Pair of Linear Equations in Two Variables — Class 10 Maths Notes

Topics covered in Pair of Linear Equations in Two Variables

1. 1.Introduction of Pair of Linear Equations in Two Variables

   Imagine you and a friend are trying to guess the number of chocolates and cookies in a box without opening it. You make a guess, like, "If there are 2 chocolates, then there must be 3 cookies." Your friend makes another guess, like, "If there are 5 chocolates, then there are 4 cookies." These guesses are like linear equations because they relate two things (chocolates and cookies) in a simple way.

   In mathematics, a linear equation in two variables looks like this: +=ax+by=c, where a, b, and c are numbers, and x and y are the variables (like chocolates and cookies in our example).

   When you have two such equations, it's called a "Pair of Linear Equations in Two Variables." For example:

   1. 2+3=62x+3y=6
   2. 5+4=205x+4y=20

   The exciting part is finding out the exact number of chocolates and cookies (or the values of x and y) that make both your guess and your friend's guess true. In math, this is like solving the equations to find the values of x and y.

   Real-Life Example:

   Suppose you're at a shop where they sell pens and pencils in packs. One pack with 2 pens and 3 pencils costs 50 rupees, and another pack with 5 pens and 4 pencils costs 110 rupees. To find out the price of one pen and one pencil, you can set up two equations and solve them. This is like solving a pair of linear equations.

   Use in Careers:

   This concept is used in various fields like economics (to calculate supply and demand), in engineering (for designing systems or solving problems), and even in everyday life for budgeting or planning.

2. 2.Solving Pair of Linear Equation Graphically

   Solving a pair of linear equations graphically is like drawing two paths and seeing where they meet or if they ever meet at all. This method visually represents the solutions of the equations. Let's look at the types of solutions we can encounter:

   Types of Solutions:

   1. Unique Solution (Intersecting Lines):

      • What Happens: The lines representing the equations intersect at a single point.
      • Meaning: The coordinates of this point of intersection are the solution to both equations.
      • Example: Equations like =2+3y=2x+3 and =−+1y=−x+1 might intersect at a point, say (2,7). This means =2x=2 and =7y=7 is the solution.

   2. No Solution (Parallel Lines):

      • What Happens: The lines are parallel and never intersect.
      • Meaning: There is no point that satisfies both equations simultaneously.
      • Example: Equations like =2+3y=2x+3 and =2−4y=2x−4 are parallel because they have the same slope but different y-intercepts.

   3. Infinite Solutions (Coincident Lines):

      • What Happens: The lines representing the equations are coincident, meaning they completely overlap each other.
      • Meaning: Every point on the line is a solution to both equations.
      • Example: Equations like =2+3y=2x+3 and 2=4+62y=4x+6 are essentially the same line, so they have infinitely many solutions.

   Real-Life Example:

   Imagine you're trying to find a meeting point with a friend in a park. The paths you take are like the equations.

   1. Unique Solution: You both agree to meet at a specific spot (the intersection).
   2. No Solution: You both walk on paths that never cross (parallel lines).
   3. Infinite Solutions: You both walk on the same path (coincident lines), so there are many spots where you could meet.

   Graphically solving linear equations provides a clear visual representation, making it easier to understand the nature of the solutions.

3. 3.Check graphically whether the pair of equations x + 3y = 6 (1) and 2x – 3y = 12 (2)

   To check graphically whether the pair of equations +3=6x+3y=6 and 2−3=122x−3y=12 have a solution, we need to plot both equations on the same set of axes and see where the lines intersect.

   Here's how we would do it step by step:

   1. Rearrange the equations into slope-intercept form (y = mx + b):

      • For equation (1), +3=6x+3y=6, solve for y: 3=−+63y=−x+6 =−13+2y=−31​x+2

      • For equation (2), 2−3=122x−3y=12, solve for y: −3=−2+12−3y=−2x+12 =23−4y=32​x−4

   2. Plot the y-intercept of each equation:

      • For equation (1), the y-intercept is at =2y=2.
      • For equation (2), the y-intercept is at =−4y=−4.

   3. Use the slope to find another point for each line:

      • For equation (1), the slope is −13−31​, which means for every step left or right on the x-axis, the line goes down or up by 1/3 step on the y-axis, respectively.
      • For equation (2), the slope is 2332​, which means for every step right on the x-axis, the line goes up 2/3 step on the y-axis.

   4. Draw the lines:

      • Draw a line through the points found for each equation.

   5. Look for the intersection:

      • The point where the two lines intersect will be the solution to the system of equations. If they do not intersect, then there is no solution.

        Let's create the graph for these two equations to see their intersection.

        The graph shows the lines representing the equations +3=6x+3y=6 and 2−3=122x−3y=12. We can observe that the lines intersect at a specific point on the graph. The coordinates of this point are the solution to the pair of equations, meaning that there is exactly one solution to this system, and thus the pair of equations is consistent and independent.

4. 4.Champa went to a ‘Sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased......

   Question:- Champa went to a ‘Sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased”. Help her friends to find how many pants and skirts Champa bought. Let's solve the problem step by step and then represent the solution graphically.

   Step 1: Translate the Statements into Equations

   Let's denote the number of pants Champa bought as p and the number of skirts as s.

   From Champa's statements, we get the following two equations:

   1. The number of skirts is two less than twice the number of pants: =2−2s=2p−2 (Equation 1)

   2. The number of skirts is four less than four times the number of pants: =4−4s=4p−4 (Equation 2)

   Step 2: Solve the System of Equations

   Now, we will solve these equations to find the values of p and s.

   Since both equations equal s, we can set them equal to each other: 2−2=4−42p−2=4p−4

   Solving for p: 2−4=−4+2
   2p−4p=−4+2 −2=−2
   −2p=−2 =1
   p=1

   Now we'll substitute =1p=1 into either equation to find s. Using Equation 1: =2(1)−2
   s=2(1)−2 =2−2s=2−2 =0s=0

   This suggests Champa bought 1 pant and 0 skirts, but this doesn't seem correct as it implies she didn't buy any skirts when she actually did. So, let's use Equation 2 to see if we made a mistake: =4(1)−4
   s=4(1)−4 =4−4s=4−4 =0s=0

   We get the same answer, but we know Champa bought skirts, so the equations or the statements might be incorrect or misinterpreted.

   Step 3: Graphical Representation

   Despite the logical inconsistency, we can still graph the equations to see their intersection. Let's plot the graphs of the two equations.

   The graph depicts the equations derived from Champa's statements:

   • The blue line represents the equation =2−2s=2p−2 (Equation 1).
   • The orange line represents the equation =4−4s=4p−4 (Equation 2).

   The red dot indicates the point of intersection based on our algebraic solution, where =1p=1 and =0s=0.

   However, as previously discussed, the solution that Champa bought 1 pant and 0 skirts does not align with the context of the problem, since we are expecting her to have purchased at least one skirt. This suggests a potential issue with the information provided in the problem statement. To resolve the contradiction, we would need additional or corrected information about Champa's purchase. ​

5. 5.On comparing the ratios a1/a2, b1/b2 and c1/c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

   (i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0 (ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
   

   To find out whether the lines representing the pairs of linear equations intersect at a point, are parallel, or coincident, we can compare the ratios of the coefficients of x (which is 1/2a1​/a2​), y (which is 1/2b1​/b2​), and the constant terms (which is 1/2c1​/c2​).

   For the pair (i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0:

   • The ratio of the coefficients of x is 1/2=5/7a1​/a2​=5/7.
   • The ratio of the coefficients of y is 1/2=−4/6=−2/3b1​/b2​=−4/6=−2/3.
   • Since 1/2≠1/2a1​/a2​=b1​/b2​, the lines are not parallel and not coincident. Therefore, they will intersect at a point.

   For the pair (ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0:

   • The ratio of the coefficients of x is 1/2=9/18=1/2a1​/a2​=9/18=1/2.
   • The ratio of the coefficients of y is 1/2=3/6=1/2b1​/b2​=3/6=1/2.
   • The ratio of the constant terms is 1/2=12/24=1/2c1​/c2​=12/24=1/2.
   • Since 1/2=1/2=1/2a1​/a2​=b1​/b2​=c1​/c2​, the lines are coincident, which means they lie on top of each other, and hence have infinitely many solutions.

   Now, let's represent these pairs of equations graphically to visually confirm our findings.

   For the first pair, we expect to see the lines intersect at a point, and for the second pair, we expect the lines to coincide. Let's plot them:

   The graphical representation shows two sets of linear equations:

   • For Pair 1 (on the left), the lines corresponding to the equations 5–4+8=05x–4y+8=0 and 7+6–9=07x+6y–9=0 are depicted. They intersect at a point, confirming that there is a unique solution to this pair of equations.

   • For Pair 2 (on the right), the lines corresponding to the equations 9+3+12=09x+3y+12=0 and 18+6+24=018x+6y+24=0 are depicted. They are coincident, which means they lie on top of each other, confirming that there are infinitely many solutions to this pair of equations.

   In the graphical representation, the intersection point for Pair 1 is where the unique solution for that system of equations is located. The coinciding lines for Pair 2 illustrate that every point on the line is a solution for the system of equations, thereby having infinitely many solutions.

6. 6.On comparing the ratios a1/a2, b1/b2 and c1/c2 , find out whether the following pair of linear equations are consistent, or inconsistent.

   (i) 3x + 2y = 5; 2x – 3y = 7
   (ii) 2x – 3y = 8; 4x – 6y = 9
   (iii) 3/2 x + 5/3 y = 7; 9x – 10y = 14
   
   

   To determine whether the given pairs of linear equations are consistent or inconsistent, we can compare the ratios of the coefficients of x (denoted by a), y (denoted by b), and the constant term (denoted by c) from both equations. Let's analyze each pair:

   Pair 1:

   Equation 1: 3+2=5Equation 1: 3x+2y=5 Equation 2: 2−3=7Equation 2: 2x−3y=7

   Ratios: 12=32,12=2−3,12=57a2​a1​​=23​,b2​b1​​=−32​,c2​c1​​=75​

   Since 12≠12a2​a1​​=b2​b1​​, the lines are not parallel and thus will intersect at a point. This pair of equations is consistent because they have a unique solution.

   Pair 2:

   Equation 1: 2−3=8Equation 1: 2x−3y=8 Equation 2: 4−6=9Equation 2: 4x−6y=9

   Ratios: 12=24,12=−3−6,12=89a2​a1​​=42​,b2​b1​​=−6−3​,c2​c1​​=98​

   Here, 12=12a2​a1​​=b2​b1​​ but 12≠12c2​c1​​=a2​a1​​, which means the lines are parallel and do not intersect. This pair of equations is inconsistent because there is no solution that satisfies both equations.

   Pair 3:

   Equation 1: 32+53=7Equation 1: 23​x+35​y=7 Equation 2: 9−10=14Equation 2: 9x−10y=14

   For simplicity in comparison, let's multiply the coefficients of Equation 1 by 6 to remove the fractions: Equation 1 (modified): 9+10=42Equation 1 (modified): 9x+10y=42

   Ratios: 12=99,12=10−10,12=4214a2​a1​​=99​,b2​b1​​=−1010​,c2​c1​​=1442​

   Here, 12=12=−1a2​a1​​=b2​b1​​=−1 and 12=3c2​c1​​=3. The ratios of the coefficients of x and y are equal but opposite in sign, and the ratio of the constants does not match the ratio of the coefficients. This means the lines are not parallel and not coincident; thus, they will intersect at a point. This pair of equations is consistent and has a unique solution.

7. 7.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.

   To find the dimensions of the garden, we can use the given information to set up an equation. The problem states that half the perimeter of the garden is 36 meters, and the length of the garden is 4 meters more than its width.

   Let's denote the width of the garden as w meters. Therefore, the length of the garden would be +4w+4 meters.

   The perimeter of a rectangle is given by the formula: Perimeter=2×(length+width)Perimeter=2×(length+width)

   Half of the perimeter is given as 36 meters, so: 12×Perimeter=3621​×Perimeter=36 Perimeter=36×2Perimeter=36×2 Perimeter=72Perimeter=72

   Now substitute the expressions for length and width into the perimeter formula: 72=2×(+(+4))72=2×(w+(w+4))

   Solving this equation will give us the width, and we can then find the length. Let's do the math:

   The dimensions of the garden are as follows:

   • Width (w) = 16 meters
   • Length (+4w+4) = 20 meters

   So, the garden is 16 meters wide and 20 meters long.

8. 8.Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

   
   

   The graph above shows the equations −+1=0x−y+1=0 and 3+2−12=03x+2y−12=0, along with the triangular region they form with the x-axis. The shaded area represents this triangular region.

   The coordinates of the vertices of the triangle are:

   1. (0,0)(0,0) - Origin, where both lines intersect the x-axis.
   2. (2,0)(2,0) - The x-coordinate of the intersection point of the two lines with the x-axis.
   3. (2,3)(2,3) - The intersection point of the two lines.

   This graphically illustrates the triangular area formed by these two lines and the x-axis.

9. 9.Algebraic Methods of Solving a Pair of Linear Equations.

   Algebraic methods are used to solve a pair of linear equations. These equations look like this:

   +=ax+by=c +=dx+ey=f

   Where a, b, c, d, e, and f are constants, and x and y are variables we want to find.

   There are three main algebraic methods:

   1. Substitution Method:

      • Solve one equation for one variable (say x).
      • Substitute this expression for x in the other equation.
      • Solve the substituted equation for y.
      • Use the value of y to find x.

   2. Elimination Method:

      • Multiply each equation by a suitable number so that the coefficients of one variable (either x or y) become the same.
      • Add or subtract the equations to eliminate one variable.
      • Solve for the remaining variable.
      • Substitute this value into one of the original equations to find the other variable.

   3. Cross-Multiplication Method:

      • Used when equations are in the form 1+1=1xa1​​+yb1​​=c1​ and 2+2=2xa2​​+yb2​​=c2​.
      • Multiply crosswise and find the values of x and y.

   Example with Verification:

   Consider the equations: +=10x+y=10 2−=32x−y=3

   Let's solve this using the substitution method:

   1. From the first equation, =10−x=10−y.
   2. Substitute this in the second equation: 2(10−)−=32(10−y)−y=3
   3. Simplify and solve for y: 20−2−=320−2y−y=3 20−3=320−3y=3 3=173y=17 =173y=317​
   4. Substitute y in =10−x=10−y: =10−173=30−173=133x=10−317​=330−17​=313​

   So, =133x=313​ and =173y=317​.

   Verification:

   Substitute these values back into the original equations:

   1. +=133+173=10x+y=313​+317​=10 (True)
   2. 2−=2×133−173=32x−y=2×313​−317​=3 (True)

   Since both equations hold true, our solution is correct.

   Imagine you're planning a party. You want to buy some chairs and tables. One chair costs ₹100, and one table costs ₹300. You have a budget of ₹7000 and need to figure out how many of each you can buy. This can be modeled as a pair of linear equations:

   Let x be the number of chairs, and y be the number of tables. So, we have: 100+300=7000100x+300y=7000
   += specific number based on space
   x+y=a specific number based on space

   You can use algebraic methods to find out the number of chairs and tables you can afford.

   Career Applications:

   1. Economics and Business:

      • Economists use linear equations to model supply and demand, pricing strategies, and financial forecasting.
      • In business, linear equations help in budgeting, financial analysis, and market research.

   2. Engineering:

      • Engineers use linear equations to design and analyze systems and structures, like calculating forces in a bridge or the electrical current in circuits.

   3. Computer Science:

      • Algorithms often involve solving linear equations. They're used in machine learning for predictions and in game development for creating realistic movements.

   4. Architecture and Construction:

      • Architects and construction managers use linear equations for designing buildings and planning construction projects, ensuring the right balance of materials and costs.

   5. Healthcare:

      • In healthcare, linear equations are used for calculating medicine dosages and predicting the spread of diseases.

   Understanding linear equations thus opens up a world of possibilities in various careers, showing the practical application of mathematics in real-world scenarios.

10. 10.Solve the following pair of equations by substitution method: 7x – 15y = 2 (1), x + 2y = 3 (2)

    Let's solve these equations using the substitution method:

    Given equations:

    1. 7−15=27x−15y=2 (Equation 1)
    2. +2=3 x+2y=3 (Equation 2)

    Step 1: Solve one equation for one variable. Let's solve Equation 2 for x: =3−2x=3−2y

    Step 2: Substitute this expression into the other equation. Substitute =3−2x=3−2y into Equation 1: 7(3−2)−15=27(3−2y)−15y=2

    Step 3: Simplify and solve for y. 21−14−15=221−14y−15y=2
    −29=2−21−29y=2−21
    −29=−19−29y=−19
    =−19−29y=−29−19​ =1929y=2919​

    Step 4: Substitute y back into the expression for x. =3−2×1929x=3−2×2919​ =3−3829
    x=3−2938​ =8729−3829
    x=2987​−2938​ =4929
    x=2949​

    So, the solution is =4929x=2949​ and =1929y=2919​.

    Verification:

    1. Substitute x and y into Equation 1: 7×4929−15×1929=27×2949​−15×2919​=2
       34329−28529=229343​−29285​=2
       5829=22958​=2
       2=22=2 (True)

    2. Substitute x and y into Equation 2: 4929+2×1929=32949​+2×2919​=3
       4929+3829=32949​+2938​=3
       8729=32987​=3
       3=33=3 (True)

    Since both equations are satisfied, our solution is correct.

11. 11.In a shop the cost of 2 pencils and 3 erasers is Rs.9 and the cost of 4 pencils and 6 erasers is Rs.18. Find the cost of each pencil and each eraser.

    To solve this problem, we'll use a system of linear equations. Let's define:

    • The cost of one pencil as p.
    • The cost of one eraser as e.

    The problem gives us two conditions:

    1. The cost of 2 pencils and 3 erasers is ₹9. So, we can write this as: 2+3=92p+3e=9.
    2. The cost of 4 pencils and 6 erasers is ₹18. This gives us the equation: 4+6=184p+6e=18.

    Now, let's solve these equations:

    Equation 1: 2+3=92p+3e=9 Equation 2: 4+6=184p+6e=18

    We can notice that Equation 2 is just twice Equation 1. This means that both equations actually represent the same relationship between the cost of pencils and erasers. Therefore, we need another independent equation to find a unique solution for p and e.

    However, we can still try to find a relationship between the cost of pencils and erasers from the given equation. Let's simplify Equation 2 by dividing it by 2:

    2+3=92p+3e=9

    This is the same as Equation 1, indicating that any combination of p and e that satisfies one equation will satisfy the other.

    To find specific values for p and e, we would need additional information, such as the cost of either the pencil or the eraser separately. Without this, there are infinitely many solutions that satisfy these equations. For instance, if a pencil is ₹1, an eraser would be ₹3; if a pencil is ₹2, an eraser would be ₹2, and so on, as long as the equation 2+3=92p+3e=9 is satisfied.

12. 12.Solve the following pair of linear equations by the substitution method.

    (i) x + y = 14, x – y = 4 (ii) s – t = 3, s/3 t/2 = 6 (iii) √ 2x + √3y = 0, √3x - √8y = 0
    

    Let's solve these pairs of linear equations using the substitution method:

    (i) +=14,−=4x+y=14,x−y=4

    1. Solve the first equation for x: =14− x=14−y

    2. Substitute this into the second equation: 14−−=414−y−y=4 14−2=414−2y=4 2=102y=10 =5y=5

    3. Now substitute =5y=5 back into =14−x=14−y: =14−5=9x=14−5=9

    So, =9x=9 and =5y=5.

    (ii) −=3,3+2=6s−t=3,3s​+2t​=6

    1. Solve the first equation for s: =+3s=t+3

    2. Substitute this into the second equation: +33+2=63t+3​+2t​=6 Multiply through by 6 (the least common multiple of 2 and 3) to clear the fractions: 2(+3)+3=362(t+3)+3t=36 2+6+3=362t+6+3t=36 5=305t=30 =6t=6

    3. Substitute =6t=6 back into =+3s=t+3: =6+3=9s=6+3=9

    So, =9s=9 and =6t=6.

    (iii) 2+3=0,3−8=02​x+3​y=0,3​x−8​y=0

    1. Solve the first equation for x: 2=−32​x=−3​y =−32
       x=−2​3​​y

    2. Substitute this into the second equation: 3(−32)−8=03​(−2​3​​y)−8​y=0
       −32−8=0−2​3​y−8​y=0 Multiply through by 22​ to clear the fraction: −3−22=0−3y−22​y=0 (−3−22)=0y(−3−22​)=0

    3. Since (−3−22)=0y(−3−22​)=0, either =0y=0 or the term in parentheses is zero. The latter can't be true, so =0y=0.

    4. Substitute =0y=0 back into the first equation: 2+3(0)=02​x+3​(0)=0
       2=02​x=0
       =0x=0

    So, =0x=0 and =0y=0.

13. 13.Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

    Let's solve the given pair of linear equations first, and then find the value of 'm' in the equation =+3y=mx+3.

    Solving the Equations

    Given equations are:

    1. 2+3=112x+3y=11 (Equation 1)
    2. 2−4=−242x−4y=−24 (Equation 2)

    Let's use the elimination method to solve them:

    Step 1: Make the coefficients of x equal in both equations.

    The coefficients of x are already equal, so we can proceed to the next step.

    Step 2: Subtract Equation 2 from Equation 1:

    (2+3)−(2−4)=11−(−24)(2x+3y)−(2x−4y)=11−(−24) 2+3−2+4=11+242x+3y−2x+4y=11+24 7=357y=35 =5y=5

    Step 3: Substitute the value of y into one of the original equations to find x:

    Using Equation 1: 2+3(5)=112x+3(5)=11 2+15=112x+15=11 2=−42x=−4 =−2x=−2

    So, the solution of the given system of equations is =−2x=−2 and =5y=5.

    Finding the Value of 'm'

    Now, we need to find 'm' in the equation =+3y=mx+3 using our found values of x and y.

    Substituting =−2x=−2 and =5y=5 into =+3y=mx+3: 5=(−2)+35=m(−2)+3 5=−2+35=−2m+3 −2=5−3−2m=5−3 −2=2−2m=2 =−1m=−1

    Hence, the value of 'm' for which =+3y=mx+3 is =−1m=−1.

14. 14.Form the pair of linear equations for the following problems and find their solution by substitution method.

    (i) The difference between two numbers is 26 and one number is three times the other. Find them. Difference Between Two Numbers

    Let the two numbers be x and y, with x being the larger number. We have:

    1. Their difference is 26: −=26x−y=26.
    2. One number is three times the other: =3x=3y.

    Substitution Method:

    Substitute =3x=3y into −=26x−y=26: 3−=263y−y=26 2=262y=26 =13y=13

    Now, substitute =13y=13 back into =3x=3y: =3(13)=39x=3(13)=39

    So, the two numbers are 39 and 13.

    (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. Supplementary Angles

    Let the two angles be a and b, with a being the larger angle. Supplementary angles sum up to 180 degrees, so:

    1. +=180a+b=180.
    2. The larger angle exceeds the smaller by 18 degrees: =+18a=b+18.

    Substitution Method:

    Substitute =+18a=b+18 into +=180a+b=180: +18+=180b+18+b=180 2+18=1802b+18=180 2=1622b=162 =81b=81

    Now, substitute =81b=81 back into =+18a=b+18: =81+18=99a=81+18=99

    So, the two angles are 99 degrees and 81 degrees. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.

    Cost of Bats and Balls

    Let the cost of one bat be p and the cost of one ball be q. The given conditions are:

    1. 7 bats and 6 balls cost ₹3800: 7+6=38007p+6q=3800.
    2. 3 bats and 5 balls cost ₹1750: 3+5=17503p+5q=1750.

    Substitution Method:

    Let's solve the first equation for p: =3800−67p=73800−6q​

    Substitute this into the second equation: 3(3800−67)+5=1750
    3(73800−6q​)+5q=1750
    11400−187+5=1750711400−18q​+5q=1750
    11400−18+35=1225011400−18q+35q=12250
    17=85017q=850
    =50q=50

    Now, substitute =50q=50 back into the first equation: 7+6(50)=38007p+6(50)=3800 7+300=38007p+300=3800 7=35007p=3500 =500p=500

    So, the cost of each bat is ₹500 and the cost of each ball is ₹50.

15. 15.The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs. 2000 per month, find their monthly incomes. By Elimination Method

    To solve this problem, let's first set up the equations based on the given ratios and then solve them using the elimination method. We'll also verify our solution at the end.

    Let the incomes of the two persons be 99x and 77x (since their ratio is 9:7), and their expenditures be 44y and 33y (since their expenditure ratio is 4:3). We are given that each saves ₹2000 per month.

    The savings can be calculated as income minus expenditure. So, for each person, we have:

    1. For the first person: Income - Expenditure = Savings 9−4=20009x−4y=2000

    2. For the second person: Income - Expenditure = Savings 7−3=20007x−3y=2000

    Solving by Elimination Method

    To eliminate one variable, we can multiply each equation by a suitable number so that the coefficients of either x or y in both equations become the same. Let's eliminate y:

    Multiply the first equation by 3 and the second by 4:

    1. 3×(9−4)=3×20003×(9x−4y)=3×2000 27−12=600027x−12y=6000... (Equation 3)

    2. 4×(7−3)=4×20004×(7x−3y)=4×2000 28−12=800028x−12y=8000... (Equation 4)

    Now subtract Equation 3 from Equation 4: 28−12−(27−12)=8000−600028x−12y−(27x−12y)=8000−6000 28−12−27+12=200028x−12y−27x+12y=2000 =2000x=2000

    Since =2000x=2000, substitute this back into one of the original equations to find y. Using 9−4=20009x−4y=2000: 9(2000)−4=20009(2000)−4y=2000 18000−4=200018000−4y=2000 −4=2000−18000−4y=2000−18000 −4=−16000−4y=−16000 =4000y=4000

    Monthly Incomes

    1. First person's income = 9=9×2000=180009x=9×2000=18000 Rs.
    2. Second person's income = 7=7×2000=140007x=7×2000=14000 Rs.

    Verification

    Verify the solution by checking if the expenditures and savings are consistent:

    • First person's expenditure = 4=4×4000=160004y=4×4000=16000 Rs, so savings = 18000−16000=200018000−16000=2000 Rs.
    • Second person's expenditure = 3=3×4000=120003y=3×4000=12000 Rs, so savings = 14000−12000=200014000−12000=2000 Rs.

    The savings in both cases are ₹2000, which matches the given information, verifying our solution.

16. 16.Use elimination method to find all possible solutions of the following pair of linear equations : 2x + 3y = 8 (1), 4x + 6y = 7 (2)

    To solve the given pair of linear equations using the elimination method, let's first look at the equations:

    1. 2+3=82x+3y=8... (Equation 1)
    2. 4+6=74x+6y=7... (Equation 2)

    The goal of the elimination method is to eliminate one of the variables by making the coefficients of that variable the same in both equations and then subtracting one equation from the other. However, it's important to note the relationship between these two equations before proceeding.

    Notice that Equation 2 is not a multiple of Equation 1. If it were, they would be equivalent, representing the same line. However, in this case, they are not multiples of each other but the coefficients of x and y in the second equation are exactly double those in the first equation.

    Let's try to eliminate one of the variables:

    Multiply Equation 1 by 2: 2×(2+3)=2×82×(2x+3y)=2×8 4+6=164x+6y=16... (Equation 3)

    Now we have:

    • Equation 3: 4+6=164x+6y=16
    • Equation 2: 4+6=74x+6y=7

    Subtracting Equation 2 from Equation 3 yields: (4+6)−(4+6)=16−7(4x+6y)−(4x+6y)=16−7 0=90=9

    This is a contradiction (0 does not equal 9), which indicates that the two equations are inconsistent and do not have a solution. In graphical terms, these two equations represent two parallel lines that never intersect, and thus there are no points (x, y) that satisfy both equations simultaneously. Therefore, there are no possible solutions to this pair of linear equations.

17. 17.The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

    To solve this problem, let's define the two digits of the number. Let the tens digit be x and the units digit be y. The number is then 10+10x+y. When the digits are reversed, the number becomes 10+10y+x.

    Given information:

    1. The sum of the original number and the number with its digits reversed is 66: (10+)+(10+)=66(10x+y)+(10y+x)=66

    2. The digits of the number differ by 2: −=2x−y=2 or −=2y−x=2 (since we don't know which digit is larger)

    Let's solve these equations.

    Equation Formation:

    From the first condition: 10++10+=6610x+y+10y+x=66 11+11=6611x+11y=66 +=6x+y=6... (Equation 1)

    From the second condition, we have two possibilities: −=2x−y=2... (Equation 2a) −=2y−x=2... (Equation 2b)

    Solving the Equations:

    Case 1: Using Equation 2a:

    Add Equation 1 and Equation 2a: +=6x+y=6 −=2x−y=2 2=82x=8 =4x=4

    Substitute =4x=4 into Equation 1: 4+=64+y=6 =2y=2

    So, one number could be 4242.

    Case 2: Using Equation 2b:

    Add Equation 1 and Equation 2b: +=6x+y=6 −=2y−x=2 2=82y=8 =4y=4

    Substitute =4y=4 into Equation 1: +4=6x+4=6 =2x=2

    So, another number could be 2424.

    Conclusion:

    The two numbers that satisfy the conditions are 42 and 24. There are two such numbers.

18. 18.Solve the following pair of linear equations by the elimination method and the substitution method :

    (i) x + y = 5 and 2x – 3y = 4 (ii) 3x + 4y = 10 and 2x – 2y = 2 (iii) x/2 + 2y/3 = -1 and x - y/3 = 3
    

    Let's solve each pair of linear equations using both the elimination method and the substitution method.

    (i) +=5x+y=5 and 2−3=42x−3y=4

    Elimination Method:

    1. Multiply the first equation by 3: 3(+)=3×53(x+y)=3×5 which gives 3+3=153x+3y=15.
    2. Now, add this to the second equation: 3+3+2−3=15+43x+3y+2x−3y=15+4 which simplifies to 5=195x=19, so =195x=519​.
    3. Substitute =195x=519​ into +=5x+y=5: 195+=5519​+y=5, which gives =5−195=65y=5−519​=56​.

    Substitution Method:

    1. From +=5x+y=5, get =5−x=5−y.
    2. Substitute into 2−3=42x−3y=4: 2(5−)−3=42(5−y)−3y=4, which simplifies to 10−2−3=410−2y−3y=4, so 5=65y=6, thus =65y=56​.
    3. Substitute =65y=56​ into =5−x=5−y: =5−65=195x=5−56​=519​.

    (ii) 3+4=103x+4y=10 and 2−2=22x−2y=2

    Elimination Method:

    1. Multiply the second equation by 2: 2(2−2)=2×22(2x−2y)=2×2 which gives 4−4=44x−4y=4.
    2. Now, add this to the first equation: 3+4+4−4=10+43x+4y+4x−4y=10+4 which simplifies to 7=147x=14, so =2x=2.
    3. Substitute =2x=2 into 3+4=103x+4y=10: 3(2)+4=103(2)+4y=10, which gives 4=44y=4, thus =1y=1.

    Substitution Method:

    1. From 3+4=103x+4y=10, get =10−43x=310−4y​.
    2. Substitute into 2−2=22x−2y=2: 2(10−43)−2=22(310−4y​)−2y=2, which simplifies to 20−83−2=2320−8y​−2y=2, solve for y to get =1y=1.
    3. Substitute =1y=1 into =10−43x=310−4y​: =10−4(1)3=2x=310−4(1)​=2.

    (iii) 2+23=−12x​+32y​=−1 and −3=3x−3y​=3

    Elimination Method:

    1. Multiply the first equation by 3 and the second by 2: 3(2+23)=3×(−1)3(2x​+32y​)=3×(−1) and 2(−3)=2×32(x−3y​)=2×3 which gives 32+2=−323x​+2y=−3 and 2−23=62x−32y​=6.
    2. Now, add these equations: 32+2+2−23=−3+623x​+2x+2y−32y​=−3+6, solve for x.
    3. Substitute the value of x into one of the original equations to find y.

    Substitution Method:

    1. From 2+23=−12x​+32y​=−1, express x in terms of y.
    2. Substitute into −3=3x−3y​=3 and solve for y.
    3. Substitute the value of y into the equation from step 1 to find x.

    Solve these steps to find the exact values of x and y for each case.

19. 19.Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

    (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
    
    (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
    
    (iii) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ` 27 for a book kept for seven days, while Susy paid ` 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
    
    
    

    Let's form and solve the pair of linear equations for each problem using the elimination method.

    (i) Fraction Problem

    Let the numerator of the fraction be x and the denominator be y. According to the problem, we have:

    1. Adding 1 to the numerator and subtracting 1 from the denominator gives 1: +1−1=1y−1x+1​=1

    2. Adding 1 to the denominator gives 1/2: +1=12y+1x​=21​

    Let's turn these into linear equations:

    1. +1=−1x+1=y−1... (Equation 1)
    2. 2=+12x=y+1... (Equation 2)

    Solution by Elimination Method:

    1. Subtract Equation 2 from Equation 1: (+1)−2=(−1)−(+1)(x+1)−2x=(y−1)−(y+1) −=−2−x=−2 =2x=2

    2. Substitute =2x=2 into Equation 2: 2(2)=+12(2)=y+1 4=+14=y+1 =3y=3

    The fraction is =23yx​=32​.

    (ii) Age Problem

    Let's assume Nuri's current age is x years and Sonu's current age is y years. According to the problem, we have:

    1. Five years ago, Nuri was thrice as old as Sonu: −5=3(−5)x−5=3(y−5)

    2. Ten years later, Nuri will be twice as old as Sonu: +10=2(+10)x+10=2(y+10)

    Solution by Elimination Method:

    1. Expand and simplify both equations:

       • Equation 1: −5=3−15x−5=3y−15 or −3=−10x−3y=−10... (Equation 3)
       • Equation 2: +10=2+20x+10=2y+20 or −2=10x−2y=10... (Equation 4)

    2. Subtract Equation 4 from Equation 3: (−3)−(−2)=−10−10(x−3y)−(x−2y)=−10−10 −=−20−y=−20 =20y=20

    3. Substitute =20y=20 into Equation 3 or 4: −3(20)=−10x−3(20)=−10 =60−10x=60−10 =50x=50

    Nuri is 50 years old, and Sonu is 20 years old.

    (iii) Library Charges

    Let the fixed charge be x and the charge per day after the first three days be y. According to the problem, we have:

    1. Saritha paid ₹27 for a book kept for seven days: +4=27x+4y=27... (Equation 5)

    2. Susy paid ₹21 for the book she kept for five days: +2=21x+2y=21... (Equation 6)

    Solution by Elimination Method:

    1. Subtract Equation 6 from Equation 5: (+4)−(+2)=27−21(x+4y)−(x+2y)=27−21 2=62y=6 =3y=3

    2. Substitute =3y=3 into Equation 5 or 6: +4(3)=27x+4(3)=27 +12=27x+12=27 =15x=15

    The fixed charge is ₹15, and the charge for each extra day is ₹3.

20. 20.Quick Revision

    1. Introduction of Pair of Linear Equations in Two Variables: A pair of linear equations in two variables looks like this:

    • Equation 1: +=ax+by=c
    • Equation 2: +=dx+ey=f Here, x and y are the variables, and a, b, c, d, e, f are constants. These two equations together form a system that we can solve to find x and y.

    2. Solving Pair of Linear Equation Graphically: To solve the equations graphically, plot both equations on the same graph. The point where the two lines intersect is the solution (x, y) to both equations.

    3. Algebraic Methods of Solving a Pair of Linear Equations: Algebraically, you can solve a pair of linear equations using:

    • Substitution method: Solve one equation for one variable and substitute it into the other equation.
    • Elimination method: Add or subtract the equations to eliminate one variable, making it easier to solve for the other.
    • Cross-multiplication method: Applicable for equations in the form of =xa​=yb​; multiply crosswise to find the variables.

    4. Solve the Following Pair of Linear Equations by the Substitution Method: For this, you need specific equations. However, here's how you do it:

    • Solve one of the equations for one variable (say x).
    • Substitute this expression for x in the other equation.
    • Solve for y.
    • Substitute the value of y back into the first equation to find x.