Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n 3x \u2013 5y \u2013 4 = 0<\/strong> On putting $y=-\\frac{5}{13}$ in Eq. (ii), we get Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n 3x + 4y = 10<\/strong> Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n x + y = 5<\/strong> On putting $y=\\frac{6}{5}$ in Eq. (i), we get Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n 2x + 3y = 8<\/strong> Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n 8x + 5y = 9<\/strong> Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n 2x + 3y = 46<\/strong> Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n 0.4x \u2013 1.5y = 6.5<\/strong> On putting y = \u2013 3 in Eq. (ii), we get Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n \u221a2x \u2013 \u221a3y = 0<\/strong> Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n 2x + 5y = 1<\/strong> Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n $3 x-\\frac{8}{y}=5$<\/strong> On adding Eq. (i) and Eq. (ii), we get On putting x=2 in Eq. (ii), we get Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n $\\frac{x}{6}+\\frac{y}{15}=4$<\/strong> On subtracting Eq. (ii) from Eq. (iii), we get<\/p>\n\n\n\n $\\frac{x}{6}-\\frac{y}{24}-\\frac{x}{6}-\\frac{y}{15}=\\frac{19}{8}-4$<\/p>\n\n\n\n $\\Rightarrow-\\frac{y}{24}-\\frac{y}{15}=\\frac{19-32}{8}$<\/p>\n\n\n\n $\\Rightarrow \\frac{-5 y-8 y}{120}=\\frac{19-32}{8}$<\/p>\n\n\n\n $\\Rightarrow \\frac{-13 y}{120}=\\frac{-13}{8}$<\/p>\n\n\n\n $\\Rightarrow \\mathrm{y}=\\frac{13}{8} \\times \\frac{120}{13}$ On putting y = 15 in Eq. (ii), we get Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n $x+\\frac{6}{y}=6$<\/strong> On subtracting Eq. (ii) from Eq. (iii), we get On putting y = 2 in Eq. (i), we get Solve the following equations by elimination method:<\/strong><\/p>\n\n\n\n 37x + 43y = 123<\/strong> On subtracting Eq. (iii) from Eq. (iv), we get On putting y = 2 in Eq. (ii), we get Solve the following equations by elimination method:<\/strong><\/p>\n\n\n\n 217x + 131y = 913<\/strong> Solve the following equations by elimination method:<\/strong><\/p>\n\n\n\n 99x + 101y = 499<\/strong> On subtracting Eq. (iii) from Eq. (iv), we get On putting y = 2 in Eq. (i), we get Solve the following equations by elimination method:<\/strong><\/p>\n\n\n\n 29x \u2013 23y = 110<\/strong> Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n $\\frac{1}{x}-\\frac{1}{y}=1$<\/strong> Adding Eq. (i) and Eq. (ii), we get On putting $\\mathrm{X}=\\frac{1}{4}$ in Eq. (ii), we get $\\Rightarrow 4+\\frac{1}{\\mathrm{y}}=7$<\/p>\n\n\n\n $\\Rightarrow \\frac{1}{y}=7-4=3$ Hence, $x=\\frac{1}{4}$ and $=\\frac{1}{3}$ , which is the required solution.<\/p>\n\n\n\n Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n $\\frac{2}{x}+\\frac{3}{y}=13$<\/strong> On multiplying Eq. (i) by 5 and Eq. (ii) by 2 to make the coefficients of x equal, we get the equation as On subtracting Eq. (iii) from Eq. (iv), we get On putting $y=\\frac{1}{2}$ in Eq. (ii), we get $\\Rightarrow \\frac{5}{x}-8=7$<\/p>\n\n\n\n $\\Rightarrow \\frac{5}{x}=15$<\/p>\n\n\n\n $\\Rightarrow x=\\frac{1}{3}$<\/p>\n\n\n\n Hence, $x=\\frac{1}{3}$ and $=\\frac{1}{2}$ , which is the required solution.<\/p>\n\n\n\n Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n $\\frac{3 a}{x}-\\frac{2 b}{y}+5=0$ , $\\frac{a}{x}+\\frac{3 b}{y}-2=0$, $(\\mathrm{x} \\neq 0, \\mathrm{y} \\neq 0)$<\/strong><\/p>\n\n\n\n Sol : On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of x equal, we get the equation as On subtracting Eq. (iii) from Eq. (iv), we get<\/p>\n\n\n\n $\\frac{12}{x}-\\frac{20}{y}-\\frac{12}{x}-\\frac{21}{y}=-8-33$ \u21d2y=1<\/p>\n\n\n\n On putting y=1 in Eq. (ii), we get $\\Rightarrow \\frac{3}{x}=-2+5$<\/p>\n\n\n\n $\\Rightarrow \\frac{3}{x}=3$<\/p>\n\n\n\n \u21d2x=1<\/p>\n\n\n\n Hence, $x=\\frac{1}{3}$ and $=\\frac{1}{2}$ , which is the required solution.<\/p>\n\n\n\n Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n $\\frac{3 a}{x}-\\frac{2 b}{y}+5=0$ , $\\frac{a}{x}+\\frac{3 b}{y}-2=0$, $(\\mathrm{x} \\neq 0, \\mathrm{y} \\neq 0)$<\/strong><\/p>\n\n\n\n Sol : On multiplying Eq. (ii) by 3 to make the coefficients of x equal, we get the equation as On subtracting Eq. (i) from Eq. (iii), we get On putting y = b in Eq. (ii), we get $\\Rightarrow \\frac{a}{x}=2-3$<\/p>\n\n\n\n $\\Rightarrow \\frac{\\mathrm{a}}{\\mathrm{x}}=-1$ Hence, x = \u2013 a and y = b , which is the required solution.<\/p>\n\n\n\n Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n $\\frac{2 x+5 y}{x y}=6$, $\\frac{4 x-5 y}{x y}=-3$<\/strong>, where x \u2260 0 and y \u2260 0<\/strong><\/p>\n\n\n\n Sol : Or 2x + 5y = 6xy \u2026(i) On adding Eq. (i) and Eq. (ii), we get On putting y = 2 in Eq. (ii), we get On putting x = 0 , we get y = 0 Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n x + y = 2xy<\/strong> $\\Rightarrow \\mathrm{y}=\\frac{1}{4}$ and x=0<\/p>\n\n\n\n On putting $y=\\frac{1}{4}$ in Eq. (ii), we get On putting x = 0 , we get y = 0 Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n 5x + 3y = 19xy<\/strong> On putting $y=\\frac{1}{2}$ in Eq. (ii), we get On putting x = 0 , we get y = 0<\/p>\n\n\n\n Hence, $x=\\frac{1}{3}, 0$ and $=\\frac{1}{2}, 0$ , which is the required solution.<\/p>\n\n\n\n Solve the following system of equations by elimination method:<\/strong><\/p>\n\n\n\n x + y = 7xy<\/strong> $\\Rightarrow x=\\frac{1}{3}$ and y=0<\/p>\n\n\n\n On putting $x=\\frac{1}{3}$ in Eq. (ii), we get On putting x = 0 , we get x = 0 Solve for x and y the following system of equations:<\/strong><\/p>\n\n\n\n $\\frac{1}{2(2 x+3 y)}+\\frac{12}{7(3 x-2 y)}=\\frac{1}{2}$ , $\\frac{7}{(2 x+3 y)}+\\frac{4}{(3 x-2 y)}=2$<\/strong><\/p>\n\n\n\n Where (2x + 3y) \u2260 0 and (3x \u2013 2y) \u2260 0<\/strong> On multiplying Eq. (i) by 7 and Eq. (ii) by $\\frac{1}{2}$ to make the coefficients equal of first term, we get the equation as $\\Rightarrow \\frac{7}{2(2 x+3 y)}+\\frac{2}{(3 x-2 y)}=1$ \u2026(iv)<\/p>\n\n\n\n On substracting Eq. (iii) from Eq. (iv), we get \u21d23x-2y=4<\/p>\n\n\n\n \u21d23x=4+2y On multiplying Eq. (ii) by $\\frac{3}{7}$ to make the coefficients equal of second term, we get the equation as On substracting Eq. (i) from Eq. (iv), we get \u21d22x+3y=7<\/p>\n\n\n\n \u21d2$\\mathrm{X}=\\frac{7-3 \\mathrm{y}}{2}$ \u2026(b)<\/p>\n\n\n\n From Eq. (a) and (b), we get On putting the value of y = 1 in Eq. (b), we get Hence, x = 2 and y = 1 , which is the required solution.<\/p>\n\n\n\n Solve for x and y the following system of equations:<\/strong><\/p>\n\n\n\n $\\frac{6}{x-1}-\\frac{3}{y-2}=1, x \\neq 1, y \\neq 1$<\/strong> On multiplying Eq. (i) by 3 and Eq. (ii) by 2 to make the coefficients equal of first term, we get the equation as $\\frac{6}{x-1}+\\frac{4}{y+1}=\\frac{13}{3}$ \u2026(iv)<\/p>\n\n\n\n On substracting Eq. (iii) from Eq. (iv), we get On putting the value of y = 2 in Eq. (ii), we get Solve for x and y the following system of equations:<\/strong><\/p>\n\n\n\n $\\frac{44}{x+y}+\\frac{30}{x-y}=10$<\/strong> On multiplying Eq. (i) by 4 and Eq. (ii) by 3 to make the coefficients equal of second term, we get the equation as $\\frac{165}{x+y}+\\frac{120}{x-y}=39$ \u2026(iv)<\/p>\n\n\n\n On substracting Eq. (iii) from Eq. (iv), we get On putting the value of x + y = 11 in Eq. (1), we get Adding Eq. (a) and (b), we get On putting value of x = 8 in eq. (a), we get Solve for x and y the following system of equations:<\/strong><\/p>\n\n\n\n $\\frac{5}{x-1}+\\frac{1}{y-2}=2$ ,$\\frac{6}{x-1}-\\frac{3}{y-2}=1$<\/strong> On multiplying Eq. (i) by 3 to make the coefficients equal of second term, we get the equation as On adding Eq. (ii) and Eq. (iii), we get On putting the value of x = 4 in Eq. (ii), we get Form the pair of linear equations for the following problems and find their solution by elimination method:<\/strong><\/p>\n\n\n\n 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.” Find their present ages.<\/strong> Form the pair of linear equations for the following problems and find their solution by elimination method:<\/strong><\/p>\n\n\n\n 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?<\/strong> According to the question, On subtracting Eq. (ii) from (i) we get On putting y = 20 in Eq. (i) we get Form the pair of linear equations for the following problems and find their solution by elimination method:<\/strong><\/p>\n\n\n\n The difference between two numbers is 26 and one number is three times the other. Find them.<\/strong>
9x = 2y + 7<\/strong>
Sol :
Given pair of linear equations is
3x \u2013 5y \u2013 4 = 0 \u2026(i)
And 9x = 2y + 7 \u2026(ii)
On multiplying Eq. (i) by 3 to make the coefficients of x equal, we get the equation as
9x \u2013 15y \u2013 12 = 0 \u2026(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
9x \u2013 15y \u2013 12 \u2013 9x = 0 \u2013 2y \u2013 7
\u21d2 \u2013 15y + 2y = \u2013 7 + 12
\u21d2 \u2013 13y = 5
$\\Rightarrow \\mathrm{y}=-\\frac{5}{13}$<\/p>\n\n\n\n
$9 x=2\\left(-\\frac{5}{13}\\right)+7$
$\\Rightarrow 9 x=-\\frac{10}{13}+7$
$\\Rightarrow 9 x=\\frac{-10+91}{13}$
$\\Rightarrow 9 x=\\frac{81}{13}$
$\\Rightarrow x=\\frac{81}{13 \\times 9}$
$\\Rightarrow x=\\frac{9}{13}$
Hence, $x=\\frac{9}{13}$ and $=-\\frac{5}{13}$ , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 1 B <\/h4>\n\n\n\n
2x \u2013 2y = 2<\/strong>
Sol :
Given pair of linear equations is
3x + 4y = 10 \u2026(i)
And 2x \u2013 2y = 2 \u2026(ii)
On multiplying Eq. (i) by 2 and Eq. (ii) by 3 to make the coefficients of x equal, we get the equation as
6x + 8y = 20 \u2026(iii)
6x \u2013 6y = 6 \u2026(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
6x \u2013 6y \u2013 6x \u2013 8y = 6 \u2013 20
\u21d2 \u2013 14y = \u2013 14
\u21d2 y = 1
On putting y = 1 in Eq. (ii), we get
2x \u2013 2(1) = 2
\u21d2 2x \u2013 2 = 2
\u21d2 x = 2
Hence, x = 2 and y = 1 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 1 C <\/h4>\n\n\n\n
2x \u2013 3y = 4<\/strong>
Sol :
Given pair of linear equations is
x + y = 5 \u2026(i)
And 2x \u2013 3y = 4 \u2026(ii)
On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as
2x + 2y = 10 \u2026(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
2x + 2y \u2013 2x + 3y = 10 \u2013 4
\u21d2 5y = 6
$\\Rightarrow \\mathrm{y}=\\frac{6}{5}$<\/p>\n\n\n\n
x + y = 5
$\\Rightarrow x+\\frac{6}{5}=5$
$\\Rightarrow x=\\frac{25-6}{5}$
$\\Rightarrow x=\\frac{19}{5}$
Hence, $x=\\frac{19}{5}$ and $=\\frac{6}{5}$ , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 1 D <\/h4>\n\n\n\n
4x + 6y = 7<\/strong>
Sol :
Given pair of linear equations is
2x + 3y = 8 \u2026(i)
And 4x + 6y = 7 \u2026(ii)
On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as
4x + 6y = 16 \u2026(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
4x + 6y \u2013 4x \u2013 6y = 16 \u2013 7
\u21d2 0 = 9
Which is a false equation involving no variable.
So, the given pair of linear equations has no solution i.e. this pair of linear equations is inconsistent.<\/p>\n\n\n\n
\n\n\n\nQuestion 1 E <\/h4>\n\n\n\n
3x + 2y = 4<\/strong>
Sol :
Given pair of linear equations is
8x + 5y = 9 \u2026(i)
And 3x + 2y = 4 \u2026(ii)
On multiplying Eq. (i) by 2 and Eq. (ii) by 5 to make the coefficients of y equal, we get the equation as
16x + 10y = 18 \u2026(iii)
15x + 10y = 20 \u2026(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
15x + 10y \u2013 16x \u2013 10y = 20 \u2013 18
\u21d2 \u2013 x = 2
\u21d2 x = \u2013 2
On putting x = \u2013 2 in Eq. (ii), we get
3x + 2y = 4
\u21d2 3( \u2013 2) + 2y = 4
\u21d2 \u2013 6 + 2y = 4
\u21d2 2y = 4 + 6
\u21d2 2y = 10
$\\Rightarrow \\mathrm{y}=\\frac{10}{2}=5$
Hence, x=-2 and =5, which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 1 F <\/h4>\n\n\n\n
3x + 5y = 74<\/strong>
Sol :
Given pair of linear equations is
2x + 3y = 46 \u2026(i)
And 3x + 5y = 74 \u2026(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 2 to make the coefficients of x equal, we get the equation as
6x + 9y = 138 \u2026(iii)
6x + 10y = 148 \u2026(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
6x + 10y \u2013 6x \u2013 9y = 148 \u2013 138
\u21d2 y = 10
On putting y = 10 in Eq. (ii), we get
3x + 5y = 74
\u21d2 3x + 5(10) = 74
\u21d2 3x + 50 = 74
\u21d2 3x = 74 \u2013 50
\u21d2 3x = 24
\u21d2 x = 8
Hence, x = 8 and y = 10 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 1 G <\/h4>\n\n\n\n
0.3x + 0.2y = 0.9<\/strong>
Sol :
Given pair of linear equations is
0.4x \u2013 1.5y = 6.5 \u2026(i)
And 0.3x + 0.2y = 0.9 \u2026(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of x equal, we get the equation as
1.2x \u2013 4.5y = 19.5 \u2026(iii)
1.2x + 0.8y = 3.6 \u2026(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
1.2x + 0.8y \u2013 1.2x + 4.5y = 3.6 \u2013 19.5
\u21d2 5.3y = \u2013 15.9
$\\Rightarrow \\mathrm{y}=-\\frac{15.9}{5.3}$
\u21d2 y = \u2013 3<\/p>\n\n\n\n
0.3x + 0.2y = 0.9
\u21d2 0.3x + 0.2( \u2013 3) = 0.9
\u21d2 0.3x \u2013 0.6 = 0.9
\u21d2 0.3x = 1.5
\u21d2 x = 1.5\/0.3
\u21d2 x = 5
Hence, x = 5 and y = \u2013 3 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 1 H <\/h4>\n\n\n\n
\u221a5x + \u221a2y = 0<\/strong>
Sol :
Given pair of linear equations is
\u221a2 x \u2013 \u221a3 y = 0 \u2026(i)
And \u221a5 x + \u221a2 y = 0 \u2026(ii)
On multiplying Eq. (i) by \u221a2 and Eq. (ii) by \u221a3 to make the coefficients of y equal, we get the equation as
2x \u2013 \u221a6 y = 0 \u2026(iii)
\u221a15 x + \u221a6 y = 0 \u2026(iv)
On adding Eq. (iii) and (iv), we get
2x \u2013 \u221a6 y + \u221a15 x + \u221a6 y = 0
\u21d2 2x + \u221a15 x = 0
\u21d2 x(2 + \u221a15) = 0
\u21d2 x = 0
On putting x = 0 in Eq. (i), we get
\u221a2 x \u2013 \u221a3 y = 0
\u21d2 \u221a2(0) \u2013 \u221a3 y = 0
\u21d2 \u2013 \u221a3 y = 0
\u21d2 y = 0
Hence, x = 0 and y = 0 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 1 I <\/h4>\n\n\n\n
2x + 3y = 3<\/strong>
Sol :
Given pair of linear equations is
2x + 5y = 1 \u2026(i)
And 2x + 3y = 3 \u2026(ii)
On subtracting Eq. (ii) from Eq. (i), we get
2x + 3y \u2013 2x \u2013 5y = 3 \u2013 1
\u21d2 \u2013 2y = 2
\u21d2 y = \u2013 1
On putting y = \u2013 1 in Eq. (ii), we get
2x + 3( \u2013 1) = 3
\u21d2 2x \u2013 3 = 3
\u21d2 2x = 6
\u21d2 x = 6\/2
\u21d2 x = 3
Hence, x = 3 and y = \u2013 1 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 2 A <\/h4>\n\n\n\n
$x-\\frac{y}{3}=3$
Sol :
Given pair of linear equations is
$\\frac{x}{2}+\\frac{2 y}{3}=-1$ \u2026(i)
And $x-\\frac{y}{3}=3$ \u2026(ii)
On multiplying Eq. (ii) by 2 to make the coefficients of y equal, we get the equation as
$2 x-\\frac{2 y}{3}=6$ \u2026(iii)<\/p>\n\n\n\n
$\\frac{x}{2}+\\frac{2 y}{3}+x-\\frac{2 y}{3}=-1+6$
$\\Rightarrow \\frac{\\mathrm{x}}{2}+2 \\mathrm{x}=5$
$\\Rightarrow \\frac{x+4 x}{2}=5$
$\\Rightarrow \\frac{5 x}{2}=5$
\u21d2 x = 2<\/p>\n\n\n\n
$x-\\frac{y}{3}=3$
$\\Rightarrow 2-\\frac{y}{3}=3$
$\\Rightarrow-\\frac{y}{3}=3-2$
$\\Rightarrow-\\frac{y}{3}=1$
\u21d2 y = \u2013 3
Hence, x = 2 and y = \u2013 3 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 2 B <\/h4>\n\n\n\n
$\\frac{x}{3}-\\frac{y}{12}=\\frac{19}{4}$<\/strong>
Sol :
Given pair of linear equations is
$\\frac{x}{6}+\\frac{y}{15}=4$ \u2026(i)
And $\\frac{x}{3}-\\frac{y}{12}=\\frac{19}{4}$ \u2026(ii)
On multiplying Eq. (ii) by $\\frac{1}{2}$ to make the coefficients of x equal, we get the equation as
$\\frac{x}{6}-\\frac{y}{24}=\\frac{19}{8}$ \u2026(iii)<\/p>\n\n\n\n
\u21d2 y = 15<\/p>\n\n\n\n
$\\frac{x}{6}+\\frac{y}{15}=4$
$\\Rightarrow \\frac{x}{6}+\\frac{15}{15}=4$
$\\Rightarrow \\frac{x}{6}=4-1$
$\\Rightarrow \\frac{x}{6}=3$
\u21d2 x = 18
Hence, x = 18 and y = 15 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 2 C <\/h4>\n\n\n\n
$3 x-\\frac{8}{y}=5$<\/strong>
Sol :
Given pair of linear equations is
$x+\\frac{6}{y}=6$ \u2026(i)
And $3 x-\\frac{8}{y}=5$ \u2026(ii)
On multiplying Eq. (i) by 3 to make the coefficients of x equal, we get the equation as
$3 x+\\frac{18}{y}=18$ \u2026(iii)<\/p>\n\n\n\n
$3 x+\\frac{18}{y}-3 x+\\frac{8}{y}=18-5$
$\\Rightarrow \\frac{26}{\\mathrm{y}}=13$
\u21d2 y = 2<\/p>\n\n\n\n
$x+\\frac{6}{y}=6$
$\\Rightarrow x+\\frac{6}{2}=6$
\u21d2 x = 3
Hence, x = 3 and y = 2 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 3 A <\/h4>\n\n\n\n
43x + 37y = 117<\/strong>
Sol :
Given pair of linear equations is
37x + 43y = 123 \u2026(i)
And 43x + 37y = 117 \u2026(ii)
On multiplying Eq. (i) by 43 and Eq. (ii) by 37 to make the coefficients of x equal, we get the equation as
1591x + 1849y = 5289 \u2026(iii)
1591x + 1369y = 4329 \u2026(iv)<\/p>\n\n\n\n
\u21d2 1591x + 1369y \u2013 1591x \u2013 1849y = 4329 \u2013 5289
\u21d2 \u2013 480y = \u2013 960
$\\Rightarrow \\mathrm{y}=\\frac{960}{480}$
\u21d2 y = 2<\/p>\n\n\n\n
\u21d2 43x + 37(2) = 117 \u21d2 43x + 74 = 117
\u21d2 43x = 117 \u2013 74
\u21d2 43x = 43
\u21d2 x = 1
Hence, x = 1 and y = 2 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 3 B <\/h4>\n\n\n\n
131x + 217y = 827<\/strong>
Sol :
Given pair of linear equations is
217x + 131y = 913 \u2026(i)
And 131x + 217y = 827 \u2026(ii)
On multiplying Eq. (i) by 131 and Eq. (ii) by 217 to make the coefficients of x equal, we get the equation as
28427x + 17161y = 119603 \u2026(iii)
28427x + 47089y = 179459 \u2026(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
\u21d2 28427x + 47089y \u2013 28427x \u2013 17161y = 179459 \u2013 119603
\u21d2 47089y \u2013 17161y = 179459 \u2013 119603
\u21d2 29928y = 59856
$\\Rightarrow \\mathrm{y}=\\frac{59856}{29928}$
\u21d2 y = 2
On putting y = 2 in Eq. (ii), we get
\u21d2 131x + 217(2) = 827 \u21d2 131x + 434 = 827
\u21d2 131x = 393
$\\Rightarrow x=\\frac{393}{131}$
\u21d2 x = 3
Hence, x = 3 and y = 2 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 3 C <\/h4>\n\n\n\n
101x + 99y = 501<\/strong>
Sol :
Given pair of linear equations is
99x + 101y = 499 \u2026(i)
And 101x + 99y = 501 \u2026(ii)
On multiplying Eq. (i) by 101 and Eq. (ii) by 99 to make the coefficients of x equal, we get the equation as
9999x + 10201y = 50399 \u2026(iii)
9999x + 9801y = 49599 \u2026(iv)<\/p>\n\n\n\n
\u21d2 9999x + 9801y \u2013 9999x \u2013 10201y = 49599 \u2013 50399
\u21d2 9801y \u2013 10201y = 49599 \u2013 50399
\u21d2 \u2013 400y = \u2013 800
$\\Rightarrow \\mathrm{y}=\\frac{800}{400}$
\u21d2 y = 2<\/p>\n\n\n\n
\u21d2 99x + 101(2) = 499 \u21d2 99x + 202 = 499
\u21d2 99x = 297
\u21d2 x = 297\/99
\u21d2 x = 3
Hence, x = 3 and y = 2 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 3 D <\/h4>\n\n\n\n
23x \u2013 29y = 98<\/strong>
Sol :
Given pair of linear equations is
29x \u2013 23y = 110 \u2026(i)
And 23x \u2013 29y = 98 \u2026(ii)
On multiplying Eq. (i) by 23 and Eq. (ii) by 29 to make the coefficients of x equal, we get the equation as
667x \u2013 529y = 2530 \u2026(iii)
667x \u2013 841y = 2842 \u2026(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
\u21d2 667x \u2013 841y \u2013 667x + 529y = 2842 \u2013 2530
\u21d2 \u2013 312y = 312
\u21d2 y = \u2013 1
On putting y = 2 in Eq. (ii), we get
\u21d2 29x \u2013 23( \u2013 1) = 110 \u21d2 29x + 23 = 110
\u21d2 29x = 110 \u2013 23
\u21d2 29x = 87
\u21d2 x = 3
Hence, x = 3 and y = \u2013 1 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 4 A <\/h4>\n\n\n\n
$\\frac{1}{\\mathrm{x}}+\\frac{1}{\\mathrm{y}}=7, \\mathrm{x} \\neq 0, \\mathrm{y} \\neq 0$<\/strong>
Sol :
Given pair of linear equations is
$\\frac{1}{x}-\\frac{1}{y}=1$ \u2026(i)
And $\\frac{1}{x}+\\frac{1}{y}=7$ \u2026(ii)<\/p>\n\n\n\n
$\\frac{1}{x}-\\frac{1}{y}+\\frac{1}{x}+\\frac{1}{y}=1+7$
$\\Rightarrow \\frac{1}{x}+\\frac{1}{x}=8$
$\\Rightarrow \\frac{2}{x}=8$
$\\Rightarrow \\frac{2}{8}=x$
$\\Rightarrow x=\\frac{1}{4}$<\/p>\n\n\n\n
$\\frac{1}{x}+\\frac{1}{y}=7$
$\\Rightarrow \\frac{1}{\\frac{1}{4}}+\\frac{1}{y}=7$<\/p>\n\n\n\n
$\\Rightarrow \\mathrm{y}=\\frac{1}{3}$<\/p>\n\n\n\n
\n\n\n\nQuestion 4 B <\/h4>\n\n\n\n
$\\frac{5}{x}-\\frac{4}{y}=-2, x \\neq 0, y \\neq 0$<\/strong>
Sol :
Given pair of linear equations is
$\\frac{2}{x}+\\frac{3}{y}=13$ \u2026(i)
And $\\frac{5}{x}-\\frac{4}{y}=-2$ \u2026(ii)<\/p>\n\n\n\n
$\\frac{10}{x}+\\frac{15}{y}=65$ \u2026(iii)
$\\frac{10}{x}-\\frac{8}{y}=-4$ \u2026(iv)<\/p>\n\n\n\n
$\\frac{10}{x}-\\frac{8}{y}-\\frac{10}{x}-\\frac{15}{y}=-4-65$
$\\Rightarrow-\\frac{8}{y}-\\frac{15}{y}=-69$
$\\Rightarrow-\\frac{23}{\\mathrm{y}}=-69$
$\\Rightarrow \\frac{23}{69}=\\mathrm{y}$
$\\Rightarrow \\mathrm{y}=\\frac{1}{2}$<\/p>\n\n\n\n
$\\frac{5}{x}-\\frac{4}{y}=-2$
$\\Rightarrow \\frac{5}{x}-\\frac{4}{\\frac{1}{2}}=7$<\/p>\n\n\n\n
\n\n\n\nQuestion 4 C <\/h4>\n\n\n\n
Given pair of linear equations is
$\\frac{4}{x}+\\frac{7}{y}=11$ \u2026(i)
And $\\frac{3}{x}-\\frac{5}{y}=-2$ \u2026(ii)<\/p>\n\n\n\n
$\\frac{12}{x}+\\frac{21}{y}=33$ \u2026(iii)
$\\frac{12}{x}-\\frac{20}{y}=-8$ \u2026(iv)<\/p>\n\n\n\n
$\\Rightarrow-\\frac{20}{\\mathrm{y}}-\\frac{21}{\\mathrm{y}}=-41$
$\\Rightarrow-\\frac{41}{\\mathrm{y}}=-41$
$\\Rightarrow \\frac{41}{41}=\\mathrm{y}$<\/p>\n\n\n\n
$\\frac{3}{x}-\\frac{5}{y}=-2$
$\\Rightarrow \\frac{3}{x}-5=-2$<\/p>\n\n\n\n
\n\n\n\nQuestion 4 D <\/h4>\n\n\n\n
Given pair of linear equations is
$\\frac{3 a}{x}-\\frac{2 b}{y}=-5$ \u2026(i)
And $\\frac{a}{x}+\\frac{3 b}{y}=2$ \u2026(ii)<\/p>\n\n\n\n
$\\frac{3 a}{x}+\\frac{9 b}{y}=6$ \u2026(iii)<\/p>\n\n\n\n
$\\frac{3 a}{x}+\\frac{9 b}{y}-\\frac{3 a}{x}+\\frac{2 b}{y}=6-(-5)$
$\\Rightarrow \\frac{9 b}{y}+\\frac{2 b}{y}=11$
$\\Rightarrow \\frac{11 \\mathrm{b}}{\\mathrm{y}}=11$
\u21d2 y = b<\/p>\n\n\n\n
$\\frac{a}{x}+\\frac{3 b}{y}=2$
$\\Rightarrow \\frac{a}{x}+\\frac{3 b}{b}=2$<\/p>\n\n\n\n
\u21d2 x = \u2013 a<\/p>\n\n\n\n
\n\n\n\nQuestion 5 A <\/h4>\n\n\n\n
Given pair of linear equations is
$\\frac{2 x+5 y}{x y}=6$<\/p>\n\n\n\n
And $\\frac{4 x-5 y}{x y}=-3$<\/p>\n\n\n\n
2x + 5y + 4x \u2013 5y = 6xy \u2013 3xy
\u21d2 6x = 3xy
$\\Rightarrow \\frac{6 x}{3 x}=y$
\u21d2 y = 2 and x = 0<\/p>\n\n\n\n
2x + 5(2) = 6xy
\u21d2 2x + 10 = 6x(2)
\u21d2 2x + 10 = 12x \u21d2 2x \u2013 12x = \u2013 10
\u21d2 \u2013 10x = \u2013 10
\u21d2 x = 1<\/p>\n\n\n\n
Hence, x = 0,1 and y = 0,2 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 5 B <\/h4>\n\n\n\n
x \u2013 y = 6xy<\/strong>
Sol :
Given pair of linear equations is
x + y = 2xy \u2026(i)
And x \u2013 y = 6xy \u2026(ii)
On adding Eq. (i) and Eq. (ii), we get
x + y + x \u2013 y = 2xy + 6xy
\u21d2 2x = 8xy
$\\Rightarrow \\frac{2 \\mathrm{x}}{8 \\mathrm{x}}=\\mathrm{y}$<\/p>\n\n\n\n
$x-\\frac{1}{4}=6 x y$
$\\Rightarrow \\frac{4 x-1}{4}=6 x\\left(\\frac{1}{4}\\right)$
\u21d2 4x \u2013 1 = 6x \u21d2 \u2013 1 = 6x \u2013 4x
\u21d2 \u2013 1 = 2x
$\\Rightarrow x=-\\frac{1}{2}$<\/p>\n\n\n\n
Hence, $x=-\\frac{1}{2}, 0$ and $=\\frac{1}{4}, 0$ , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 5 C <\/h4>\n\n\n\n
7x \u2013 2y = 8xy<\/strong>
Sol :
Given pair of linear equations is
5x + 3y = 19xy \u2026(i)
And 7x \u2013 2y = 8xy \u2026(ii)
On multiplying Eq. (i) by 2 and Eq. (ii) by 3 to make the coefficients of y equal, we get the equation as
10x + 6y = 38xy \u2026(iii)
And 21x \u2013 6y = 24xy \u2026(iv)
On adding Eq. (i) and Eq. (ii), we get
10x + 6y + 21x \u2013 6y = 38xy + 24xy
\u21d2 31x = 62xy
$\\Rightarrow \\frac{31 \\mathrm{x}}{62 \\mathrm{x}}=\\mathrm{y}$
$\\Rightarrow \\mathrm{y}=\\frac{1}{2}$ and x=0<\/p>\n\n\n\n
7x \u2013 2y = 8xy
$\\Rightarrow 7 x-2\\left(\\frac{1}{2}\\right)=8 x\\left(\\frac{1}{2}\\right)$
\u21d2 7x \u2013 1 = 4x \u21d2 \u2013 1 = 4x \u2013 7x
\u21d2 \u2013 1 = \u2013 3x
$\\Rightarrow x=\\frac{1}{3}$<\/p>\n\n\n\n
\n\n\n\nQuestion 5 D <\/h4>\n\n\n\n
2x \u2013 3y = \u2013 xy<\/strong>
Sol :
Given pair of linear equations is
x + y = 7xy \u2026(i)
And 2x \u2013 3y = \u2013 xy \u2026(ii)
On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as
2x + 2y = 14xy \u2026(iii)
On subtracting Eq. (ii) and Eq. (iii), we get
2x + 2y \u2013 2x + 3y = 14xy + xy
\u21d2 2y + 3y = 15xy
\u21d2 5y = 15xy
$\\Rightarrow \\frac{5 y}{15 y}=x$<\/p>\n\n\n\n
2x \u2013 3y = \u2013 xy
$\\Rightarrow 2\\left(\\frac{1}{3}\\right)-3 y=6\\left(\\frac{1}{3}\\right) y$
$\\Rightarrow\\left(\\frac{2}{3}\\right)-3 y=2 y \\Rightarrow\\left(\\frac{2}{3}\\right)=5 y$
$\\Rightarrow \\mathrm{y}=\\frac{2}{15}$<\/p>\n\n\n\n
Hence, $x=\\frac{1}{3}, 0$ and $=\\frac{2}{15}, 0$ , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 6 A <\/h4>\n\n\n\n
Sol :
Given pair of linear equations is
$\\frac{1}{2(2 x+3 y)}+\\frac{12}{7(3 x-2 y)}=\\frac{1}{2}$ \u2026(i)
And $\\frac{7}{(2 x+3 y)}+\\frac{4}{(3 x-2 y)}=2$ \u2026(ii)<\/p>\n\n\n\n
$\\frac{7}{2(2 x+3 y)}+\\frac{7 \\times 12}{7(3 x-2 y)}=\\frac{7}{2}$
$\\Rightarrow \\frac{7}{2(2 x+3 y)}+\\frac{12}{(3 x-2 y)}=\\frac{7}{2}$ \u2026(iii)
$\\frac{7}{2(2 x+3 y)}+\\frac{4}{2(3 x-2 y)}=\\frac{2}{2}$<\/p>\n\n\n\n
$\\frac{7}{2(2 x+3 y)}+\\frac{2}{(3 x-2 y)}-\\frac{7}{2(2 x+3 y)}-\\frac{12}{(3 x-2 y)}=1-\\frac{7}{2}$
$\\Rightarrow \\frac{2}{(3 x-2 y)}-\\frac{12}{(3 x-2 y)}=1-\\frac{7}{2}$
$\\Rightarrow \\frac{-10}{(3 x-2 y)}=\\frac{2-7}{2}$
$\\Rightarrow \\frac{-10}{(3 x-2 y)}=\\frac{-5}{2}$
$\\Rightarrow \\frac{1}{(3 x-2 y)}=\\frac{5}{2 \\times 10}$<\/p>\n\n\n\n
$\\Rightarrow \\mathrm{x}=\\frac{4+2 \\mathrm{y}}{3}$ \u2026(a)<\/p>\n\n\n\n
$\\frac{7 \\times 3}{7(2 x+3 y)}+\\frac{4 \\times 3}{7(3 x-2 y)}=\\frac{2 \\times 3}{7}$
$\\Rightarrow \\frac{3}{(2 x+3 y)}+\\frac{12}{7(3 x-2 y)}=\\frac{6}{7}$ \u2026(v)<\/p>\n\n\n\n
$\\frac{3}{(2 x+3 y)}+\\frac{12}{7(3 x-2 y)}-\\frac{1}{2(2 x+3 y)}-\\frac{12}{7(3 x-2 y)}=\\frac{6}{7}-\\frac{1}{2}$
$\\Rightarrow \\frac{3}{(2 x+3 y)}-\\frac{1}{2(2 x+3 y)}=\\frac{6}{7}-\\frac{1}{2}$
$\\Rightarrow \\frac{6-1}{2(2 x+3 y)}=\\frac{12-7}{14}$
$\\Rightarrow \\frac{5}{2(2 x+3 y)}=\\frac{5}{14}$
$\\Rightarrow \\frac{1}{(2 x+3 y)}=\\frac{2}{14}$<\/p>\n\n\n\n
$\\frac{4+2 y}{3}=\\frac{7-3 y}{2}$
\u21d2 2(4 + 2y) = 3(7 \u2013 3y)
\u21d2 8 + 4y = 21 \u2013 9y
\u21d2 4y + 9y = 21 \u2013 8
\u21d2 13y = 13
\u21d2 y = 1<\/p>\n\n\n\n
$\\Rightarrow x=\\frac{7-3(1)}{2}$
$\\Rightarrow x=\\frac{4}{2}=2$<\/p>\n\n\n\n
\n\n\n\nQuestion 6 B <\/h4>\n\n\n\n
$\\frac{3}{x-1}+\\frac{2}{y+1}=\\frac{13}{6}, x \\neq 1, y \\neq 1$
Sol :
Given pair of linear equations is
$\\frac{2}{x-1}+\\frac{3}{y+1}=2$ \u2026(i)
And $\\frac{3}{x-1}+\\frac{2}{y+1}=\\frac{13}{6}$ \u2026(ii)<\/p>\n\n\n\n
$\\frac{6}{x-1}+\\frac{9}{y+1}=6$ \u2026(iii)<\/p>\n\n\n\n
$\\frac{6}{x-1}+\\frac{4}{y+1}-\\frac{6}{x-1}-\\frac{9}{y+1}=\\frac{13}{3}-6$
$\\Rightarrow \\frac{4}{y+1}-\\frac{9}{y+1}=\\frac{13}{3}-6$
$\\Rightarrow \\frac{-5}{y+1}=\\frac{13-18}{3}$
$\\Rightarrow \\frac{-5}{y+1}=\\frac{-5}{3}$
$\\Rightarrow \\frac{1}{y+1}=\\frac{1}{3}$
\u21d2 y + 1 = 3
\u21d2 y = 3 \u2013 1
\u21d2 y = 2<\/p>\n\n\n\n
$\\frac{6}{x-1}+\\frac{4}{2+1}=\\frac{13}{3}$
$\\Rightarrow \\frac{6}{x-1}+\\frac{4}{3}=\\frac{13}{3}$
$\\Rightarrow \\frac{6}{x-1}=\\frac{13}{3}-\\frac{4}{3}$
$\\Rightarrow \\frac{6}{x-1}=\\frac{9}{3}$
$\\Rightarrow \\frac{1}{x-1}=\\frac{9}{3 \\times 6}$
\u21d2 x \u2013 1 = 2
\u21d2 x = 3
Hence, x = 3 and y = 2 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 6 C <\/h4>\n\n\n\n
$\\frac{55}{x+y}+\\frac{40}{x-y}=13, x+y \\neq 0, x-y \\neq 0$<\/strong>
Sol :
Given pair of linear equations is
$\\frac{44}{x+y}+\\frac{30}{x-y}=10$ \u2026(i)
And $\\frac{55}{x+y}+\\frac{40}{x-y}=13$ \u2026(ii)<\/p>\n\n\n\n
$\\frac{176}{x+y}+\\frac{120}{x-y}=40$ \u2026(iii)<\/p>\n\n\n\n
$\\frac{176}{x+y}+\\frac{120}{x-y}-\\frac{165}{x+y}-\\frac{120}{x-y}=40-39$
$\\Rightarrow \\frac{176-165}{x+y}=1$
$\\Rightarrow \\frac{11}{x+y}=1$
\u21d2 x + y = 11 \u2026(a)<\/p>\n\n\n\n
$\\Rightarrow \\frac{44}{11}+\\frac{30}{x-y}=10$
$\\Rightarrow \\frac{30}{x-y}=10-4$
\u21d2 6(x \u2013 y) = 30
\u21d2 x \u2013 y = 5 \u2026(b)<\/p>\n\n\n\n
\u21d2 2x = 16
\u21d2 x = 8<\/p>\n\n\n\n
8 + y = 11
\u21d2 y = 3
Hence, x = 8 and y = 3 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 6 D <\/h4>\n\n\n\n
Sol :
Given pair of linear equations is
$\\frac{5}{x-1}+\\frac{1}{y-2}=2$ \u2026(i)
And $\\frac{6}{x-1}-\\frac{3}{y-2}=1$ \u2026(ii)<\/p>\n\n\n\n
$\\frac{15}{x-1}+\\frac{3}{y-2}=6$ \u2026(iii)<\/p>\n\n\n\n
$\\frac{6}{x-1}-\\frac{3}{y-2}+\\frac{15}{x-1}+\\frac{3}{y-2}=6+1$
$\\Rightarrow \\frac{6}{x-1}-\\frac{15}{x-1}=7$
$\\Rightarrow \\frac{21}{x-1}=7$
\u21d2 x \u2013 1 = 3
\u21d2 x = 3 + 1
\u21d2 x = 4<\/p>\n\n\n\n
$\\frac{6}{4-1}-\\frac{3}{y-2}=1$
$\\Rightarrow 2-\\frac{3}{y-2}=1$
$\\Rightarrow-\\frac{3}{y-2}=1-2$
$\\Rightarrow-\\frac{3}{y-2}=-1$
\u21d2 (y \u2013 2) = 3
\u21d2 y = 3 + 2
\u21d2 y = 5
Hence, x = 4 and y = 5 , which is the required solution.<\/p>\n\n\n\n
\n\n\n\nQuestion 7 A <\/h4>\n\n\n\n
Sol :
Let the present age of father i.e. Aftab = x yr
And the present age of his daughter = y yr
Seven years ago,
Aftab\u2019s age = (x \u2013 7)yr
Daughter\u2019s age = (y \u2013 7)yr
According to the question,
(x \u2013 7) = 7(y \u2013 7)
\u21d2 x \u2013 7 = 7y \u2013 49
\u21d2 x \u2013 7y = \u2013 42 \u2026(i)
After three years,
Aftab\u2019s age = (x + 3)yr
Daughter\u2019s age = (y + 3)yr
According to the question,
(x + 3) = 3(y + 3)
\u21d2 x + 3 = 3y + 9
\u21d2 x \u2013 3y = 6 \u2026(ii)
Now, we can solve this by an elimination method
On subtracting Eq. (ii) from (i) we get
x \u2013 3y \u2013 x + 7y = 6 \u2013 ( \u2013 42)
\u21d2 \u2013 3y + 7y = 6 + 42
\u21d2 4y = 48
\u21d2 y = 12
On putting y = 12 in Eq. (ii) we get
x \u2013 3(12) = 6
\u21d2 x \u2013 36 = 6
\u21d2 x = 42
Hence, the age of Aftab is 42years and age of his daughter is 12years.<\/p>\n\n\n\n
\n\n\n\nQuestion 7 B <\/h4>\n\n\n\n
Sol :
Let the present age of Nuri = x yr
And the present age of Sonu = y yr
Five years ago,
Nuri\u2019s age = (x \u2013 5)yr
Sonu\u2019s age = (y \u2013 5)yr
According to the question,
(x \u2013 5) = 3(y \u2013 5)
\u21d2 x \u2013 5 = 3y \u2013 15
\u21d2 x \u2013 3y = \u2013 10 \u2026(i)
After ten years,
Aftab\u2019s age = (x + 10)yr
Daughter\u2019s age = (y + 10)yr<\/p>\n\n\n\n
(x + 10) = 2(y + 10)
\u21d2 x + 10 = 2y + 20
\u21d2 x \u2013 2y = 10 \u2026(ii)
Now, we can solve this by an elimination method<\/p>\n\n\n\n
x \u2013 2y \u2013 x + 3y = 10 \u2013 ( \u2013 10)
\u21d2 \u2013 2y + 3y = 10 + 10
\u21d2 y = 20<\/p>\n\n\n\n
x \u2013 3(20) = \u2013 10
\u21d2 x \u2013 60 = \u2013 10
\u21d2 x = 50
Hence, the age of Nuri is 50 years and age of Sonu is 20 years.<\/p>\n\n\n\n
\n\n\n\nQuestion 7 C <\/h4>\n\n\n\n
Sol :
Let the one number = x
And the other number = y
According to the question,
x \u2013 y = 26 \u2026(i)
and x = 3y
or x \u2013 3y = 0 \u2026(ii)
Now, we can solve this by an elimination method
On subtracting Eq. (ii) from (i) we get
x \u2013 3y \u2013 x + y = 0 \u2013 26
\u21d2 \u2013 3y + y = \u2013 26
\u21d2 \u2013 2y = \u2013 26
\u21d2 y = 13
On putting y = 13 in Eq. (ii) we get
x \u2013 3(13) = 0
\u21d2 x \u2013 39 = 0
\u21d2 x = 39
Hence, the two numbers are 39 and 13.<\/p>\n\n\n\n