### Pair Of Linear Equations In Two Variables Class 10th Mathematics AP Board Solution

##### Question 1.By comparing the ratios  find out whether the lines represented by the following pairs of linear equations intersect at a point, are parallel or are coincident.5x – 4y + 8 = 07x + 6y – 9 = 0Answer:We have,5x – 4y + 8 = 07x + 6y – 9 = 0Here, a1 = 5, b2 = -4, c1 = 8a1 = 7, b2 = 6, c1 = -9∴ ∴ Two line are intersecting with each at a point.Question 2.By comparing the ratios  find out whether the lines represented by the following pairs of linear equations intersect at a point, are parallel or are coincident.9x + 3y + 12 = 018x + 6y – 24 = 0Answer:We have,9x + 3y + 12 = 018x + 6y – 24 = 0Here, a1 = 9, b2 = 3, c1 = 12a1 = 18, b2 = 6, c1 = -24∴ ∴ Both the lines will coincide.Question 3.By comparing the ratios  find out whether the lines represented by the following pairs of linear equations intersect at a point, are parallel or are coincident.6x – 3y + 10 = 02x - y + 9 = 0Answer:We have,6x – 3y + 10 = 02x - y + 9 = 0Here, a1 = 6, b2 = -3, c1 = 10a1 = 2, b2 = -1, c1 = 9∴ ∴ Two line are parallel to each other.Question 4.Check whether the following equation are consistent or inconsistent. Solve them graphically.3x + 2y = 52x -3y = 7Answer:3x + 2y = 5 2x -3y = 7Here, a1 = 3, b2 = 2, c1 = 5a1 = 2, b2 = -3, c1 = 7∴ ∴ two equations are consistent.Question 5.Check whether the following equation are consistent or inconsistent. Solve them graphically.2x - 3y = 84x -6y = 9Answer:2x - 3y = 8 4x -6y = 9Here, a1 = 2, b2 = -3, c1 = 8a1 = 4, b2 = -6, c1 = 9∴ ∴ both lines are inconsistentQuestion 6.Check whether the following equation are consistent or inconsistent. Solve them graphically.9x -10y = 12Answer:9x -10y = 12Here, a1 = , b2 = , c1 = 7a1 = 9, b2 = -10, c1 = 12∴ ∴ both lines are consistentQuestion 7.Check whether the following equation are consistent or inconsistent. Solve them graphically.5x - 3y = 11-10x + 6y = -22Answer:5x - 3y = 11 -10x + 6y = -22Here, a1 = 5, b2 = -3, c1 = 11a1 = -10, b2 = 6, c1 = -22∴ ∴ two lines are consistentQuestion 8.Check whether the following equation are consistent or inconsistent. Solve them graphically.2x + 3y = 12Answer:2x + 3y = 12Here, a1 = , b2 = 2, c1 = 8a1 = 2, b2 = 3, c1 = 12∴ ∴ two equations are consistentQuestion 9.Check whether the following equation are consistent or inconsistent. Solve them graphically.x + y = 52x + 2y = 10Answer:x + y = 5 2x + 2y = 10Here, a1 = 1, b2 = 1, c1 = 5a1 = 2, b2 = 2, c1 = 10∴ ∴ two lines are inconsistentQuestion 10.Check whether the following equation are consistent or inconsistent. Solve them graphically.x - y = 83x + 3y = 16Answer:x - y = 8 3x + 3y = 16Here, a1 = 1, b2 = -1, c1 = 8a1 = 3, b2 = 3, c1 = 12∴ ∴ two lines are inconsistentQuestion 11.Check whether the following equation are consistent or inconsistent. Solve them graphically.2x + y =6 4x - 2y = 4Answer:2x + y = 6 4x - 2y = 4Here, a1 = 2, b2 = 1, c1 = 6a1 = 4, b2 = -2, c1 = 4∴ ∴ two lines are consistentQuestion 12.Check whether the following equation are consistent or inconsistent. Solve them graphically.2x - 2y =2 4x - 4y = 5Answer:2x - 2y = 2 4x - 4y = 5Here, a1 = 2, b2 = 2, c1 = 2a1 = 4, b2 = 4, c1 = 5∴ ∴ two lines are inconsistent.Question 13.Neha went to a ‘sale’ to purchase some pants and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts are 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 friend to find how many pants and skirts Neha bought.Answer:Let the number of pants purchased by Neha be ‘x’ and no. of skirts purchased be ‘y’.Twice the number of pants = 2xFrom the given condition,y = 2x – 2 (I)i.e. number of skirts are 2 less than 2 twice the number of pants.y = 4x – 4 (II)i.e. number of skirts are 4 less than four times the number of pants.Subtracting eq. I from II, we get2x = 2⇒ x = 1So number of pants purchased = 1 and number of skirts = 0Graphical Representation: Question 14.10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys then, find the number of boys and the number of girls who took part in the quiz.Answer:Let the number of girls and boys in the class be x and y respectivelyAccording to given condition, we have:x + y = 10x-y = 4x + y = 10 ⇒ x = 10-y x-y = 4 ⇒ x = 4 + yFrom the graph it can be observed that the two lines intersect each other at the point (7, 3)So, x = 7 and y = 3Number of Girls = 7;Number of boys = 3Question 15.5 pencils and 7 pens together cost D50 whereas 7 pencils and 5 pens together cost D46. Find the cost of one pencil and that of one pen.Answer:Let the cost of one pencil and one pen be ₹x and ₹y respectively.According to given situation, we have :5x + 7y = 507x + 5y = 465x + 7y = 50 7x + 5y = 46From the graph. It can be observed that the two lines are intersecting each other at (3,5)x = 3 and y = 5Cost of pencil =  3;Cost of pen =  5Question 16.Half the perimeter of a rectangular garden is 36 m. If the length is 4m more than its width, is 36 m. Find the dimensions of the garden.Answer:Let l be the length and b be the breadthl = b + 4l-b = 4Perimeter = 2(l + b)× 2(l + b) = 36 (given)l-b = 4 ……Il + b = 36 ….IIadding eq. I and II⇒ 2l = 40⇒ l =  = 20mSubstitute l = 20 in(1)l-b = 420-b = 4b = 20-4 = 16mLength = 20 mWidth = 16 mQuestion 17.We have a linear equation 2x + 3y-8 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines.Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.Answer:2x + 3y-8 = 03x + 2y-7 = 0(intersecting)2x + 3y + 12 = 0 (parallel lines)4x + 6y-16 = 0 (coincident lines)Question 18.The area of a rectangle gets reduced by 80 sq uints if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will increase by 50 sq units. Find the length and breadth of the rectangle.Answer:Let the length be x and breadth be yAccording to question,xy-80 = (x-5)(x + 2)⇒xy-80 = xy + 2x-5y-10⇒-80 + 10 = 2x-5y⇒ 2x-5y = -70 ….I(x + 10)(y-5) = xy + 50⇒ xy-5x + 10y-50 = xy + 50⇒ -5x + 10y = 100⇒ -x + 2y = 20 ….IIMultiplying Eq. II by 22x-5y = -70-2x + 4y = 40-y = -30y = 30substituting y = 30 in eq. II-x + 2× 30 = 20-x = 20-60x = 40length = 40units and breadth = 30 unitsQuestion 19.In X class, if three students sit on each bench, one student will be left. If four students sit on each bench, one bench will be left. Find the number of students and the number of benches in that class.Answer:Let the number no. of students be xLet the number no. of benches be yAccording to given condition, we havex = 3y + 1 ⇒ x-3y = 1 …..Ix = 4(y-1) ⇒ x = 4y-4 ⇒ x-4y = -4 ….IIEquating eq. I and IIx-3y = 1-x + 4y = 4y = 5Substituting y = 5 in eq. Ix-3×5 = 1⇒ x-15 + 1 ⇒x = 16Number of students = 16;Number of benches = 5
###### Exercise 4.3

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