{"id":6452,"date":"2023-02-22T07:40:10","date_gmt":"2023-02-22T02:10:10","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=6452"},"modified":"2023-03-29T16:07:41","modified_gmt":"2023-03-29T10:37:41","slug":"ap-ssc-10th-class-maths-solutions-chapter-4-intext-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-ssc-10th-class-maths-solutions-chapter-4-intext-questions\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables InText Questions"},"content":{"rendered":"

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables InText Questions<\/h2>\n

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 4 Pair of Linear Equations in Two Variables InText Questions and Answers.<\/p>\n

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables InText Questions and Answers<\/h3>\n

Question 1.
\nSolve the following systems of equations: i) x – 2y = 0; 3x + 4y = 20\u00a0 \u00a0 \u00a0(Page No. 79)
\nAnswer:
\ni) x – 2y = 0;\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a03x + 4y = 20
\n-2y = -x\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a04y = 20 – 3x
\ny = \\(\\frac{x}{2}\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 y = \\(\\frac{20-3x}{4}\\)
\n\"AP\"AP
\n\"AP
\nThe two lines meet at (4, 2).
\nThe solution set is {(4, 2)}<\/p>\n

\"AP<\/p>\n

ii) x + y = 2
\n2x + 2y = 4
\nAnswer:
\nx + y = 2
\n2x + 2y = 4
\n\"AP
\n\"AP
\nThese two are coincident lines.
\n\u2234 There are infinitely many solutions.<\/p>\n

iii) 2x – y = 4
\n4x – 2y = 6
\nAnswer:
\n2x – y = 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 4x – 2y = 6
\n\u21d2 y = 2x – 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u21d2 2y = 4x – 6 \u21d2 y = 2x – 3<\/p>\n

\"AP
\n\"AP
\nThese two are parallel lines.
\n\u2234 The pair of linear equations has no solution.<\/p>\n

\"AP<\/p>\n

Question 2.
\nTwo rails of a railway track are represented by the equations.
\nx + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation graphically.\u00a0 \u00a0 (Page No. 79)
\nAnswer:
\nx + 2y – 4 = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a02x + 4y – 12 = 0
\n2y = 4 – x\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a04y = 12 – 2x (or) 4y = 2 (6 – x)
\ny = \\(\\frac{4-x}{2}\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 y = \\(\\frac{6-x}{2}\\)
\nx + 2y – 4 = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a02x + 4y – 12 = 0
\n\"AP
\n\"AP
\nThese lines are parallel and hence no solution.<\/p>\n

\"AP<\/p>\n

Question 3.
\nCheck each of the given systems of equations to see if it has a unique solution, infinitely many solutions or no solution. Solve them graphically. (Page No. 83)
\ni) 2x + 3y = 1
\n3x – y = 7
\nAnswer:
\nLet a1<\/sub>x + b1<\/sub>y – c1<\/sub> = 0 \u2243 2x + 3y – 1 = 0
\na2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0 \u2243 3x – y – 7 = 0
\nNow comparing their coefficients i.e., \\(\\frac{a_{1}}{a_{2}}\\) and \\(\\frac{b_{1}}{b_{2}}\\)
\n\u21d2 \\(\\frac{2}{3}\\) \u2260 \\(\\frac{3}{-1}\\)
\nThe given lines are intersecting lines.
\n2x + 3y = 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a03x – y = 7
\n3y = 1 – 2x\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 y – 3x = 7
\ny = \\(\\frac{1-2x}{3}\\)
\n\"AP
\n\"AP
\nThe system of equations has a unique solution (2, – 1).<\/p>\n

\"AP<\/p>\n

ii) x + 2y = 6
\n2x + 4y = 12
\nAnswer:
\nFrom the given pair of equations,
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{6}{12}\\) = \\(\\frac{1}{2}\\)
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{c_{1}}{c_{2}}\\)
\n\u2234 The lines are dependent and have infinitely many solutions.
\nx + 2y = 6\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 2x + 4y = 12
\n2y = 6 – x\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 4y = 12 – 2x (or) 4y = 2(6 – x)
\ny = \\(\\frac{6-x}{2}\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0y = \\(\\frac{12-2x}{4}\\) or y = \\(\\frac{6-x}{2}\\)
\n\"AP
\n\"AP<\/p>\n

iii) 3x + 2y = 6
\n6x + 4y = 18
\nAnswer:
\nFrom the given pair of equations,
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{3}{6}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{6}{18}\\) = \\(\\frac{1}{3}\\)
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\)
\n\u2234 The lines are parallel and hence no solution.
\n3x + 2y = 6\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a06x + 4y = 18
\n2y = 6 – 3x\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 4y = 18 – 6x
\ny = \\(\\frac{6-3x}{2}\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 y = \\(\\frac{18-6x}{4}\\)
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

Try these<\/span><\/p>\n

(Page No. 75, 76)<\/p>\n

Question 1.
\nMark the correct option in the following questions:
\nWhich of the following equations is not a linear equation?
\na) 5 + 4x = y + 3
\nb) x + 2y = y – x
\nc) 3 – x = y2<\/sup> + 4
\nd) x + y = 0
\nAnswer:
\n[ c ]<\/p>\n

Question 2.
\nWhich of the following is a linear equation in one variable?
\na) 2x + 1 = y – 3
\nb) 2t – 1= 2t + 5
\nc) 2x – 1 = x2<\/sup>
\nd) x2<\/sup> – x + 1 =0
\nAnswer:
\n[ b ]<\/p>\n

Question 3.
\nWhich of the following numbers is a solution for the equation 2(x + 3) = 18?
\na) 5
\nb) 6
\nc) 13
\nd) 21
\nAnswer:
\n[b]<\/p>\n

Question 4.
\nThe value of x which satisfies the equation 2x – (4 – x) = 5 – x is
\na) 4.5
\nb) 3
\nc) 2.25
\nd) 0.5
\nAnswer:
\n[ c ]<\/p>\n

\"AP<\/p>\n

Question 5.
\nThe equation x – 4y = 5 has
\na) no solution
\nb) unique solution
\nc) two solutions
\nd) infinitely many solutions
\nAnswer:
\n[ d ]<\/p>\n

Question 6.
\nIn the example given above, can you find the cost of each bat and ball?\u00a0 \u00a0 (Page No. 79)
\nAnswer:
\nWe can’t find the exact values for the costs of bat and ball as there are infinitely many possibilities.<\/p>\n

Question 7.
\nFor what value of ‘p’ the following pair of equations has a unique solution.\u00a0 \u00a0 \u00a0(Page No. 83)
\n2x + py = – 5 and 3x + 3y = – 6
\nAnswer:
\nGiven: 2x + py = – 5
\n3x 4- 3y = – 6
\nTo have an unique solution we should have
\n\"AP
\n\u2234 The pair has unique solution when p \u2260 2.<\/p>\n

Question 8.
\nFind the value of \u2018k\u2019 for which the pair of equations 2x – ky + 3 = 0, 4x + 6y – 5 = 0 represent parallel lines.\u00a0 \u00a0(Page No. 83)
\nAnswer:
\nGiven: 2x – ky + 3 = 0
\n4x + 6y – 5 = 0
\nIf the above lines are to be parallel, then
\n\"AP
\n\u2234 k = – 3 is the required value for which lines are parallel.<\/p>\n

\"AP<\/p>\n

Question 9.
\nFor what value of ‘k’, the pair of equation 3x + 4y + 2 = 0 and 9x + 12y + k = 0 represent coincident lines.\u00a0 \u00a0(Page No. 83)
\nAnswer:
\nGiven: 3x + 4y + 2 = 0
\n9x + 12y + k = 0
\nIf the lines are to be coincident with each other, then
\n\"AP
\n\u2234 k = 2 \u00d7 3 = 6<\/p>\n

Question 10.
\nFor what positive values of ‘p’ the following pair of linear equations have infinitely many solutions?\u00a0 \u00a0(Page No. 83)
\npx + 3y – (p – 3) = 0
\n12x + py – p = 0
\nAnswer:
\nGiven: px + 3y – (p – 3) = 0
\n12x + py – p = 0
\nThe above equations to have infinitely many solutions
\n\"AP
\np.p = 12 \u00d7 3
\n\u21d2 p2<\/sup> = 36
\n\u21d2 p = \u00b16<\/p>\n

Think & Discuss<\/span><\/p>\n

Question 1.
\nTwo situations are given below:
\ni) The cost of 1 kg potatoes and 2 kg tomatoes was Rs. 30 on a certain day. After two days, the cost of 2 kg potatoes and 4 kg tomatoes was found to be Rs. 66.
\nii) The coach of a cricket team of M.K. Nagar High School buys 3 bats and 6 balls for Rs. 3900. Later he buys one more bat and 2 balls for Rs. 1300.
\nIdentify the unknowns in each situation. We observe that there are two unknowns in each case. (Page No. 73)
\nAnswer:
\ni) The unknowns in the first problem are
\na) cost of 1 kg tomatoes
\nb) cost of 1 kg potatoes
\nii) In the second problem, the unknowns are
\na) cost of each bat
\nb) cost of each ball<\/p>\n

\"AP<\/p>\n

Question 2.
\nIs a dependent pair of linear equations always consistent? Why or why not? (Page No. 79)
\nAnswer:
\nReason: \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{c_{1}}{c_{2}}\\) always holds. In other words, they have infinitely many solutions.<\/p>\n

Do these<\/span><\/p>\n

Solve each pair of equations by using the substitution method. (Page No. 88)
\nQuestion 1.
\n3x – 5y = -1
\nx – y = -1
\nAnswer:
\nGiven: 3x – 5y = -1 ……. (1)
\nx – y = -1 …….. (2)
\nFrom equation (2), x – y = – 1
\nx = y – 1
\nSubstituting x = y – 1 in equation (1)
\nwe get
\n3 (y – 1) – 5y = – 1
\n\u21d2 3y – 3 – 5y = r 1
\n\u21d2 – 2y = – 1 + 3
\n\u21d2 2y = – 2
\n\u21d2 y = -1
\nSubstituting y = – 1 in equation (1) we get
\n3x – 5 (- 1) = -1
\n3x + 5 = – 1
\n3x = – 1 – 5
\nx = \\(\\frac{-6}{3}\\) = -2
\n\u2234 The solution is (-2, -1)<\/p>\n

Question 2.
\nx + 2y = – 1
\n2x – 3y = 12
\nAnswer:
\nGiven: x + 2y = -1 ……. (1)
\n2x – 3y = 12 …….. (2)
\nFrom equation (1)x + 2y = -l
\n\u21d2 x = – 1 – 2y
\nSubstituting x = – 1 – 2y in equation (2), we get
\n2 (- 1 – 2y) – 3y = 12
\n– 2 – 4y – 3y = 12
\n– 2 – 7y = 12
\n7y = – 2 – 12
\n\u2234 y = \\(\\frac{-14}{7}\\) = -2
\nSubstituting y = – 2 in equation (1), we get
\nx + 2 (- 2) = – 1
\nx = – 1 + 4
\nx = 3
\n\u2234 The solution is (3, – 2)<\/p>\n

\"AP<\/p>\n

Question 3.
\n2x + 3y = 9
\n3x + 4y = 5
\nAnswer:
\nGiven: 2x + 3y – 9 …….. (1)
\n3x + 4y = 5 ……. (2)
\nFrom equation (1);
\n2x = 9 – 3y
\nx = \\(\\frac{9-3y}{2}\\)
\nSubstituting x = \\(\\frac{9-3y}{2}\\) in equation (2)
\nwe get
\n\"AP
\nSubstituting y = + 17 in equation (1) we get
\n2x + 3 (+ 17) = 9
\n\u21d2 2x = 9 – 51
\n\u21d2 2x = -42
\n\u21d2 x = -21
\n\u2234 The solution is (-21, 17)<\/p>\n

Question 4.
\nx + \\(\\frac{6}{y}\\) = 6
\n3x – \\(\\frac{8}{y}\\) = 5
\nAnswer:
\nGiven:
\nx + \\(\\frac{6}{y}\\) = 6 …….. (1)
\n3x – \\(\\frac{8}{y}\\) = 5 …….. (2)
\nFrom equation (1) x = 6 – \\(\\frac{6}{y}\\)
\nSubstituting x = 6 – \\(\\frac{6}{y}\\) in equation (2)
\nwe get
\n\"AP
\nSubstituting y = 2 in equation (1) we get
\nx + \\(\\frac{6}{2}\\) = 6 \u21d2 x + 3 = 6
\n\u2234 x = 3
\n\u2234 The solution is (3, 2)<\/p>\n

\"AP<\/p>\n

Question 5.
\n0.2x + 0.3y =1.3
\n0.4x + 0.5y = 2.3
\nAnswer:
\nGiven:
\n0.2x + 0.3y = 1.3
\n\u21d2 2x + 3y = 13 …… (1)
\n0.4x + 0.5y = 2.3
\n\u21d2 4x + 5y = 23 …… (2)
\nFrom equation (1)
\n2x = 13 – 3y
\n\u21d2 x = \\(\\frac{13-3y}{2}\\)
\nSubstituting x = \\(\\frac{13-3y}{2}\\) equation (2) we get
\n\\(\\frac{13-3y}{2}\\) + 5y = 23
\n\u21d2 26 – 6y + 5y = 23
\n\u21d2 -y + 26 = 23
\n\u21d2 y = 26 \u2014 23 = 3
\nSubstituting y = 3 in equaion (1) we get
\n2x + 3(3) = 13
\n\u21d2 2x + 9 = 13
\n\u21d2 2x = 13 – 9
\n\u21d2 2x = 4
\n\u21d2 x = \\(\\frac{4}{2}\\) = 2
\n\u2234 The solution is (2, 3)<\/p>\n

Question 6.
\n\u221a2x + \u221a3y = 0
\n\u221a3x – \u221a8y = 0
\nAnswer:
\nGiven:
\n\u221a2x + \u221a3y = 0 ……. (1)
\n\u221a3x – \u221a8y = 0 ……. (2)
\nSubstitute x = 0 in (1),
\n\u221a2(0) + \u221a3y = 0
\n\u221a3y = 0
\n\u2234 y = 0
\n\u2234 The solution is x = 0, y = 0
\nNote: a1<\/sub>x + b1<\/sub>y + c1<\/sub> = 0
\na2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0, if c1<\/sub> = c2<\/sub> = 0
\nthen, x = 0, y = 0 is a solution.<\/p>\n

\"AP<\/p>\n

Solve each of the following pairs of equations by the elimination method. (Page No. 89)
\nQuestion 7.
\n8x + 5y = 9
\n3x + 2y = 4
\nAnswer:
\nGiven: 8x + 5y = 9 ……. (1)
\n3x + 2y = 4 …….. (2)
\n\"AP
\n\u2234 y = 5
\nSubstituting y = 5 in equation (1) we get
\n8x + 5 \u00d7 5 = 9
\n\u21d2 8x = 9 – 25
\nx = \\(\\frac{-16}{8}\\) = -2
\n\u2234 The solution is (- 2, 5)<\/p>\n

Question 8.
\n2x + 3y = 8
\n4x + 6y = 7
\nAnswer:
\nGiven: 2x + 3y = 8 ……. (1)
\n4x + 6y = 7 …….. (2)
\n\"AP
\nThe lines are parallel.
\n\u2234 The pair of lines has no solution.<\/p>\n

Question 9.
\n3x + 4y = 25
\n5x – 6y = -9
\nAnswer:
\nGiven: 3x + 4y = 25 ……. (1)
\n5x – 6y = -9 …….. (2)
\n\"AP
\nSubstituting y = 4 in equation (1) we get
\n3x + 4 \u00d7 4 = 25
\n3x = 25 – 16
\n\u21d2 x = \\(\\frac{9}{3}\\) = 3
\n\u2234 (3,4) is the solution for given pair of lines.<\/p>\n

\"AP<\/p>\n

Question 10.
\nIn a competitive exam, 3 marks are awarded for every correct answer and for every wrong answer, 1 mark will be deducted. Madhu scored 40 marks in this exam. Had 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer, Madhu would have scored 50 marks. How many questions were there in the test? (Madhu attempted all the questions) .
\nNow use the elimination method to solve the above example – 9.
\nAnswer:
\nThe equations formed are
\n3x – y = 40 ……. (1)
\n4x – 2y = 50 …….. (2)
\nSubstituting y = 5 in equation (1) we get
\n3x – 5 = 40
\n\u21d2 3x = 40 + 5
\n\u21d2 x = \\(\\frac{45}{3}\\) = 15
\nTotal number of questions = Number of correct questions + Number of wrong answers
\n= x + y
\n= 15 + 5 = 20<\/p>\n

Question 11.
\nMary told her 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.” Find the present age of Mary and her daughter. Solve example – 10 by the substitution method.
\nAnswer:
\nThe equations formed are
\nThe equations formed are
\nx – 7y + 42 = 0 ……. (1)
\nx – 3y – 6 = 0 …….. (2)
\nFrom (1), x = – 42 + 7y
\nSubstituting x = – 42 + 7y in equation (2) we get
\n-42 + 7y – 3y – 6 = 0
\n\u21d2 4y – 48 = 0
\n\u21d2 y = \\(\\frac{48}{4}\\) = 12
\nSubstituting y = 12 in equation (2) we get
\nx – 3 \u00d7 12 – 6 = 0
\nx – 36 – 6 = 0
\nx = 42<\/p>\n

\"AP<\/p>\n

Try this<\/span><\/p>\n

Question 1.
\nSolve the given pair of linear equations, (a – b)x + (a + b)y = a2<\/sup> – 2ab – b2<\/sup>
\n(a + b) (x + y) = a2<\/sup> + b2<\/sup> (Page No. 89)
\nAnswer:
\n\"AP
\nx(-2b) = – 2b(a + b)
\n\u21d2 x = (a + b)
\nPut this value of ‘x’ in eq (1) we get
\n(a – b) (a + b) + (a + b)y = a2<\/sup> – 2ab – b2<\/sup>
\na2<\/sup> – b2<\/sup> + (a + b)y = a2<\/sup> – 2ab – b2<\/sup>
\n\u21d2 y = \\(\\frac{-2ab}{a+b}\\)
\n\u2234 Solution to given pair of linear equations x = a + b, y = \\(\\frac{-2ab}{a+b}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables InText Questions AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables InText Questions and Answers. 10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables InText … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":35170,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2023\/02\/Chapter-4-Pair-of-Linear-Equations-in-Two-Variables-InText-Questions.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/6452"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=6452"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/6452\/revisions"}],"predecessor-version":[{"id":34303,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/6452\/revisions\/34303"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/35170"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=6452"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=6452"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=6452"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}