{"id":608,"date":"2020-09-15T11:40:13","date_gmt":"2020-09-15T06:10:13","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=608"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-3-ex-3-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-3-ex-3-2\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2<\/h2>\n

Question 1.
\nForm the pair of linear equations in the following problems, and find their solutions graphically.
\n(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.
\n(ii) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
\nSolution:
\n(i) Strength of the X Std. is 10
\nNumber of boys be ‘y’, then
\nnumber of girls be ‘x’ .
\nx + y = 10 …….. (i)
\nx = y + 4
\n\u2234 x – y = 4 ………. (ii)
\nFrom equations (i) + (ii)
\n\"KSEEB
\n\\(\\quad x=\\frac{14}{2}=7\\)
\nSubstituting the value of ‘x’ in eqn. (i),
\nx + y = 10
\n7 + y = 10
\ny = 10 – 7
\ny = 4.
\n\u2234 Number of girls, x = 7
\nNumber of boys, y = 4.<\/p>\n

(ii) Cost of each pencil be Rs. ‘x’
\nCost of pen be Rs. ‘y’
\n5x + 7y = 50 ………. (i)
\n7x + 5y = 46 ……….. (ii)
\nFrom equation (i) + equation (ii)
\n\"KSEEB
\n\u2234 x + y = 8 …………… (iii)
\nfrom Eqn. (ii) – Eqn. (i),
\n\"KSEEB
\n\u2234 -x + y = 2 …………. (iv)
\nEqn. (iii) + Eqn. (iv)
\n\"KSEEB
\n\u2234 y = 5
\nSubstituting the value of ‘y’ in eqn. (i)
\nx + y = 8
\nx + 5 = 8
\n\u2234 x = 8 – 5 x = 3
\n\u2234Cost of each pencil is Rs. 3.
\nCost of each pen is Rs. 5.<\/p>\n

Question 2.
\nOn comparing the ratios \\(\\frac{a_{1}}{a_{2}}, \\frac{b_{1}}{b_{2}} \\text { and } \\frac{c_{1}}{c_{2}}\\) find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
\ni) 5x – 4y + 8 = 0
\n7x + 6y – 9 = 0
\nii) 9x + 3y + 12 = 0
\n18x + 6y + 24 = 0
\niii) 6x – 3y + 10 = 0
\n2x – y + 9 = 0
\nSolution:
\ni) 5x – 4y + 8 = 0 a1<\/sub>= 5, b1<\/sub> = -4, C1<\/sub>= 8
\n7x + 6y – 9 = 0 a2<\/sub> = 7, b2<\/sub>= 6, c2<\/sub>= -9
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{5}{7} \\quad \\frac{b_{1}}{b_{2}}=\\frac{-4}{6} \\quad \\frac{c_{1}}{c_{2}}=\\frac{8}{-9}\\)
\nHere, \\(\\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}} \\neq \\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}}\\)
\n\u2234 Lines representing the pairs of linear equations intersect at a point.<\/p>\n

ii) 9x + 3y + 12 = 0
\n18x + 6y + 24 = 0
\na1<\/sub> =9, b1<\/sub> = 3, c1<\/sub> = 12
\na2<\/sub> = 18, b2<\/sub> = 6, c2<\/sub> = 24
\n\\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{9}{18}=\\frac{1}{2} \\quad \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}=\\frac{3}{6}=\\frac{1}{2}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{12}{24}=\\frac{1}{2}\\)
\nHere, \\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}=\\frac{\\mathrm{c}_{1}}{\\mathrm{c}_{2}}\\)
\n\u2234 Representation of lines graphically are coincident.<\/p>\n

iii) 6x – 3y + 10 = 0
\n2x – y + 9 = 0
\nHere, a1<\/sub> = 6, b1<\/sub> = -3, c1<\/sub> = 10
\na2<\/sub> = 2, b2<\/sub> = -1, c2<\/sub> = 9
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{6}{2}=\\frac{3}{1} \\quad \\frac{b_{1}}{b_{2}}=\\frac{-3}{-1}=\\frac{3}{1}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{10}{9}\\)
\nHere, \\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}} \\neq \\frac{c_{1}}{c_{2}}\\)
\n\u2234 Representation of lines graphically is parallel.<\/p>\n

Question 3.
\nOn compairing the ratios \\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}, \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}\\) and \\(\\frac{c_{1}}{c_{2}}\\) find out whether the following pair of linear equations are consistent, or inconsistent.
\ni) 3x + 2y = 5; 2x – 3y = 7
\nii) 2x – 3y = 8; 4x-6y = 9
\niii) \\(\\frac{3}{2} x+\\frac{5}{3} y=7 : 9 x-10 y=14\\)
\niv) 5x – 3y = 11: -10x – 6y = -22
\nv) \\(\\frac{4}{3} x+2 y=8 ; 2 x+3 y=12\\)
\nSolution:
\ni) 3x + 2y = 5 \u21d2 3x + 2y – 5 = 0
\n2x – 3y = 7 \u21d2 2x – 3y – 7 = 0
\nHere, a1<\/sub> = 3, b1<\/sub> = 2, c1<\/sub> = -5
\na2<\/sub> = 2, b2<\/sub> = -3, c2<\/sub> = -7
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{3}{2} \\quad \\frac{b_{1}}{b_{2}}=-\\frac{2}{3} \\quad \\frac{c_{1}}{c_{2}}=\\frac{-5}{-7}=\\frac{5}{7}\\)
\nHere, \\(\\frac{a_{1}}{a_{2}} \\neq \\frac{b_{1}}{b_{2}}\\)
\n\u2234 Graphical representation is intersecting lines.
\n\u2234 Pair of linear equations are consistent.<\/p>\n

ii) 2x – 3y = 8 \u21d2 2x – 3y – 8 = 0
\n4x – 6y = 9 \u21d2 4x – 6y – 9 = 0
\na1<\/sub> = 2, b1<\/sub> = -3, c1<\/sub> = -8
\na2<\/sub> = 4, b2<\/sub> = -6, c2<\/sub> = -9
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{2}{4}=\\frac{1}{2} \\quad \\frac{b_{1}}{b_{2}}=\\frac{-3}{-6}=\\frac{1}{2}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{-8}{-9}=\\frac{8}{9}\\)
\nHere, \\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}} \\neq \\frac{\\mathrm{c}_{1}}{\\mathrm{c}_{2}}\\)
\n\u2234 Graphical representation is parallel lines.
\n\u2234 Equations are inconsistent.<\/p>\n

iii) \\(\\frac{3}{2} x+\\frac{5}{3} y=7 \\quad \\frac{3}{2} x+\\frac{5}{3} y-7=0\\)
\n9x – 10y = 14 \u21d2 9x – 10y – 14 = 0
\n\\(a_{1}=\\frac{3}{2}, \\quad b_{1}=\\frac{5}{3}, \\quad c_{1}=-7\\)
\na2<\/sub> = 9, b2<\/sub> = -10, c2<\/sub> = -17
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{3}{2} \\times \\frac{1}{9} \\quad \\frac{b_{1}}{b_{2}}=\\frac{5}{3} \\times \\frac{-1}{6}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{-7}{-14}=\\frac{7}{14}=\\frac{1}{2}\\)
\nHere, \\(\\frac{a_{1}}{a_{2}} \\neq \\frac{b_{1}}{b_{2}}\\)
\n\u2234 Pair of equations are consistent<\/p>\n

iv) 5x – 3y = 11 \u21d2 5x – 3y – 11 = 0
\n-10x + 6y = – 22 \u21d2 -10x + 6y + 22 = 0
\na1<\/sub> = 5, b1<\/sub> = -3, c1<\/sub> = -11
\na2<\/sub> = -10, b2<\/sub> = 6, c2<\/sub> = -22
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{5}{-10}=-\\frac{1}{2} \\quad \\frac{b_{1}}{b_{2}}=\\frac{-3}{6}=-\\frac{1}{2}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{-11}{22}=-\\frac{1}{2}\\)
\nHere, \\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}\\)
\n\u2234 Pair of equations are consistent
\n\u2234 Graphical representation is coninciding.<\/p>\n

v) \\(\\frac{4}{3} x+2 y=8 \\quad \\frac{4}{3} x+26-8=0\\)
\n2x + 3y = 12 \u21d2 2x + 3y – 12 = 0
\n\\(a_{1}=\\frac{4}{3}, \\quad b_{1}=2, \\quad c_{1}=-8\\)
\na2<\/sub> = 2, b2<\/sub> = 3, c2<\/sub> = -12
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{4}{3} \\times \\frac{1}{2}=\\frac{1}{6} \\quad \\frac{b_{1}}{b_{2}}=\\frac{2}{3}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{-8}{-12}=\\frac{2}{3}\\)
\nHere, \\(\\frac{\\mathbf{a}_{1}}{\\mathbf{a}_{2}} \\neq \\frac{\\mathbf{b}_{1}}{\\mathbf{b}_{2}}\\)
\n\u2234 Pair of equations are consistent<\/p>\n

Question 4.
\nWhich of the following pairs of linear equations are consistent\/inconsistent? If consistent obtain the solution graphically
\n(i) x + y = 5, 2x + 2y = 10
\n(ii) x-y = 8 3x-3y= 16
\n(iii) 2x + y – 6 = 0 4x – 2y – 4 = 0
\n(iv) 2x – 2y – 2 = 0 4x – 3y – 5 = 0
\nSolution:
\n(i) x + y = 5 \u21d2 x + y – 5 = 0
\n2x + 2y = 10 \u21d2 2x + 2y – 10 = 0
\na1<\/sub> = 1, b1<\/sub> = 1, c1<\/sub> = -5
\na2<\/sub> = 2, b2<\/sub> = 2, c2<\/sub> = -10
\n\\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{1}{2} \\quad \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}=\\frac{1}{2} \\quad \\frac{\\mathrm{c}_{1}}{\\mathrm{c}_{2}}=\\frac{-5}{-10}=\\frac{1}{2}\\)
\nHere, \\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}=\\frac{\\mathrm{c}_{1}}{\\mathrm{c}_{2}}\\)
\n\u2234 Pair of equations are consistent
\n(i) x + y = 5
\ny = 5 – x<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n2<\/td>\n4<\/td>\n<\/tr>\n
y = 5 – x<\/td>\n5<\/td>\n3<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(ii) 2x + 2y = 10
\nx + y = 5
\ny = 5 – x<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n2<\/td>\n5<\/td>\n<\/tr>\n
y = 5 – x<\/td>\n5<\/td>\n3<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u2234 We can give any value for ‘x’, i.e., solutions are infinite.
\n\u2234 Solution, P (5, 0) x = 5, y = 0
\n\"KSEEB<\/p>\n

(ii) x – y = 8 \u21d2 x – y – 8 = 0
\n3x – 3y = 16 \u21d2 3x – 3y – 16 = 0
\nHere, \\(\\frac{a_{1}}{a_{2}}=\\frac{1}{3} \\quad \\frac{b_{1}}{b_{2}}=\\frac{-1}{-3}=\\frac{1}{3}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{-8}{-16}=\\frac{1}{2}\\)
\n\\(\\quad \\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}} \\neq \\frac{\\mathrm{c}_{1}}{\\mathrm{c}_{2}}\\)
\n\u2234 Linear equations are inconsistenttent.
\n\u2234 Algebraically it has no solution.
\nGraphical representation \u2192 Parallel Lines.
\n(i) x – y = 8
\n-y = 8 – x
\ny = -8 + x<\/p>\n\n\n\n\n
x<\/td>\n8<\/td>\n10<\/td>\n9<\/td>\n<\/tr>\n
y = -8+x<\/td>\n0<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(ii) 3x – 3y = 16
\n-3y = 16 – 3x
\n3y = -16 + 3x
\n\\(\\quad y=\\frac{-16+3 x}{3}\\)<\/p>\n\n\n\n\n
x<\/td>\n6<\/td>\n8<\/td>\n<\/tr>\n
\\(y=\\frac{-16+3 x}{3}\\)<\/td>\n0.8<\/td>\n2.6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

No solution because it is inconsistent
\n\"KSEEB<\/p>\n

(iii) 2x + y – 6 = 0
\n4x – 2y – 4 = 0
\nHere a1<\/sub> = 2, b1<\/sub> = 1, c1<\/sub> = -6
\na2<\/sub> = 4, b2<\/sub> = -2, c2<\/sub> = -4
\n\\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{2}{4}=\\frac{1}{2} \\quad \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}=\\frac{1}{-2}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{-6}{-4}=\\frac{3}{2}\\)
\nHere, \\(\\frac{a_{1}}{a_{2}} \\neq \\frac{b_{1}}{b_{2}}\\)
\nPair of equations are consistent. Algebraically both lines intersect.
\nGraphical Representation :
\n(i) 2x + y = 6
\ny = 6 – 2x<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n2<\/td>\n<\/tr>\n
y = 6 – 2x<\/td>\n6<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(ii) 4x – 2y – 4 = 0
\n4x – 2y = 4
\n-2y = 4 – 4x
\n2y = -4 + 4x
\n\\(\\quad y=\\frac{-4+4 x}{2}\\)<\/p>\n\n\n\n\n
x<\/td>\n1<\/td>\n3<\/td>\n<\/tr>\n
\\(y=\\frac{-4+4 x}{2}\\)<\/td>\n0<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Solution: intersecting point, P (2, 2) i.e., x = 2, y = 2
\n\"KSEEB<\/p>\n

(iv) 2x – 2y – 2 = 0
\n4x – 3y – 5 = 0
\na1<\/sub> = 2, b1<\/sub> = -2, c1<\/sub> = -2
\na2<\/sub> = 4, b2<\/sub> = -3, c2<\/sub> = -5
\n\\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{2}{4}=\\frac{1}{2} \\quad \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}=\\frac{-2}{-3}=\\frac{2}{3}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{-2}{-5}=\\frac{2}{5}\\)
\nHere, \\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}} \\neq \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}\\)
\nPair of equations are consistent.
\n\u2234 Algebraically both lines intersect.
\nGraphical Representation :
\n(i) 2x – 2y – 2 =0
\n2x – 2y = 2
\n-2y = 2 – 2x
\n2y = -2 + 2x
\n\\(\\quad y=\\frac{-2+2 x}{2}\\)
\n\u2234 y = – 1 + x<\/p>\n\n\n\n\n
x<\/td>\n2<\/td>\n4<\/td>\n<\/tr>\n
y= -1+ x<\/td>\n1<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(ii) 4x – 3y – 5 = 0
\n4x – 3y = 5
\n-3y = 5 – 4x
\n3y = -5 + 4x
\n\\(\\quad y=\\frac{-5+4 x}{3}\\)<\/p>\n\n\n\n\n
x<\/td>\n2<\/td>\n5<\/td>\n<\/tr>\n
\\(y=\\frac{-5+4 x}{3}\\)<\/td>\n1<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Solution: P(2, 1) i.e., x = 2, y = 1
\n\"KSEEB<\/p>\n

Question 5.
\nHalf the perimeter of a rectangular garden, whose length is 4m more than its width, is 36m. Find the dimensions of the garden.
\n\"KSEEB
\nSolution:
\nLength of rectangular garden be ‘x’ m.
\nBreadth of rectangular garden be ‘y’ m, then
\nLength of the garden is 4m more than its width.
\nx = y + 4 ……….. (i)
\nx – y = 4
\nHalf of the circumference is 36 m.
\n\\(\\frac{2 x+2 y}{2}=36\\)
\n2x + 2y = 72
\nx + y = 36 …………… (ii)
\n\u2234 x – y = 4 (i)
\nx + y = 36 (ii)
\nFrom eqn. (i) + eqn. (ii)
\n\"KSEEB
\n\\(\\quad x=\\frac{40}{2}\\)
\n\u2234 x = 20 m.
\nSubstituting the value of \u2019x’ in eqn. (i)
\nx – y = 4
\n20 – y = 4
\n-y = 4 – 20
\n-y = -16
\ny = 16 m.
\n\u2234 Length of the garden = 20 m.
\nBreadth of the garden = 16 m.<\/p>\n

Question 6.
\nGiven 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 :
\n(i) intersecting lines
\n(ii) parallel lines
\n(iii) coincident lines.
\nSolution:
\nLinear equation is 2x + 3y – 8 = 0.
\n(i) Linear equations for intersecting lines:
\n2x + 3y – 8 = 0
\n3x + 2y – 7 = 0
\na1<\/sub> = 2, b1<\/sub> = 3, c1<\/sub> = -8
\na2 <\/sub>= 3, b2<\/sub> = 2, c2<\/sub> = -7
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{2}{3} \\quad \\frac{b_{1}}{b_{2}}=\\frac{3}{2}\\)
\nHere, when \\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}} \\neq \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}\\) Geometrical representation is intersecting lines.<\/p>\n

(ii) For Parallel lines :
\n2x + 3y – 8 = 0
\n2x + 3y – 12 = 0
\na1<\/sub> = 2, b1<\/sub> = 3, c1<\/sub>= -8
\na1<\/sub> = 2, b1<\/sub> = 3, c1<\/sub> = -12
\n\\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{2}{2}=\\frac{1}{1} \\quad \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}=\\frac{3}{3}=\\frac{1}{1}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{-8}{-12}=\\frac{8}{12}=\\frac{2}{3}\\)
\nHere, \\(\\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}} \\neq \\frac{\\mathrm{c}_{1}}{\\mathrm{c}_{2}}\\) hence
\nGraphical representation is parallel lines.<\/p>\n

(iii) For Intersecting lines :
\n2x + 3y – 8 = 0
\n4x + 6y – 16 = 0
\na1<\/sub> = 2, b1<\/sub> = 3, c1<\/sub> = -8
\na1<\/sub> = 4, b1<\/sub> = 6, c1<\/sub> = -16
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{2}{4}=\\frac{1}{2} \\quad \\frac{b_{1}}{b_{2}}=\\frac{3}{6}=\\frac{1}{2}\\)
\n\\(\\frac{c_{1}}{c_{2}}=\\frac{-8}{-16}=\\frac{1}{2}\\)
\nHere,
\n\u2234 Lines are intersecting.<\/p>\n

Question 7.
\nDraw 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.
\nSolution:
\nx – y + 1 = 0 ………. (i)
\n3x – 2y – 12 = 0 ……….. (ii)
\nx – y + 1 = 0
\n-y = -x -1
\ny = x + 1<\/p>\n\n\n\n\n
x<\/td>\n2<\/td>\n4<\/td>\n<\/tr>\n
y = x + 1<\/td>\n3<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

3x + 2y – 12 = 0
\n2y = -3x + 12
\n\\(\\quad y=\\frac{-3 x+12}{2}\\)<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n2<\/td>\n<\/tr>\n
\\(y=\\frac{-3 x+12}{2}\\)<\/td>\n6<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB
\nGraphs of these two equations intersect at A. Vertices formed for \u2206ABC are,
\nA (2, 3), B (-1, 0), C (4, 0).<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2. Karnataka SSLC Class 10 Maths Solutions Chapter 3 …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/608"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=608"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/608\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=608"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=608"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=608"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}