Pair of linear equation in 2 variables

29問 • 1年前

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Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Let the present age of Aftab be ‘x’. And, the present age of his daughter be ‘y’. Now, we can write, seven years ago, Age of Aftab = x-7 Age of his daughter = y-7 According to the question, x−7 = 7(y−7) ⇒x−7 = 7y−49 ⇒x−7y = −42 ………………………(i) Also, three years from now or after three years, Age of Aftab will become = x+3. Age of his daughter will become = y+3 According to the situation given, x+3 = 3(y+3) ⇒x+3 = 3y+9 ⇒x−3y = 6 …………..…………………(ii) Subtracting equation (i) from equation (ii) we have (x−3y)−(x−7y) = 6−(−42) ⇒−3y+7y = 6+42 ⇒4y = 48 ⇒y = 12 The algebraic equation is represented by x−7y = −42 x−3y = 6 For, x−7y = −42 or x = −42+7y The solution table is Ncert solutions class 10 chapter 3-1 For, x−3y = 6 or x = 6+3y The solution table is The graphical representation is:

The coach of a cricket team buys 3 bats and 6 balls for Rs.3900. Later, she buys another bat and 3 more balls of the same kind for Rs.1300. Represent this situation algebraically and geometrically.

Let us assume that the cost of a bat be ‘Rs x’ And,the cost of a ball be ‘Rs y’ According to the question, the algebraic representation is 3x+6y = 3900 And x+3y = 1300 For, 3x+6y = 3900 Or x = (3900-6y)/3 The solution table is Ncert solutions class 10 chapter 3-4 For, x+3y = 1300 Or x = 1300-3y The solution table is The graphical representation is as follows.

The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically and geometrically.

Let the cost of 1 kg of apples be ‘Rs. x’ And, cost of 1 kg of grapes be ‘Rs. y’ According to the question, the algebraic representation is 2x+y = 160 And 4x+2y = 300 For, 2x+y = 160 or y = 160−2x, the solution table is; Ncert solutions class 10 chapter 3-7 For 4x+2y = 300 or y = (300-4x)/2, the solution table is. The graphical representation is as follows;

Form the pair of linear equations in the following problems, and find their solutions graphically. (i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. (ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.

(i)Let there are x number of girls and y number of boys. As per the given question, the algebraic expression can be represented as follows. x +y = 10 x– y = 4 Now, for x+y = 10 or x = 10−y, the solutions are; Ncert solutions class 10 chapter 3-10 For x – y = 4 or x = 4 + y, the solutions are; Ncert solutions class 10 chapter 3-11 The graphical representation is as follows; Ncert solutions class 10 chapter 3-12 From the graph, it can be seen that the given lines cross each other at point (7, 3). Therefore, there are 7 girls and 3 boys in the class. (ii) Let 1 pencil costs Rs.x and 1 pen costs Rs.y. According to the question, the algebraic expression cab be represented as; 5x + 7y = 50 7x + 5y = 46 For, 5x + 7y = 50 or x = (50-7y)/5, the solutions are; Ncert solutions class 10 chapter 3-13 For 7x + 5y = 46 or x = (46-5y)/7, the solutions are; Ncert solutions class 10 chapter 3-14 Hence, the graphical representation is as follows; From the graph, it is can be seen that the given lines cross each other at point (3, 5). So, the cost of a pencil is 3/- and cost of a pen is 5/-.

On comparing the ratios a1/a2 , b1/b2 , 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 (iii) 6x – 3y + 10 = 0 2x – y + 9 = 0

(i) Given expressions; 5x−4y+8 = 0 7x+6y−9 = 0 Comparing these equations with a1x+b1y+c1 = 0 And a2x+b2y+c2 = 0 We get, a1 = 5, b1 = -4, c1 = 8 a2 = 7, b2 = 6, c2 = -9 (a1/a2) = 5/7 (b1/b2) = -4/6 = -2/3 (c1/c2) = 8/-9 Since, (a1/a2) ≠ (b1/b2) So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point. (ii) Given expressions; 9x + 3y + 12 = 0 18x + 6y + 24 = 0 Comparing these equations with a1x+b1y+c1 = 0 And a2x+b2y+c2 = 0 We get, a1 = 9, b1 = 3, c1 = 12 a2 = 18, b2 = 6, c2 = 24 (a1/a2) = 9/18 = 1/2 (b1/b2) = 3/6 = 1/2 (c1/c2) = 12/24 = 1/2 Since (a1/a2) = (b1/b2) = (c1/c2) So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident. (iii) Given Expressions; 6x – 3y + 10 = 0 2x – y + 9 = 0 Comparing these equations with a1x+b1y+c1 = 0 And a2x+b2y+c2 = 0 We get, a1 = 6, b1 = -3, c1 = 10 a2 = 2, b2 = -1, c2 = 9 (a1/a2) = 6/2 = 3/1 (b1/b2) = -3/-1 = 3/1 (c1/c2) = 10/9 Since (a1/a2) = (b1/b2) ≠ (c1/c2) So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.

On comparing the ratio, (a1/a2) , (b1/b2) , (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 (iv) 5x – 3y = 11 ; – 10x + 6y = –22 (v)(4/3)x+2y = 8 ; 2x + 3y = 12

(i) Given : 3x + 2y = 5 or 3x + 2y -5 = 0 and 2x – 3y = 7 or 2x – 3y -7 = 0 Comparing these equations with a1x+b1y+c1 = 0 And a2x+b2y+c2 = 0 We get, a1 = 3, b1 = 2, c1 = -5 a2 = 2, b2 = -3, c2 = -7 (a1/a2) = 3/2 (b1/b2) = 2/-3 (c1/c2) = -5/-7 = 5/7 Since, (a1/a2) ≠ (b1/b2) So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent. (ii) Given 2x – 3y = 8 and 4x – 6y = 9 Therefore, a1 = 2, b1 = -3, c1 = -8 a2 = 4, b2 = -6, c2 = -9 (a1/a2) = 2/4 = 1/2 (b1/b2) = -3/-6 = 1/2 (c1/c2) = -8/-9 = 8/9 Since , (a1/a2) = (b1/b2) ≠ (c1/c2) So, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent. (iii)Given (3/2)x + (5/3)y = 7 and 9x – 10y = 14 Therefore, a1 = 3/2, b1 = 5/3, c1 = -7 a2 = 9, b2 = -10, c2 = -14 (a1/a2) = 3/(2×9) = 1/6 (b1/b2) = 5/(3× -10)= -1/6 (c1/c2) = -7/-14 = 1/2 Since, (a1/a2) ≠ (b1/b2) So, the equations are intersecting each other at one point and they have only one possible solution. Hence, the equations are consistent. (iv) Given, 5x – 3y = 11 and – 10x + 6y = –22 Therefore, a1 = 5, b1 = -3, c1 = -11 a2 = -10, b2 = 6, c2 = 22 (a1/a2) = 5/(-10) = -5/10 = -1/2 (b1/b2) = -3/6 = -1/2 (c1/c2) = -11/22 = -1/2 Since (a1/a2) = (b1/b2) = (c1/c2) These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent. (v)Given, (4/3)x +2y = 8 and 2x + 3y = 12 a1 = 4/3 , b1= 2 , c1 = -8 a2 = 2, b2 = 3 , c2 = -12 (a1/a2) = 4/(3×2)= 4/6 = 2/3 (b1/b2) = 2/3 (c1/c2) = -8/-12 = 2/3 Since (a1/a2) = (b1/b2) = (c1/c2) These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) x + y = 5, 2x + 2y = 10 (ii) x – y = 8, 3x – 3y = 16 (iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0 (iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

(i)Given, x + y = 5 and 2x + 2y = 10 (a1/a2) = 1/2 (b1/b2) = 1/2 (c1/c2) = 1/2 Since (a1/a2) = (b1/b2) = (c1/c2) ∴The equations are coincident and they have infinite number of possible solutions. So, the equations are consistent. For, x + y = 5 or x = 5 – y Ncert solutions class 10 chapter 3-16 For 2x + 2y = 10 or x = (10-2y)/2 Ncert solutions class 10 chapter 3-17 So, the equations are represented in graphs as follows: Ncert solutions class 10 chapter 3-18 From the figure, we can see, that the lines are overlapping each other. Therefore, the equations have infinite possible solutions. (ii) Given, x – y = 8 and 3x – 3y = 16 (a1/a2) = 1/3 (b1/b2) = -1/-3 = 1/3 (c1/c2) = 8/16 = 1/2 Since, (a1/a2) = (b1/b2) ≠ (c1/c2) The equations are parallel to each other and have no solutions. Hence, the pair of linear equations is inconsistent. (iii) Given, 2x + y – 6 = 0 and 4x – 2y – 4 = 0 (a1/a2) = 2/4 = ½ (b1/b2) = 1/-2 (c1/c2) = -6/-4 = 3/2 Since, (a1/a2) ≠ (b1/b2) The given linear equations are intersecting each other at one point and have only one solution. Hence, the pair of linear equations is consistent. Now, for 2x + y – 6 = 0 or y = 6 – 2x Ncert solutions class 10 chapter 3-19 And for 4x – 2y – 4 = 0 or y = (4x-4)/2 Ncert solutions class 10 chapter 3-20 So, the equations are represented in graphs as follows: Ncert solutions class 10 chapter 3-21 From the graph, it can be seen that these lines are intersecting each other at only one point,(2,2). (iv) Given, 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0 (a1/a2) = 2/4 = ½ (b1/b2) = -2/-4 = 1/2 (c1/c2) = 2/5 Since, a1/a2 = b1/b2 ≠ c1/c2 Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.

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.

Let us consider. The width of the garden is x and length is y. Now, according to the question, we can express the given condition as; y – x = 4 and y + x = 36 Now, taking y – x = 4 or y = x + 4 Ncert solutions class 10 chapter 3-22 For y + x = 36, y = 36 – x Ncert solutions class 10 chapter 3-23 The graphical representation of both the equation is as follows; Ncert solutions class 10 chapter 3-24 From the graph you can see, the lines intersects each other at a point(16, 20). Hence, the width of the garden is 16 and length is 20.

Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) Intersecting lines (ii) Parallel lines (iii) Coincident lines

(i) Given the linear equation 2x + 3y – 8 = 0. To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition; (a1/a2) ≠ (b1/b2) Thus, another equation could be 2x – 7y + 9 = 0, such that; (a1/a2) = 2/2 = 1 and (b1/b2) = 3/-7 Clearly, you can see another equation satisfies the condition. (ii) Given the linear equation 2x + 3y – 8 = 0. To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition; (a1/a2) = (b1/b2) ≠ (c1/c2) Thus, another equation could be 6x + 9y + 9 = 0, such that; (a1/a2) = 2/6 = 1/3 (b1/b2) = 3/9= 1/3 (c1/c2) = -8/9 Clearly, you can see another equation satisfies the condition. (iii) Given the linear equation 2x + 3y – 8 = 0. To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition; (a1/a2) = (b1/b2) = (c1/c2) Thus, another equation could be 4x + 6y – 16 = 0, such that; (a1/a2) = 2/4 = 1/2 ,(b1/b2) = 3/6 = 1/2, (c1/c2) = -8/-16 = 1/2 Clearly, you can see another equation satisfies the condition.

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.

Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0. For, x – y + 1 = 0 or x = -1+y Ncert solutions class 10 chapter 3-25 For, 3x + 2y – 12 = 0 or x = (12-2y)/3 Ncert solutions class 10 chapter 3-26 Hence, the graphical representation of these equations is as follows; Ncert solutions class 10 chapter 3-27 From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

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) 3x – y = 3 9x – 3y = 9 (iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 (v) √2 x+√3 y = 0 √3 x-√8 y = 0 (vi) (3x/2) – (5y/3) = -2 (x/3) + (y/2) = (13/6)

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

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. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. (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. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? (v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

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) 3x – 5y – 4 = 0 and 9x = 2y + 7 (iv) x/2+ 2y/3 = -1 and x-y/3 = 3

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 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) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. (iv) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received. (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method. (i) x – 3y – 3 = 0 and 3x – 9y – 2 = 0 (ii) 2x + y = 5 and 3x + 2y = 8 (iii) 3x – 5y = 20 and 6x – 10y = 40 (iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0

(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7 (a – b) x + (a + b) y = 3a + b – 2 (ii) For which value of k will the following pair of linear equations have no solution? 3x + y = 1 (2k – 1) x + (k – 1) y = 2k + 1

Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5y = 9 3x + 2y = 4

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method: (i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the cost of food per day. (ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction. (iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test? (iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars? (v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solve the following pairs of equations by reducing them to a pair of linear equations: (i) 1/2x + 1/3y = 2 1/3x + 1/2y = 13/6 (ii) 2/√x + 3/√y = 2 4/√x + 9/√y = -1 (iii) 4/x + 3y = 14 3/x -4y = 23 (iv) 5/(x-1) + 1/(y-2) = 2 6/(x-1) – 3/(y-2) = 1 (v) (7x-2y)/ xy = 5 (8x + 7y)/xy = 15 (vi) 6x + 3y = 6xy 2x + 4y = 5xy (vii) 10/(x+y) + 2/(x-y) = 4 15/(x+y) – 5/(x-y) = -2 (viii) 1/(3x+y) + 1/(3x-y) = 3/4 1/2(3x+y) – 1/2(3x-y) = -1/8

Formulate the following problems as a pair of equations, and hence find their solutions: (i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone. (iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)].

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

Solve the following pair of linear equations: (i) px + qy = p – q qx – py = p + q (ii) ax + by = c bx + ay = 1 + c (iii) x/a – y/b = 0 ax + by = a2 + b2 (iv) (a – b)x + (a + b) y = a2 – 2ab – b2 (a + b)(x + y) = a2 + b2 (v) 152x – 378y = – 74 –378x + 152y = – 604

ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.

Arithmetic progressions - Textbook

Nabiha · 35問 · 1年前