8x + 5y = 9<\/strong> On taking I and III ratio, we get On taking II and III ratio, we get Solve the following pair of linear equation by cross – multiplication method:<\/strong><\/p>\n\n\n\n 2x + 3y = 46<\/strong> $\\Rightarrow \\frac{x}{-222+230}=\\frac{y}{-138+148}=\\frac{1}{10-9}$ On taking I and III ratio, we get On taking II and III ratio, we get Solve the following pair of linear equation by cross – multiplication method:<\/strong><\/p>\n\n\n\n x + 4y + 9 = 0<\/strong> $\\Rightarrow \\frac{x}{-4+27}=\\frac{y}{45+1}=\\frac{1}{-3-20}$<\/p>\n\n\n\n $\\Rightarrow \\begin{matrix}\\frac{x}{23}&=\\frac{y}{46}&=\\frac{1}{-23}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n On taking I and III ratio, we get On taking II and III ratio, we get Solve the following pair of linear equation by cross – multiplication method:<\/strong> $\\Rightarrow \\frac{x}{-33+35}=\\frac{y}{-42+22}=\\frac{1}{10-18}$ On taking I and III ratio, we get On taking II and III ratio, we get Solve the following pair of linear equation by cross – multiplication method:<\/strong><\/p>\n\n\n\n $\\frac{2}{x}+\\frac{3}{y}=13$<\/strong> Let $u=\\frac{1}{x}$ and $v=\\frac{1}{y}$<\/p>\n\n\n\n So, Eq. (1) and (2) reduces to By cross – multiplication method, we have<\/p>\n\n\n\n $\\Rightarrow \\frac{u}{6-52}=\\frac{v}{-65-4}=\\frac{1}{-8-15}$<\/p>\n\n\n\n $\\Rightarrow \\begin{matrix}\\frac{u}{-46}&=\\frac{v}{-69}&=\\frac{1}{-23}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n On taking I and III ratio, we get On taking II and III ratio, we get So, $u=\\frac{1}{x}=2 \\Rightarrow x=\\frac{1}{2}$ Solve the following pair of linear equation by cross – multiplication method:<\/strong><\/p>\n\n\n\n $\\frac{x}{3}-\\frac{y}{12}=\\frac{19}{4}$<\/strong> And $\\frac{x}{3}-\\frac{y}{12}=\\frac{19}{4}$<\/p>\n\n\n\n \u21d24x-y=57<\/p>\n\n\n\n By cross – multiplication method, we have<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{-114-120}=\\frac{y}{-480+285}=\\frac{1}{-5-8}$<\/p>\n\n\n\n $\\Rightarrow \\begin{matrix}\\frac{x}{-234}&=\\frac{y}{-195}&=\\frac{1}{-13}\\<\/p>\n\n\n\n \\ \\text{I}& \\text{II}& \\text{III}\\end{matrix}$<\/p>\n\n\n\n On taking I and III ratio, we get On taking II and III ratio, we get Solve the following pair of equations by cross – multiplication method.<\/strong><\/p>\n\n\n\n ax + by = a \u2013 b<\/strong> $\\Rightarrow \\frac{x}{-b(a+b)-a(a-b)}=\\frac{y}{-b(a-b)+a(a+b)}=\\frac{1}{-a^{2}-b^{2}}$ On taking I and III ratio, we get On taking II and III ratio, we get Solve the following pair of equations by cross – multiplication method.<\/strong><\/p>\n\n\n\n $a^{2}-b^{2}$<\/strong> $\\frac{x}{a}+\\frac{y}{b}=a+b$<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{a}+\\frac{y}{b}-(a+b)=0$<\/p>\n\n\n\n And $\\frac{x}{a^{2}}+\\frac{y}{b^{2}}=2$<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{a^{2}}+\\frac{y}{b^{2}}-2=0$<\/p>\n\n\n\n By cross – multiplication method, we have<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{\\frac{-2}{b}+\\frac{a+b}{b^{2}}}=\\frac{y}{-\\frac{a+b}{a^{2}}+\\frac{2}{a}}=\\frac{1}{\\frac{1}{a b^{2}}-\\frac{1}{a^{2} b}}$ On taking I and III ratio, we get On taking II and III ratio, we get Solve the following pair of equations by cross – multiplication method.<\/strong><\/p>\n\n\n\n x \u2013 y = a + b<\/strong> By cross \u2013 multiplication method, we have<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{a^{2}-b^{2}+b(a+b)}=\\frac{y}{-a(a+b)+a^{2}-b^{2}}=\\frac{1}{b+a}$ $\\Rightarrow \\begin{matrix}\\frac{x}{a(a+b)}&=\\frac{y}{-b(a+b)}&=\\frac{1}{a+b}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n On taking I and III ratio, we get On taking II and III ratio, we get Solve the following pair of equations by cross – multiplication method.<\/strong><\/p>\n\n\n\n $\\frac{2 x}{a}+\\frac{y}{b}=2$<\/strong> Given, pair of equations is<\/p>\n\n\n\n $\\frac{2 x}{a}+\\frac{y}{b}=2$<\/p>\n\n\n\n $\\Rightarrow \\frac{2 \\mathrm{x}}{\\mathrm{a}}+\\frac{\\mathrm{y}}{\\mathrm{b}}-2=0$<\/p>\n\n\n\n And $\\frac{x}{a}-\\frac{y}{b}=4$<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{a}-\\frac{y}{b}-4=0$<\/p>\n\n\n\n By cross – multiplication method, we have<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{\\frac{-4}{b}-\\frac{2}{b}}=\\frac{y}{\\frac{-2}{a}+\\frac{8}{a}}=\\frac{1}{-\\frac{2}{a b}-\\frac{1}{a b}}$ On taking I and III ratio, we get On taking II and III ratio, we get Solve the following pair of equations by cross – multiplication method.<\/strong> $\\Rightarrow \\frac{x}{-3 b(2 a+b)+2 b(a+2 b)}=\\frac{y}{-3 a(a+2 b)+2 a(2 a+b)}=\\frac{1}{4ab-9ab}$ On taking I and III ratio, we get On taking II and III ratio, we get Solve the following pair of equations by cross – multiplication method.<\/strong><\/p>\n\n\n\n $\\frac{x}{a}+\\frac{y}{b}=2$<\/strong> Given, pair of equations is<\/p>\n\n\n\n $\\frac{x}{a}+\\frac{y}{b}=2$ ax + by = a2<\/sup> \u2013b2<\/sup>\u21d2 ax + by \u2013 (a2<\/sup> \u2013b2<\/sup> ) = 0<\/p>\n\n\n\n By cross – multiplication method, we have<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{\\frac{-\\left(a^{2}+b^{2}\\right)}{b}+2 b}=\\frac{y}{-2 a+\\frac{\\left(a^{2}+b^{2}\\right)}{a}}=\\frac{1}{\\frac{b}{a}-\\frac{a}{b}}$ On taking I and III ratio, we get On taking II and III ratio, we get Solve the following system of equations by cross – multiplication method.<\/strong><\/p>\n\n\n\n a(x + y) + b(x \u2013 y) = a2<\/sup> \u2013 ab + b2<\/sup><\/strong> $\\Rightarrow \\frac{x}{-(a-b)\\left(a^{2}+a b+b^{2}\\right)+(a+b)\\left(a^{2}-a b+b^{2}\\right)}$<\/p>\n\n\n\n $=\\frac{y}{-\\left(a^{2}-a b+b^{2}\\right)(a-b)+\\left(a^{2}+a b+b^{2}\\right)(a+b)}=\\frac{1}{(a+b)(a+b)-(a-b)(a-b)}$ On taking I and III ratio, we get On taking II and III ratio, we get Which of the following pair 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:<\/strong><\/p>\n\n\n\n x \u2013 3y \u2013 7 = 0<\/strong> $\\frac{a_{1}}{a_{2}}=\\frac{1}{1}=1$, $\\frac{b_{1}}{b_{2}}=\\frac{-3}{-1}=3$ and $\\frac{c_{1}}{c_{2}}=\\frac{-7}{-5}=\\frac{7}{5}$ \u2234 Given pair of equations has unique solution<\/p>\n\n\n\n By cross – multiplication method, we have<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{15-7}=\\frac{y}{-7+5}=\\frac{1}{-1+3}$ On taking I and III ratio, we get On taking II and III ratio, we get Which of the following pair 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:<\/strong><\/p>\n\n\n\n 2x + y = 5<\/strong> $\\frac{a_{1}}{a_{2}}=\\frac{2}{3}$ , $\\frac{b_{1}}{b_{2}}=\\frac{1}{2}$ and $\\frac{c_{1}}{c_{2}}=\\frac{-5}{-8}=\\frac{5}{8}$ \u2234 Given pair of equations has unique solution<\/p>\n\n\n\n By cross – multiplication method, we have<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{-8+10}=\\frac{y}{-15+16}=\\frac{1}{4-3}$<\/p>\n\n\n\n $\\Rightarrow \\begin{matrix}\\frac{x}{2}&=\\frac{y}{1}&=\\frac{1}{1}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n On taking I and III ratio, we get On taking II and III ratio, we get Which of the following pair 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:<\/strong><\/p>\n\n\n\n 3x \u2013 5y = 20<\/strong> $\\frac{a_{1}}{a_{2}}=\\frac{3}{3}=1$, $\\frac{b_{1}}{b_{2}}=\\frac{-5}{-5}=1$ and $\\frac{c_{1}}{c_{2}}=\\frac{-20}{-20}=1$ \u2234 Given pair of equations has infinitely many solutions.<\/p>\n\n\n\n Which of the following pair 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:<\/strong><\/p>\n\n\n\n x \u2013 3y \u2013 3 = 0<\/strong> $\\frac{a_{1}}{a_{2}}=\\frac{1}{3}$ , $\\frac{b_{1}}{b_{2}}=\\frac{-3}{-9}=\\frac{1}{3}$ and $\\frac{c_{1}}{c_{2}}=\\frac{-3}{-2}=\\frac{3}{2}$<\/p>\n\n\n\n $\\Rightarrow \\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}} \\neq \\frac{c_{1}}{c_{2}}$<\/p>\n\n\n\n \u2234 Given pair of equations has no solution<\/p>\n\n\n\n Which of the following pair 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:<\/strong><\/p>\n\n\n\n x + y = 2<\/strong> $\\frac{a_{1}}{a_{2}}=\\frac{1}{1}=1$, $\\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}}=\\frac{1}{1}=3$ and $\\frac{c_{1}}{c_{2}}=\\frac{-2}{-2}=1$<\/p>\n\n\n\n $\\Rightarrow \\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}$<\/p>\n\n\n\n \u2234 Given pair of equations has infinitely many solution<\/p>\n\n\n\n Which of the following pair 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:<\/strong><\/p>\n\n\n\n x + y = 2<\/strong> $\\frac{a_{1}}{a_{2}}=\\frac{1}{1}=1$ , $\\frac{b_{1}}{b_{2}}=\\frac{1}{1}=1$ and $\\frac{c_{1}}{c_{2}}=\\frac{-2}{-3}=\\frac{2}{3}$<\/p>\n\n\n\n $\\Rightarrow \\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}} \\neq \\frac{c_{1}}{c_{2}}$<\/p>\n\n\n\n \u2234 Given pair of equations has no solution<\/p>\n\n\n\n Solve the following system of linear equations by cross – multiplication method.<\/strong><\/p>\n\n\n\n $\\frac{15}{x+y}+\\frac{7}{x-y}-10=0$<\/strong> Sol : $\\frac{15}{x+y}+\\frac{7}{x-y}-10=0$ \u2026(2)<\/p>\n\n\n\n Let $u=\\frac{1}{x+y}$ and $v=\\frac{1}{x-y}$<\/p>\n\n\n\n Now, the Eq. (1) and (2) reduces to By cross – multiplication method, we have<\/p>\n\n\n\n $\\Rightarrow \\frac{\\mathrm{u}}{-20-7}=\\frac{\\mathrm{v}}{15+50}=\\frac{1}{35-30}$ On taking I and III ratio, we get $\\Rightarrow \\mathrm{u}=\\frac{-27}{5}$<\/p>\n\n\n\n On taking II and III ratio, we get So, $u=\\frac{1}{x+y}=\\frac{-27}{5}$<\/p>\n\n\n\n $\\Rightarrow \\mathrm{x}+\\mathrm{y}=-\\frac{5}{27}$ \u2026(a)<\/p>\n\n\n\n and $v=\\frac{1}{x-y}=13$ <\/p>\n\n\n\n $\\Rightarrow \\mathrm{x}-\\mathrm{y}=\\frac{1}{13}$ \u2026(b)<\/p>\n\n\n\n On adding Eq. (a) and (b), we get On putting the value of $x=\\frac{-19}{351}$ in Eq. (a), we get Solve the following system of linear equations by cross – multiplication method.<\/strong><\/p>\n\n\n\n ax \u2013 ay = 2<\/strong> Sol : $\\Rightarrow \\frac{x}{2 a\\left(a^{2}+1\\right)+2(a+1)}=\\frac{y}{-2(a-1)+2 a\\left(a^{2}+1\\right)}=\\frac{1}{a(a+1)+a(a-1)}$<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{2 a^{3}+2 a+2 a+2}=\\frac{y}{-2 a+2+2 a^{3}+2 a}=\\frac{1}{a^{2}+a+a^{2}-a}$<\/p>\n\n\n\n $\\Rightarrow \\begin{matrix}\\frac{x}{2a^3+4a+2}&=\\frac{y}{2a^3+2}&=\\frac{1}{2a^2}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n On taking I and III ratio, we get<\/p>\n\n\n\n $\\Rightarrow \\frac{x}{2 a^{3}+4 a+2}=\\frac{1}{2 a^{2}}$<\/p>\n\n\n\n $\\Rightarrow \\mathrm{x}=\\frac{\\mathrm{a}^{3}+2 \\mathrm{a}+1}{\\mathrm{a}^{2}}$<\/p>\n\n\n\n On taking II and III ratio, we get<\/p>\n\n\n\n $\\Rightarrow \\frac{\\mathrm{y}}{2 \\mathrm{a}^{3}+2}=\\frac{1}{2 \\mathrm{a}^{2}}$<\/p>\n\n\n\n $\\Rightarrow \\mathrm{y}=\\frac{\\mathrm{a}^{3}+1}{\\mathrm{a}^{2}}$<\/p>\n\n\n\n If the cost of 2 pencils and 3 erasers is Rs. 9 and the cost of 4 pencils and 6 erasers is Rs. 18. Find the cost of each pencil and each eraser.<\/strong><\/p>\n\n\n\n Sol : $\\Rightarrow \\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}$<\/p>\n\n\n\n \u2234Given pair of linear equations has infinitely many solutions.<\/p>\n\n\n\n The paths traced by the wheels of two trains are given by equations x + 2y \u2013 4 = 0 and 2x + 4y \u2013 12 = 0. Will the paths cross each other?<\/strong><\/p>\n\n\n\n Sol : $\\frac{a_{1}}{a_{2}}=\\frac{1}{1}=1$, $\\frac{b_{1}}{b_{2}}=\\frac{2}{2}=1$ and $\\frac{c_{1}}{c_{2}}=\\frac{-4}{-6}=\\frac{2}{3}$<\/p>\n\n\n\n $\\Rightarrow \\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}}=\\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}} \\neq \\frac{\\mathrm{c}_{1}}{\\mathrm{c}_{2}}$ The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditure is 4 : 3. If each of them manages to save Rs. 2000 per month, find their monthly incomes.<\/strong><\/p>\n\n\n\n Sol : $\\Rightarrow \\frac{x}{8000-6000}=\\frac{y}{-14000+18000}=\\frac{1}{-27+28}$<\/p>\n\n\n\n $\\Rightarrow \\begin{matrix}\\frac{x}{2000}&=\\frac{y}{4000}&=\\frac{1}{1}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n On taking I and III ratios, we get On taking II and III ratios, we get Hence, the monthly incomes of two persons are 9(2000) = Rs18000 and 7(2000) = Rs14000<\/p>\n\n\n\n The sum of two – digits number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?<\/strong><\/p>\n\n\n\n Sol : $\\Rightarrow \\frac{x}{-2-6}=\\frac{y}{-6+2}=\\frac{1}{-1-1}$<\/p>\n\n\n\n $\\Rightarrow \\begin{matrix}\\frac{x}{-8}&=\\frac{y}{-4}&=\\frac{1}{-2}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n On taking I and III ratio, we get On taking II and III ratio, we get So, the original number = 10x + y If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1\/2 if we add 1 to the denominator. What is the fraction?<\/strong><\/p>\n\n\n\n Sol : So, the fraction $=\\frac{x}{y}$<\/p>\n\n\n\n According to the question,<\/p>\n\n\n\n Condition I: Condition II: $\\Rightarrow \\frac{x}{1+2}=\\frac{y}{4+1}=\\frac{1}{-1+2}$<\/p>\n\n\n\n $\\Rightarrow \\begin{matrix}\\frac{x}{3}&=\\frac{y}{5}&=\\frac{1}{1}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n On taking I and III ratio, we get On taking II and III ratio, we get<\/p>\n\n\n\n $\\Rightarrow \\frac{y}{5}=\\frac{1}{1}$ So, the numerator is 3 and the denominator is 5 The cost of 5 oranges and 3 apples is Rs. 35 and the cost of 2 oranges and 4 apples is Rs. 28. Find the cost of an orange and an apple.<\/strong><\/p>\n\n\n\n Sol : $\\Rightarrow \\frac{x}{-42+70}=\\frac{y}{-35+70}=\\frac{1}{10-3}$<\/p>\n\n\n\n $\\Rightarrow \\begin{matrix}\\frac{x}{28}&=\\frac{y}{35}&=\\frac{1}{7}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n On taking I and III ratio, we get
3x + 2y = 4<\/strong>
Sol :
Given, pair of equations is
8x + 5y \u2013 9 = 0 and 3x + 2y \u2013 4 = 0
By cross – multiplication method, we have
<\/a>
$\\Rightarrow \\frac{x}{-20+18}=\\frac{y}{-27+32}=\\frac{1}{16-15}$
$\\Rightarrow \\begin{matrix}\\frac{x}{-2}&=\\frac{y}{5}&=\\frac{1}{1}\\\\\\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{x}{-2}=\\frac{1}{1}$
\u21d2 x = \u2013 2<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{5}=\\frac{1}{1}$
\u21d2 y = 5<\/p>\n\n\n\n
\n\n\n\nQuestion 1 B <\/h4>\n\n\n\n
3x + 5y = 74<\/strong>
Sol :
Given, pair of equations is
2x + 3y \u2013 46 = 0 and 3x + 5y \u2013 74 = 0
By cross – multiplication method, we have<\/p>\n\n\n\n![\"\"\/](\"https:\/\/1.bp.blogspot.com\/-Rb7GScofME4\/X4pUnlL0cgI\/AAAAAAAAKsc\/DkOUOm5IzZwOO0FgIf1v9XyBh62XWzJ7wCPcBGAsYHg\/s263\/image.png\")
<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\begin{matrix}\\frac{x}{8}&=\\frac{y}{10}&=\\frac{1}{1}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{x}{8}=\\frac{1}{1}$
\u21d2 x = 8<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{10}=\\frac{1}{1}$
\u21d2 y = 10<\/p>\n\n\n\n
\n\n\n\nQuestion 1 C <\/h4>\n\n\n\n
5x \u2013 1 = 3y<\/strong>
Sol :
Given, pair of equations is
x + 4y + 9 = 0 and 5x \u2013 3y \u2013 1 = 0
By cross – multiplication method, we have<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{x}{23}=\\frac{1}{-23}$
\u21d2 x = \u2013 1<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{46}=\\frac{1}{-23}$
\u21d2 y = \u2013 2<\/p>\n\n\n\n
\n\n\n\nQuestion 1 D <\/h4>\n\n\n\n
2x + 3y \u2013 7 = 0<\/strong>
6x + 5y \u2013 11 = 0<\/strong>
Sol :
Given, pair of equations is
2x + 3y \u2013 7 = 0 and 6x + 5y \u2013 11 = 0
By cross – multiplication method, we have<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\begin{matrix}\\frac{x}{2}&=\\frac{y}{-20}&=\\frac{1}{-8}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{x}{2}=\\frac{1}{-8}$
$\\Rightarrow x=\\frac{1}{-4}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{-20}=\\frac{1}{-8}$
$\\Rightarrow \\mathrm{y}=\\frac{5}{2}$<\/p>\n\n\n\n
\n\n\n\nQuestion 1 E <\/h4>\n\n\n\n
$\\frac{5}{x}-\\frac{4}{y}=-2$<\/strong>
Sol :
Given, pair of equations is
$\\frac{2}{x}+\\frac{3}{y}=13$
$\\frac{5}{x}-\\frac{4}{y}=-2$<\/p>\n\n\n\n
2u + 3v \u2013 13 = 0
5u \u2013 4v + 2 = 0<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{\\mathrm{u}}{-46}=\\frac{1}{-23}$
\u21d2 u = 2<\/p>\n\n\n\n
$\\Rightarrow \\frac{\\mathrm{v}}{-69}=\\frac{1}{-23}$
\u21d2 v = 3<\/p>\n\n\n\n
and $v=\\frac{1}{y}=3 \\Rightarrow y=\\frac{1}{3}$<\/p>\n\n\n\n
\n\n\n\nQuestion 1 F <\/h4>\n\n\n\n
$\\frac{x}{3}-\\frac{y}{12}=\\frac{19}{4}$<\/strong>
Sol :
Given, pair of equations is
$\\frac{x}{6}+\\frac{y}{15}=4$<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{x}{-234}=\\frac{1}{-13}$
\u21d2 x = 18<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{-195}=\\frac{1}{-13}$
\u21d2 y = 15<\/p>\n\n\n\n
\n\n\n\nQuestion 2 A <\/h4>\n\n\n\n
bx \u2013 ay = a + b<\/strong>
Sol :
Given, pair of equations is
ax + by = a \u2013 b \u21d2 ax + by \u2013(a \u2013 b) = 0
bx \u2013 ay = a + b \u21d2 bx \u2013ay \u2013 (a + b) = 0
By cross – multiplication method, we have<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{x}{-b a-b^{2}-a^{2}+a b}=\\frac{y}{-b a+b^{2}+a^{2}+a b}=\\frac{1}{-a^{2}-b^{2}}$
$\\Rightarrow \\begin{matrix}\\frac{x}{-b^{2}-a^{2}}&=\\frac{y}{b^{2}+a^{2}}&=\\frac{1}{-a^{2}-b^{2}}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{x}{-b^{2}-a^{2}}=\\frac{1}{-a^{2}-b^{2}}$
\u21d2 x = 1<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{b^{2}+a^{2}}=\\frac{1}{-a^{2}-b^{2}}$
\u21d2 y = \u2013 1<\/p>\n\n\n\n
\n\n\n\nQuestion 2 B <\/h4>\n\n\n\n
$\\frac{x}{a^{2}}+\\frac{y}{b^{2}}=2$<\/strong>
$a \\neq 0, b \\neq 0$
Sol :
Given, pair of equations is<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{b^{2} x}{-2 b+a+b}=\\frac{a^{2} y}{-a-b+2 a}=\\frac{a^{2} b^{2}}{a-b}$
$\\Rightarrow \\begin{matrix}\\frac{b^{2} x}{a-b}&=\\frac{a^{2} y}{a-b}&=\\frac{a^{2} b^{2}}{a-b}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{b^{2} x}{a-b}=\\frac{a^{2} b^{2}}{a-b}$
$\\Rightarrow \\frac{x}{1}=\\frac{a^{2} b^{2}}{b^{2}}$
\u21d2x = a2<\/sup><\/p>\n\n\n\n
$\\Rightarrow \\frac{a^{2} y}{a-b}=\\frac{a^{2} b^{2}}{a-b}$
\u21d2 y = b2<\/sup><\/p>\n\n\n\n
\n\n\n\nQuestion 2 C <\/h4>\n\n\n\n
ax + by = a2<\/sup> \u2013 b2<\/sup><\/strong>
Sol :
Given, pair of equations is
x \u2013 y = a + b \u21d2 x \u2013 y \u2013(a + b) = 0
ax + by = a2<\/sup> \u2013b2<\/sup>\u21d2 ax + by \u2013 (a2<\/sup> \u2013b2<\/sup> ) = 0<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{x}{a^{2}+a b}=\\frac{y}{-b a-b^{2}}=\\frac{1}{b+a}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{x}{a(a+b)}=\\frac{1}{a+b}$
\u21d2 x = a<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{-b(a+b)}=\\frac{1}{a+b}$
\u21d2 y = \u2013 b<\/p>\n\n\n\n
\n\n\n\nQuestion 2 D <\/h4>\n\n\n\n
$\\frac{x}{a}-\\frac{y}{b}=4$<\/strong>
$a \\neq 0, b \\neq 0$<\/strong>
Sol :
$\\frac{2 x}{a}+\\frac{y}{b}=2$
$\\frac{x}{a}-\\frac{y}{b}=4$<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\begin{matrix}\\frac{b x}{-6}&=\\frac{a y}{6}&=\\frac{a b}{-3}\\\\\\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{b x}{-6}=\\frac{a b}{-3}$
\u21d2x = 2a<\/p>\n\n\n\n
$\\Rightarrow \\frac{\\mathrm{ay}}{6}=\\frac{\\mathrm{ab}}{-3}$
\u21d2 y = \u2013 2b<\/p>\n\n\n\n
\n\n\n\nQuestion 2 E <\/h4>\n\n\n\n
2ax + 3by = a + 2b<\/strong>
3ax + 2by = 2a + b<\/strong>
Sol :
2ax + 3by = a + 2b
3ax + 2by = 2a + b
Given, pair of equations is
2ax + 3by = a + 2b \u21d2 2ax + 3by \u2013(a + 2b) = 0
3ax + 2by = 2a + b \u21d2 3ax + 2by \u2013(2a + b) = 0
By cross – multiplication method, we have<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{x}{-4 b a-3 b^{2}+4 a b+4 b^{2}}=\\frac{y}{-3 a^{2}-6 b a+4 a^{2}+2 a b}=\\frac{1}{-5 a b}$
$\\Rightarrow \\begin{matrix}\\frac{x}{b^{2}-4 a b}&=\\frac{y}{-4 a b+a^{2}}&=\\frac{1}{-5 a b}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{x}{b^{2}-4 a b}=\\frac{1}{-5 a b}$
$\\Rightarrow x=\\frac{b(b-4 a)}{-5 a b}$
$\\Rightarrow x=\\frac{4 a-b}{5 a}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{-4 a b+a^{2}}=\\frac{1}{-5 a b}$
$\\Rightarrow \\mathrm{y}=\\frac{\\mathrm{a}(\\mathrm{a}-4 \\mathrm{b})}{-5 \\mathrm{ab}}$
$\\Rightarrow \\mathrm{y}=\\frac{4 \\mathrm{b}-\\mathrm{a}}{5 \\mathrm{b}}$<\/p>\n\n\n\n
\n\n\n\nQuestion 2 F <\/h4>\n\n\n\n
ax + by = a2<\/sup>\u2013b2<\/sup><\/strong>
Sol :
$\\frac{x}{a}+\\frac{y}{b}=2$
ax + by = a2<\/sup>\u2013b2<\/sup><\/strong><\/p>\n\n\n\n
$\\Rightarrow \\frac{\\mathrm{x}}{\\mathrm{a}}+\\frac{\\mathrm{y}}{\\mathrm{b}}-2=0$<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{b x}{-a^{2}-b^{2}+2 b^{2}}=\\frac{a y}{-2 a^{2}+a^{2}+b^{2}}=\\frac{a b}{b^{2}-a^{2}}$
$\\Rightarrow \\frac{b x}{-a^{2}+b^{2}}=\\frac{a y}{-a^{2}+b^{2}}=\\frac{a b}{b^{2}-a^{2}}$
$\\Rightarrow \\begin{matrix}\\frac{b x}{b^{2}-a^{2}}&=\\frac{a y}{b^{2}-a^{2}}&=\\frac{a b}{b^{2}-a^{2}}\\\\ \\text{I}& \\text{II}& \\text{III}\\end{matrix}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{b x}{b^{2}-a^{2}}=\\frac{a b}{b^{2}-a^{2}}$
\u21d2 x = a<\/p>\n\n\n\n
$\\Rightarrow \\frac{a y}{b^{2}-a^{2}}=\\frac{a b}{b^{2}-a^{2}}$
\u21d2 y = b<\/p>\n\n\n\n
\n\n\n\nQuestion 3 <\/h4>\n\n\n\n
a(x + y) \u2013 b(x \u2013 y) = a2<\/sup> + ab + b2<\/sup><\/strong>
Sol :
The given system of equations can be re – written as
ax + ay + bx \u2013 by \u2013 a2<\/sup> + ab \u2013 b2<\/sup> = 0
\u21d2(a + b)x + (a \u2013 b)y \u2013 (a2<\/sup> \u2013 ab + b2<\/sup> ) = 0 \u2026(1)
and ax + ay \u2013 bx + by \u2013 a2<\/sup> \u2013 ab \u2013 b2<\/sup> = 0
\u21d2 (a \u2013 b)x + (a + b)y \u2013 (a2<\/sup> + ab + b2<\/sup> ) = 0 \u2026(2)
Now, by cross \u2013 multiplication method, we have
By cross – multiplication method, we have<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{x}{-\\left(a^{3}-b^{3}\\right)+\\left(a^{3}+b^{3}\\right)}$
$=\\frac{y}{-a^{3}+a^{2} b-b^{2} a+b a^{2}-a b^{2}+b^{3}+a^{3}+a^{2} b+b^{2} a+b a^{2}+a b^{2}+b^{3}}$
$=\\frac{1}{a^{2}+2 a b+b^{2}-a^{2}+2 a b-b^{2}}$
$\\Rightarrow \\begin{matrix}\\frac{x}{2 b^{2}}&=\\frac{y}{4 a^{2} b+2 b^{2}}&=\\frac{1}{4 a b}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{x}{2 b^{3}}=\\frac{1}{4 a b}$
$\\Rightarrow x=\\frac{b^{2}}{2 a}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{4 a^{2} b+2 b^{3}}=\\frac{1}{4 a b}$
$\\Rightarrow \\mathrm{y}=\\frac{2 \\mathrm{a}^{2}+\\mathrm{b}^{2}}{2 \\mathrm{a}}$<\/p>\n\n\n\n
\n\n\n\nQuestion 4 A <\/h4>\n\n\n\n
3x \u2013 3y \u2013 15 = 0<\/strong>
Sol :
Given pair of linear equations
x \u2013 3y \u2013 7 = 0
3x \u2013 3y \u2013 15 = 0
\u21d2 x \u2013 y \u2013 5 = 0 \u2026(ii)
As we can see that a1<\/sub> = 1, b1<\/sub> = \u2013 3 and c1<\/sub> = \u2013 7
and a2 =<\/sub> 1, b2<\/sub> = \u2013 1 and c2<\/sub> = \u2013 5<\/p>\n\n\n\n
$\\Rightarrow \\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}} \\neq \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}} \\neq \\frac{\\mathrm{C}_{1}}{\\mathrm{C}_{2}}$<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\begin{matrix}\\frac{x}{8}&=\\frac{y}{-2}&=\\frac{1}{2}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{x}{8}=\\frac{1}{2}$
\u21d2 x = 4<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{-2}=\\frac{1}{2}$
\u21d2 y = \u2013 1<\/p>\n\n\n\n
\n\n\n\nQuestion 4 B <\/h4>\n\n\n\n
3x + 2y = 8<\/strong>
Sol :
Given pair of linear equations
2x + y = 5
3x + 2y = 8
As we can see that a1<\/sub> = 2, b1<\/sub> = 1 and c1<\/sub> = \u2013 5
and a2 =<\/sub> 3, b2<\/sub> = 2 and c2<\/sub> = \u2013 8<\/p>\n\n\n\n
$\\Rightarrow \\frac{\\mathrm{a}_{1}}{\\mathrm{a}_{2}} \\neq \\frac{\\mathrm{b}_{1}}{\\mathrm{b}_{2}} \\neq \\frac{\\mathrm{c}_{1}}{\\mathrm{c}_{2}}$<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{x}{2}=1$
\u21d2 x = 2<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{1}=1$
\u21d2 y = 1<\/p>\n\n\n\n
\n\n\n\nQuestion 4 C <\/h4>\n\n\n\n
6x \u2013 10y = 40<\/strong>
Sol :
Given pair of linear equations
3x \u2013 5y = 20
6x \u2013 10y = 40
\u21d2 3x \u2013 5y = 20 \u2026(ii)
As we can see that a1<\/sub> = 3, b1<\/sub> = \u2013 5 and c1<\/sub> = \u2013 20
and a2 =<\/sub> 3, b2<\/sub> = \u2013 5 and c2<\/sub> = \u2013 20<\/p>\n\n\n\n
$\\Rightarrow \\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}$<\/p>\n\n\n\n
\n\n\n\nQuestion 4 D <\/h4>\n\n\n\n
3x \u2013 9y \u2013 2 = 0<\/strong>
Sol :
Given pair of linear equations
x \u2013 3y \u2013 3 = 0
3x \u2013 9y \u2013 2 = 0
As we can see that a1<\/sub> = 1, b1<\/sub> = \u2013 3 and c1<\/sub> = \u2013 3
and a2 =<\/sub> 3, b2<\/sub> = \u2013 9 and c2<\/sub> = \u2013 2<\/p>\n\n\n\n
\n\n\n\nQuestion 4 E <\/h4>\n\n\n\n
2x + 2y = 4<\/strong>
Sol :
Given pair of linear equations
x + y = 2
2x + 2y = 4
\u21d2 x + y \u2013 2 = 0
As we can see that a1<\/sub> = 1, b1<\/sub> = 1 and c1<\/sub> = \u2013 2
and a2 =<\/sub> 1, b2<\/sub> = 1 and c2<\/sub> = \u2013 2<\/p>\n\n\n\n
\n\n\n\nQuestion 4 F <\/h4>\n\n\n\n
2x + 2y = 6<\/strong>
Sol :
Given pair of linear equations
x + y = 2
2x + 2y = 6
\u21d2 x + y \u2013 3 = 0
As we can see that a1<\/sub> = 1, b1<\/sub> = 1 and c1<\/sub> = \u2013 2
and a2 =<\/sub> 1, b2<\/sub> = 1 and c2<\/sub> = \u2013 3<\/p>\n\n\n\n
\n\n\n\nQuestion 5 A <\/h4>\n\n\n\n
$\\frac{15}{x+y}+\\frac{7}{x-y}-10=0$<\/strong>
[Hint: Let <\/strong>$u=\\frac{1}{x-y}$<\/strong> and <\/strong>$v=\\frac{1}{x-y}$]<\/strong><\/p>\n\n\n\n
Given, pair of equations is
$\\frac{5}{x+y}+\\frac{2}{x-y}+1=0$ \u2026(1)<\/p>\n\n\n\n
5u + 2v + 1 = 0
15u + 7v \u2013 10 = 0<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\begin{matrix}\\frac{u}{-27}&=\\frac{v}{65}&=\\frac{1}{5}\\\\ \\text{I}&\\text{II}&\\text{III}\\end{matrix}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{\\mathrm{u}}{-27}=\\frac{1}{5}$<\/p>\n\n\n\n
$\\Rightarrow \\frac{v}{65}=\\frac{1}{5}$
\u21d2 v = 13<\/p>\n\n\n\n
$2 x=-\\frac{5}{27}+\\frac{1}{13}$
$\\Rightarrow 2 \\mathrm{x}=\\frac{-65+27}{27 \\times 13}$
$\\Rightarrow 2 \\mathrm{x}=\\frac{-38}{351}$
$\\Rightarrow x=\\frac{-19}{351}$<\/p>\n\n\n\n
$\\frac{-19}{351}+y=-\\frac{5}{27}$
$\\Rightarrow \\mathrm{y}=-\\frac{5}{27}+\\frac{19}{351}$
$\\Rightarrow \\mathrm{y}=\\frac{-65+19}{351}$
$\\Rightarrow \\mathrm{y}=-\\frac{46}{351}$<\/p>\n\n\n\n
\n\n\n\nQuestion 5 B <\/h4>\n\n\n\n
(a \u2013 1)x + (a + 1)y = 2(a2<\/sup> + 1)<\/strong>
[Hint: Let $u = \\frac{1}{x-y}$ and $v = \\frac{1}{x-y}$]<\/strong><\/p>\n\n\n\n
Given, pair of equations is
ax \u2013 ay = 2
(a \u2013 1)x + (a + 1)y = 2(a2<\/sup> + 1)
By cross – multiplication method, we have<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
\n\n\n\nQuestion 6 <\/h4>\n\n\n\n
Let the cost of one pencil = Rs x
and cost of one eraser = Rs y
According to the question
2x + 3y = 9 \u2026(1)
4x + 6y = 18
\u21d22(2x + 3y) = 18
\u21d22x + 3y = 9 \u2026(2)
As we can see From Eq. (1) and (2)<\/p>\n\n\n\n
\n\n\n\nQuestion 7 <\/h4>\n\n\n\n
Given paths traced by the wheel of two trains are
x + 2y \u2013 4 = 0 \u2026(i)
2x + 4y \u2013 12 = 0
\u21d2 x + 2y \u2013 6 = 0 \u2026(ii)
As we can see that a1<\/sub> = 1, b1<\/sub> = 2 and c1<\/sub> = \u2013 4
and a2 =<\/sub> 1, b2<\/sub> = 2 and c2<\/sub> = \u2013 6<\/p>\n\n\n\n
\u2234 Given pair of equations has no solution
Hence, two paths will not cross each other.<\/p>\n\n\n\n
\n\n\n\nQuestion 8 <\/h4>\n\n\n\n
Given ratio of incomes = 9:7
And the ratio of their expenditures = 4:3
Saving of each person = Rs. 2000
Let incomes of two persons = 9x and 7x
And their expenditures = 4y and 3y
According to the question,
9x \u2013 4y = 2000
\u21d29x \u2013 4y \u2013 2000 = 0 \u2026(i)
7x \u2013 3y = 2000
\u21d27x \u2013 3y \u2013 2000 = 0 \u2026(ii)
By cross – multiplication method, we have<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{x}{2000}=\\frac{1}{1}$
\u21d2 x = 2000<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{4000}=\\frac{1}{1}$
\u21d2 y = 4000<\/p>\n\n\n\n
\n\n\n\nQuestion 9 <\/h4>\n\n\n\n
Let unit\u2019s digit = y
and the ten\u2019s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
reversing number = x + 10y
According to the question,
10x + y + x + 10y = 66
\u21d211x + 11y = 66
\u21d2 x + y = 6 \u2026(i)
x \u2013 y = 2 \u2026(ii)
By cross – multiplication method, we have<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{x}{-8}=\\frac{1}{-1}$
\u21d2 x = 4<\/p>\n\n\n\n
$\\Rightarrow \\frac{y}{-4}=\\frac{1}{-2}$
\u21d2 y = 2<\/p>\n\n\n\n
= 10(4) + 2
= 42
Reversing the number = x + 10y
= 24
Hence, the two digit number is 42 and 24. These are two such numbers.<\/p>\n\n\n\n
\n\n\n\nQuestion 10 <\/h4>\n\n\n\n
Let the numerator = x
and the denominator = y<\/p>\n\n\n\n
$\\frac{x+1}{y-1}=1$
\u21d2 x + 1 = y \u2013 1
\u21d2 x \u2013 y = \u2013 2
\u21d2 x \u2013 y + 2 = 0 \u2026(i)<\/p>\n\n\n\n
$\\frac{x}{y+1}=\\frac{1}{2}$
\u21d2 2x = y + 1
\u21d2 2x \u2013 y = 1
\u21d2 2x \u2013 y \u2013 1 = 0 \u2026(ii)
By cross – multiplication method, we have<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n
$\\Rightarrow \\frac{x}{3}=\\frac{1}{1}$
\u21d2 x = 3<\/p>\n\n\n\n
\u21d2 y = 5<\/p>\n\n\n\n
Hence, the fraction is $\\frac{3}{5}$<\/p>\n\n\n\n
\n\n\n\nQuestion 11 <\/h4>\n\n\n\n
Let the cost of an orange = Rs x
And the cost of an apple = Rs y
According to the question,
5x + 3y = 35
And 2x + 4y = 28
\u21d2 x + 2y = 14
By cross – multiplication method, we have<\/p>\n\n\n\n<\/a><\/figure>\n\n\n\n