. Combine like terms: \[ 10+y=13 \]
6. Subtract \(10\): \[ y=3 \]
7. Put \(y=3\) into \(x=5-y\): \[ x=5-3=2 \]

Answer: \(\boxed{(2,3)}\).

Question 3. Solve by substitution: \(3x-y=7,\;2x+y=8\).

1. Take: \[ 3x-y=7 \]
2. Make \(y\) the subject: \[ -y=7-3x \]
3. Multiply by \(-1\): \[ y=3x-7 \]
4. Substitute into \(2x+y=8\): \[ 2x+(3x-7)=8 \]
5. Remove brackets: \[ 2x+3x-7=8 \]
6. \[ 5x-7=8 \]
7. Add \(7\): \[ 5x=15 \]
8. Divide by \(5\): \[ x=3 \]
9. Put \(x=3\) into \(y=3x-7\): \[ y=3(3)-7 \]
10. \[ y=9-7=2 \]

Answer: \(\boxed{(3,2)}\).

6. Elimination Method

Basic Idea

In the elimination method, we add or subtract the two equations after making the coefficients of one variable equal in magnitude. This removes that variable.

Step-by-Step Method

1. Write both equations clearly.
2. Make the coefficients of one variable equal if necessary.
3. Add or subtract the equations to eliminate that variable.
4. Solve for the remaining variable.
5. Substitute its value into either original equation.
6. Find the other variable.
7. Verify if needed.

Three Detailed Solved Questions

Question 1. Solve by elimination: \(2x+3y=13,\;3x+2y=12\).

1. Given: \[ 2x+3y=13 \quad ...(1) \] \[ 3x+2y=12 \quad ...(2) \]
2. Multiply (1) by \(3\): \[ 6x+9y=39 \quad ...(3) \]
3. Multiply (2) by \(2\): \[ 6x+4y=24 \quad ...(4) \]
4. Subtract (4) from (3): \[ (6x+9y)-(6x+4y)=39-24 \]
5. \[ 5y=15 \]
6. Divide by \(5\): \[ y=3 \]
7. Put \(y=3\) in (1): \[ 2x+3(3)=13 \]
8. \[ 2x+9=13 \]
9. \[ 2x=4 \]
10. \[ x=2 \]

Answer: \(\boxed{(2,3)}\).

Question 2. Solve by elimination: \(5x+2y=12,\;3x-2y=4\).

1. Add the equations because the \(y\)-coefficients are \(+2\) and \(-2\): \[ (5x+2y)+(3x-2y)=12+4 \]
2. Combine: \[ 8x=16 \]
3. Divide by \(8\): \[ x=2 \]
4. Substitute \(x=2\) in the first equation: \[ 5(2)+2y=12 \]
5. \[ 10+2y=12 \]
6. Subtract \(10\): \[ 2y=2 \]
7. Divide by \(2\): \[ y=1 \]

Answer: \(\boxed{(2,1)}\).

Question 3. Solve by elimination: \(2x+5y=19,\;3x+2y=14\).

1. Given: \[ 2x+5y=19 \quad ...(1) \] \[ 3x+2y=14 \quad ...(2) \]
2. Multiply (1) by \(3\): \[ 6x+15y=57 \quad ...(3) \]
3. Multiply (2) by \(2\): \[ 6x+4y=28 \quad ...(4) \]
4. Subtract (4) from (3): \[ 11y=29 \]
5. Therefore: \[ y=\frac{29}{11} \]
6. Substitute in (2): \[ 3x+2\left(\frac{29}{11}\right)=14 \]
7. \[ 3x+\frac{58}{11}=14 \]
8. Write \(14\) with denominator \(11\): \[ 3x+\frac{58}{11}=\frac{154}{11} \]
9. Subtract: \[ 3x=\frac{154-58}{11}=\frac{96}{11} \]
10. Divide by \(3\): \[ x=\frac{96}{33}=\frac{32}{11} \]

Answer: \(\boxed{\left(\frac{32}{11},\frac{29}{11}\right)}\).

7. Cross-Multiplication Method

Formula

For the pair

\[ a_1x+b_1y+c_1=0 \] \[ a_2x+b_2y+c_2=0 \]

the cross-multiplication formula is

\[ \boxed{ \frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1} } \]

Careful: This formula is used after writing both equations in the standard form \(a_1x+b_1y+c_1=0\).

Three Detailed Solved Questions

Question 1. Solve \(2x+3y-7=0\) and \(4x-y-5=0\) by cross-multiplication.

1. Compare with standard form.
2. Therefore: \[ a_1=2,\;b_1=3,\;c_1=-7 \] \[ a_2=4,\;b_2=-1,\;c_2=-5 \]
3. Use: \[ \frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac1{a_1b_2-a_2b_1} \]
4. For \(x\): \[ b_1c_2-b_2c_1=3(-5)-(-1)(-7) \] \[ =-15-7=-22 \]
5. For \(y\): \[ c_1a_2-c_2a_1=(-7)(4)-(-5)(2) \] \[ =-28+10=-18 \]
6. Denominator: \[ a_1b_2-a_2b_1=2(-1)-4(3) \] \[ =-2-12=-14 \]
7. Thus: \[ \frac{x}{-22}=\frac{1}{-14} \]
8. Cross multiply: \[ -14x=-22 \]
9. \[ x=\frac{22}{14}=\frac{11}{7} \]
10. Similarly: \[ \frac{y}{-18}=\frac1{-14} \]
11. \[ -14y=-18 \]
12. \[ y=\frac{18}{14}=\frac97 \]

Answer: \(\boxed{\left(\frac{11}{7},\frac97\right)}\).

Question 2. Solve \(x+y-5=0\) and \(x-y-1=0\) by cross-multiplication.

1. Write: \[ a_1=1,b_1=1,c_1=-5 \] \[ a_2=1,b_2=-1,c_2=-1 \]
2. Calculate: \[ b_1c_2-b_2c_1=(1)(-1)-(-1)(-5) \] \[ =-1-5=-6 \]
3. Calculate: \[ c_1a_2-c_2a_1=(-5)(1)-(-1)(1) \] \[ =-5+1=-4 \]
4. Calculate: \[ a_1b_2-a_2b_1=(1)(-1)-(1)(1) \] \[ =-1-1=-2 \]
5. Therefore: \[ \frac{x}{-6}=\frac1{-2} \]
6. \[ -2x=-6 \Rightarrow x=3 \]
7. For \(y\): \[ \frac{y}{-4}=\frac1{-2} \]
8. \[ -2y=-4 \Rightarrow y=2 \]

Answer: \(\boxed{(3,2)}\).

Question 3. Solve \(3x+2y-8=0\) and \(x-y-1=0\) by cross-multiplication.

1. Identify: \[ a_1=3,b_1=2,c_1=-8 \] \[ a_2=1,b_2=-1,c_2=-1 \]
2. Calculate \(b_1c_2-b_2c_1\): \[ 2(-1)-(-1)(-8) =-2-8=-10 \]
3. Calculate \(c_1a_2-c_2a_1\): \[ (-8)(1)-(-1)(3) =-8+3=-5 \]
4. Calculate \(a_1b_2-a_2b_1\): \[ 3(-1)-1(2) =-3-2=-5 \]
5. Thus: \[ \frac{x}{-10}=\frac1{-5} \]
6. \[ -5x=-10\Rightarrow x=2 \]
7. And: \[ \frac{y}{-5}=\frac1{-5} \]
8. \[ -5y=-5\Rightarrow y=1 \]

Answer: \(\boxed{(2,1)}\).

8. Word Problems Based on Pair of Linear Equations

General Strategy

Word problems look difficult mainly because the information is written in words. Convert each statement into an equation carefully.

1. Let the unknown quantities be \(x\) and \(y\).
2. Translate the first condition into one equation.
3. Translate the second condition into another equation.
4. Choose substitution or elimination.
5. Solve completely.
6. Write the answer with units or a proper statement.

Three Detailed Solved Questions

Question 1. The sum of two numbers is 27 and their difference is 5. Find the numbers.

1. Let the first number be \(x\).
2. Let the second number be \(y\).
3. Sum is \(27\), so: \[ x+y=27 \quad ...(1) \]
4. Difference is \(5\), so: \[ x-y=5 \quad ...(2) \]
5. Add (1) and (2): \[ (x+y)+(x-y)=27+5 \]
6. \[ 2x=32 \]
7. \[ x=16 \]
8. Put \(x=16\) in (1): \[ 16+y=27 \]
9. \[ y=11 \]
10. Check: \[ 16+11=27 \] and \[ 16-11=5 \]

Answer: The numbers are \(\boxed{16\text{ and }11}\).

Question 2. The cost of 2 pens and 3 pencils is ₹31. The cost of 3 pens and 2 pencils is ₹29. Find the cost of one pen and one pencil.

1. Let the cost of one pen be ₹\(x\).
2. Let the cost of one pencil be ₹\(y\).
3. Cost of 2 pens and 3 pencils: \[ 2x+3y=31 \quad ...(1) \]
4. Cost of 3 pens and 2 pencils: \[ 3x+2y=29 \quad ...(2) \]
5. Multiply (1) by \(3\): \[ 6x+9y=93 \quad ...(3) \]
6. Multiply (2) by \(2\): \[ 6x+4y=58 \quad ...(4) \]
7. Subtract (4) from (3): \[ 5y=35 \]
8. \[ y=7 \]
9. Put \(y=7\) in (1): \[ 2x+3(7)=31 \]
10. \[ 2x+21=31 \]
11. \[ 2x=10 \]
12. \[ x=5 \]

Answer: One pen costs ₹\(\boxed5\), one pencil costs ₹\(\boxed7\).

Question 3. A two-digit number has digits whose sum is 9. When the digits are reversed, the new number is 27 less than the original number. Find the number.

1. Let the tens digit be \(x\).
2. Let the units digit be \(y\).
3. Sum of digits is \(9\): \[ x+y=9 \quad ...(1) \]
4. The original number is: \[ 10x+y \]
5. After reversing the digits, the new number is: \[ 10y+x \]
6. The reversed number is \(27\) less than the original: \[ 10y+x=10x+y-27 \]
7. Bring like terms together: \[ 10y-y+x-10x=-27 \]
8. \[ 9y-9x=-27 \]
9. Divide by \(9\): \[ y-x=-3 \]
10. Therefore: \[ x-y=3 \quad ...(2) \]
11. Add (1) and (2): \[ (x+y)+(x-y)=9+3 \]
12. \[ 2x=12 \]
13. \[ x=6 \]
14. Put \(x=6\) in (1): \[ 6+y=9 \]
15. \[ y=3 \]
16. Original number: \[ 10x+y=10(6)+3=63 \]
17. Reversed number is \(36\), and \(63-36=27\), so the answer is verified.

Answer: The number is \(\boxed{63}\).

9. 10 Extra Fully Detailed Board-Style Questions

Question 1. Solve \(2x+3y=11\) and \(3x-2y=4\).

1. Multiply first equation by \(2\): \[ 4x+6y=22 \]
2. Multiply second equation by \(3\): \[ 9x-6y=12 \]
3. Add: \[ 13x=34 \]
4. \[ x=\frac{34}{13} \]
5. Substitute in \(3x-2y=4\): \[ 3\left(\frac{34}{13}\right)-2y=4 \]
6. \[ \frac{102}{13}-2y=\frac{52}{13} \]
7. \[ -2y=-\frac{50}{13} \]
8. \[ y=\frac{25}{13} \]

Answer: \(\boxed{\left(\frac{34}{13},\frac{25}{13}\right)}\).

Question 2. Solve \(5x-3y=1\) and \(2x+y=9\).

1. From second equation: \[ y=9-2x \]
2. Substitute in first: \[ 5x-3(9-2x)=1 \]
3. \[ 5x-27+6x=1 \]
4. \[ 11x-27=1 \]
5. \[ 11x=28 \]
6. \[ x=\frac{28}{11} \]
7. Now: \[ y=9-2\left(\frac{28}{11}\right) \]
8. \[ y=\frac{99}{11}-\frac{56}{11} =\frac{43}{11} \]

Answer: \(\boxed{\left(\frac{28}{11},\frac{43}{11}\right)}\).

Question 3. Find \(k\) so that \(kx+3y=5\) and \(12x+ky=10\) have a unique solution.

1. For a unique solution: \[ \frac{a_1}{a_2}\neq\frac{b_1}{b_2} \]
2. Here: \[ a_1=k,\quad b_1=3,\quad a_2=12,\quad b_2=k \]
3. Therefore: \[ \frac{k}{12}\neq\frac3k \]
4. Cross multiply: \[ k^2\neq36 \]
5. Thus: \[ k\neq\pm6 \]

Answer: \(\boxed{k\neq6,-6}\).

Question 4. Find \(k\) so that \(2x+3y=7\) and \(4x+ky=14\) have infinitely many solutions.

1. For infinitely many solutions: \[ \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \]
2. Write equations in standard form: \[ 2x+3y-7=0 \] \[ 4x+ky-14=0 \]
3. Compare: \[ a_1=2,b_1=3,c_1=-7 \] \[ a_2=4,b_2=k,c_2=-14 \]
4. First ratio: \[ \frac{a_1}{a_2}=\frac24=\frac12 \]
5. For infinitely many solutions: \[ \frac3k=\frac12 \]
6. Cross multiply: \[ 6=k \]
7. Therefore: \[ \boxed{k=6} \]

Question 5. Find \(k\) so that \(3x+ky=8\) and \(6x+4y=10\) have no solution.

1. For no solution: \[ \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2} \]
2. Here: \[ a_1=3,b_1=k,c_1=-8 \] \[ a_2=6,b_2=4,c_2=-10 \]
3. Set the first two ratios equal: \[ \frac36=\frac{k}{4} \]
4. \[ \frac12=\frac{k}{4} \]
5. Cross multiply: \[ 4=2k \]
6. \[ k=2 \]
7. Check the third ratio: \[ \frac{c_1}{c_2}=\frac{-8}{-10}=\frac45 \]
8. But: \[ \frac12\neq\frac45 \]

Answer: \(\boxed{k=2}\).

Question 6. The sum of two numbers is 40 and one number is 8 more than the other. Find them.

1. Let the numbers be \(x\) and \(y\).
2. Sum: \[ x+y=40 \]
3. One is 8 more: \[ x-y=8 \]
4. Add: \[ 2x=48 \]
5. \[ x=24 \]
6. Substitute: \[ 24+y=40 \]
7. \[ y=16 \]

Answer: \(\boxed{24,16}\).

Question 7. A fraction becomes \(\frac13\) when 1 is subtracted from its numerator and it becomes \(\frac14\) when 8 is added to its denominator. Find the fraction.

1. Let numerator be \(x\) and denominator be \(y\).
2. Original fraction is: \[ \frac{x}{y} \]
3. After subtracting \(1\) from numerator: \[ \frac{x-1}{y}=\frac13 \]
4. Cross multiply: \[ 3x-3=y \]
5. Therefore: \[ 3x-y=3 \quad ...(1) \]
6. After adding \(8\) to denominator: \[ \frac{x}{y+8}=\frac14 \]
7. Cross multiply: \[ 4x=y+8 \]
8. Therefore: \[ 4x-y=8 \quad ...(2) \]
9. Subtract (1) from (2): \[ (4x-y)-(3x-y)=8-3 \]
10. \[ x=5 \]
11. Put \(x=5\) in (1): \[ 3(5)-y=3 \]
12. \[ 15-y=3 \]
13. \[ y=12 \]

Answer: The fraction is \(\boxed{\frac5{12}}\).

Question 8. The present age of a father is three times the age of his son. After 10 years, the father's age will be twice the son's age. Find their present ages.

1. Let son's present age be \(x\) years.
2. Let father's present age be \(y\) years.
3. Father is three times son: \[ y=3x \]
4. After 10 years: \[ y+10=2(x+10) \]
5. Expand: \[ y+10=2x+20 \]
6. Rearrange: \[ y-2x=10 \]
7. Substitute \(y=3x\): \[ 3x-2x=10 \]
8. \[ x=10 \]
9. Therefore: \[ y=3(10)=30 \]

Answer: Son \(=\boxed{10}\) years, Father \(=\boxed{30}\) years.

Question 9. The cost of 5 notebooks and 3 pens is ₹115, while 3 notebooks and 5 pens cost ₹101. Find the cost of each.

1. Let one notebook cost ₹\(x\).
2. Let one pen cost ₹\(y\).
3. First condition: \[ 5x+3y=115 \quad ...(1) \]
4. Second condition: \[ 3x+5y=101 \quad ...(2) \]
5. Multiply (1) by \(5\): \[ 25x+15y=575 \quad ...(3) \]
6. Multiply (2) by \(3\): \[ 9x+15y=303 \quad ...(4) \]
7. Subtract: \[ 16x=272 \]
8. \[ x=17 \]
9. Put \(x=17\) in (1): \[ 5(17)+3y=115 \]
10. \[ 85+3y=115 \]
11. \[ 3y=30 \]
12. \[ y=10 \]

Answer: Notebook ₹\(\boxed{17}\), Pen ₹\(\boxed{10}\).

Question 10. Solve \(7x-3y=11\) and \(4x+5y=3\), and verify the answer.

1. Given: \[ 7x-3y=11 \quad ...(1) \] \[ 4x+5y=3 \quad ...(2) \]
2. Multiply (1) by \(5\): \[ 35x-15y=55 \quad ...(3) \]
3. Multiply (2) by \(3\): \[ 12x+15y=9 \quad ...(4) \]
4. Add (3) and (4): \[ 47x=64 \]
5. \[ x=\frac{64}{47} \]
6. Put this in (2): \[ 4\left(\frac{64}{47}\right)+5y=3 \]
7. \[ \frac{256}{47}+5y=\frac{141}{47} \]
8. \[ 5y=\frac{141-256}{47} =-\frac{115}{47} \]
9. \[ y=-\frac{115}{235}=-\frac{23}{47} \]
10. Verify in (1): \[ 7\left(\frac{64}{47}\right)-3\left(-\frac{23}{47}\right) \]
11. \[ =\frac{448}{47}+\frac{69}{47} =\frac{517}{47} \]
12. \[ \frac{517}{47}=11 \]
13. Verify in (2): \[ 4\left(\frac{64}{47}\right)+5\left(-\frac{23}{47}\right) \]
14. \[ =\frac{256-115}{47} =\frac{141}{47}=3 \]

Answer: \(\boxed{\left(\frac{64}{47},-\frac{23}{47}\right)}\), verified.

10. Final 96% Target Revision

Most Important Forms

\[ \boxed{a_1x+b_1y+c_1=0} \qquad \boxed{a_2x+b_2y+c_2=0} \]

Number of Solutions

\[ \boxed{\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\Rightarrow\text{unique solution}} \] \[ \boxed{\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\Rightarrow\text{no solution}} \] \[ \boxed{\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\Rightarrow\text{infinitely many solutions}} \]

Graph Meaning

Graph	Solution	Type
Intersecting lines	One point	Unique solution
Parallel lines	No point	No solution
Coincident lines	Every point	Infinitely many solutions

Board-Exam Writing Checklist

1. Write both equations clearly and number them as (1) and (2).
2. For substitution, show how one variable is isolated.
3. For elimination, show the multiplication of complete equations.
4. Do not skip sign changes while subtracting equations.
5. For cross multiplication, first convert to standard form.
6. For word problems, define variables with units.
7. Write the final answer as an ordered pair or a proper statement.
8. Check the answer in the original equations whenever possible.

Common Mistakes to Avoid

• Comparing ratios without first putting equations in standard form.
• Using \(c_1/c_2\) with the wrong signs.
• Forgetting to multiply every term of an equation when using elimination.
• Making an error while subtracting negative terms.
• Giving only \(x\) and forgetting to find \(y\).
• In word problems, writing the wrong expression for a two-digit number.
• Forgetting units such as rupees or years in final answers.

96% Marks Target – Practise These Most

• Graphical solution and interpretation of intersecting, parallel and coincident lines.
• Conditions for unique, no and infinitely many solutions.
• Substitution method with complete working.
• Elimination method, especially questions requiring multiplication.
• Cross-multiplication method and sign accuracy.
• Finding unknown parameters \(k\).
• Age, number, fraction and money word problems.
• Verification of obtained solutions.

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Class 10 Mathematics • Chapter 3: Pair of Linear Equations in Two Variables

Detailed handwritten-style notes for simple learning and board-exam preparation.