\displaystyle \textbf{Question 1: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle 3x+2y+25=0,\qquad2x+y+10=0
\displaystyle \text{Answer:}
\displaystyle 3x+2y+25=0\qquad\ldots\text{(i)}
\displaystyle 2x+y+10=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{2(10)-1(25)}=\frac{-y}{3(10)-2(25)}=\frac{1}{3(1)-2(2)}
\displaystyle \Rightarrow \frac{x}{-5}=\frac{-y}{-20}=\frac{1}{-1}
\displaystyle \frac{x}{-5}=\frac{1}{-1}\Rightarrow x=5
\displaystyle \frac{-y}{-20}=\frac{1}{-1}\Rightarrow y=-20
\displaystyle \therefore x=5,\qquad y=-20.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle \frac{x+y}{xy}=2,\qquad\frac{x-y}{xy}=6
\displaystyle \text{Answer:}
\displaystyle \frac{x+y}{xy}=2
\displaystyle \Rightarrow \frac{1}{x}+\frac{1}{y}=2\qquad\ldots\text{(i)}
\displaystyle \frac{x-y}{xy}=6
\displaystyle \Rightarrow -\frac{1}{x}+\frac{1}{y}=6\qquad\ldots\text{(ii)}
\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{The equations become}
\displaystyle u+v-2=0\qquad\ldots\text{(iii)}
\displaystyle -u+v-6=0\qquad\ldots\text{(iv)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{u}{1(-6)-1(-2)}=\frac{-v}{1(-6)-(-1)(-2)}=\frac{1}{1(1)-(-1)(1)}
\displaystyle \Rightarrow \frac{u}{-4}=\frac{-v}{-8}=\frac{1}{2}
\displaystyle \frac{u}{-4}=\frac{1}{2}\Rightarrow u=-2
\displaystyle \frac{-v}{-8}=\frac{1}{2}\Rightarrow v=4
\displaystyle \therefore \frac{1}{x}=-2\Rightarrow x=-\frac{1}{2}
\displaystyle \text{and }\frac{1}{y}=4\Rightarrow y=\frac{1}{4}.
\displaystyle \therefore x=-\frac{1}{2},\qquad y=\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle ax+by=a-b,\qquad bx-ay=a+b
\displaystyle \text{Answer:}
\displaystyle ax+by-a+b=0\qquad\ldots\text{(i)}
\displaystyle bx-ay-a-b=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{b(-a-b)-(-a)(-a+b)}=\frac{-y}{a(-a-b)-b(-a+b)}=\frac{1}{a(-a)-b^2}
\displaystyle \Rightarrow \frac{x}{-(a^2+b^2)}=\frac{-y}{-(a^2+b^2)}=\frac{1}{-(a^2+b^2)}
\displaystyle \therefore x=1,\qquad -y=1
\displaystyle \Rightarrow y=-1
\displaystyle \therefore x=1,\qquad y=-1.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle ax+by=a^2,\qquad bx+ay=b^2
\displaystyle \text{Answer:}
\displaystyle ax+by-a^2=0\qquad\ldots\text{(i)}
\displaystyle bx+ay-b^2=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{b(-b^2)-a(-a^2)}=\frac{-y}{a(-b^2)-b(-a^2)}=\frac{1}{a^2-b^2}
\displaystyle \Rightarrow \frac{x}{a^3-b^3}=\frac{-y}{ab(a-b)}=\frac{1}{a^2-b^2}
\displaystyle \frac{x}{a^3-b^3}=\frac{1}{a^2-b^2}
\displaystyle \Rightarrow x=\frac{a^3-b^3}{a^2-b^2}
\displaystyle =\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}
\displaystyle \Rightarrow x=\frac{a^2+ab+b^2}{a+b}
\displaystyle \frac{-y}{ab(a-b)}=\frac{1}{(a-b)(a+b)}
\displaystyle \Rightarrow -y=\frac{ab}{a+b}
\displaystyle \Rightarrow y=-\frac{ab}{a+b}
\displaystyle \therefore x=\frac{a^2+ab+b^2}{a+b},\qquad y=-\frac{ab}{a+b}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle \frac{57}{x+y}+\frac{6}{x-y}=5,\qquad\frac{38}{x+y}+\frac{21}{x-y}=9
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{x+y}=u\text{ and }\frac{1}{x-y}=v.
\displaystyle \text{The given equations become}
\displaystyle 57u+6v-5=0\qquad\ldots\text{(i)}
\displaystyle 38u+21v-9=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{u}{6(-9)-21(-5)}=\frac{-v}{57(-9)-38(-5)}=\frac{1}{57(21)-38(6)}
\displaystyle \Rightarrow \frac{u}{51}=\frac{-v}{-323}=\frac{1}{969}
\displaystyle \frac{u}{51}=\frac{1}{969}\Rightarrow u=\frac{51}{969}=\frac{1}{19}
\displaystyle \frac{-v}{-323}=\frac{1}{969}\Rightarrow v=\frac{323}{969}=\frac{1}{3}
\displaystyle \therefore \frac{1}{x+y}=\frac{1}{19}\Rightarrow x+y=19\qquad\ldots\text{(iii)}
\displaystyle \frac{1}{x-y}=\frac{1}{3}\Rightarrow x-y=3\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 2x=22
\displaystyle \Rightarrow x=11
\displaystyle \text{Substituting }x=11\text{ in equation (iii),}
\displaystyle 11+y=19
\displaystyle \Rightarrow y=8
\displaystyle \therefore x=11,\qquad y=8.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle 5ax+6by=28,\qquad3ax+4by=18\hfill\text{[CBSE 2002C]}
\displaystyle \text{Answer:}
\displaystyle 5ax+6by-28=0\qquad\ldots\text{(i)}
\displaystyle 3ax+4by-18=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{6b(-18)-4b(-28)}=\frac{-y}{5a(-18)-3a(-28)}=\frac{1}{5a(4b)-3a(6b)}
\displaystyle \Rightarrow \frac{x}{4b}=\frac{-y}{-6a}=\frac{1}{2ab}
\displaystyle \frac{x}{4b}=\frac{1}{2ab}\Rightarrow x=\frac{2}{a}
\displaystyle \frac{-y}{-6a}=\frac{1}{2ab}\Rightarrow y=\frac{3}{b}
\displaystyle \therefore x=\frac{2}{a},\qquad y=\frac{3}{b}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle mx-ny=m^2+n^2,\qquad x+y=2m\hfill\text{[CBSE 2006C]}
\displaystyle \text{Answer:}
\displaystyle mx-ny-(m^2+n^2)=0\qquad\ldots\text{(i)}
\displaystyle x+y-2m=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{(-n)(-2m)-1[-(m^2+n^2)]}=\frac{-y}{m(-2m)-1[-(m^2+n^2)]}
\displaystyle =\frac{1}{m(1)-1(-n)}
\displaystyle \Rightarrow \frac{x}{m^2+2mn+n^2}=\frac{-y}{n^2-m^2}=\frac{1}{m+n}
\displaystyle \Rightarrow \frac{x}{(m+n)^2}=\frac{-y}{(n-m)(n+m)}=\frac{1}{m+n}
\displaystyle \frac{x}{(m+n)^2}=\frac{1}{m+n}\Rightarrow x=m+n
\displaystyle \frac{-y}{(n-m)(n+m)}=\frac{1}{m+n}\Rightarrow -y=n-m
\displaystyle \Rightarrow y=m-n
\displaystyle \therefore x=m+n,\qquad y=m-n.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle \frac{b}{a}x+\frac{a}{b}y=a^2+b^2,\qquad x+y=2ab\hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \frac{b}{a}x+\frac{a}{b}y=a^2+b^2
\displaystyle \Rightarrow b^2x+a^2y-ab(a^2+b^2)=0\qquad\ldots\text{(i)}
\displaystyle x+y-2ab=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{a^2(-2ab)-1[-ab(a^2+b^2)]}=\frac{-y}{b^2(-2ab)-1[-ab(a^2+b^2)]}
\displaystyle =\frac{1}{b^2(1)-a^2(1)}
\displaystyle \Rightarrow \frac{x}{ab(b^2-a^2)}=\frac{-y}{ab(a^2-b^2)}=\frac{1}{b^2-a^2}
\displaystyle \frac{x}{ab(b^2-a^2)}=\frac{1}{b^2-a^2}\Rightarrow x=ab
\displaystyle \frac{-y}{ab(a^2-b^2)}=\frac{1}{b^2-a^2}\Rightarrow -y=-ab
\displaystyle \Rightarrow y=ab
\displaystyle \therefore x=ab,\qquad y=ab.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle \frac{ax}{b}-\frac{by}{a}=a+b,\qquad ax-by=2ab\hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \frac{ax}{b}-\frac{by}{a}=a+b
\displaystyle \Rightarrow a^2x-b^2y-ab(a+b)=0\qquad\ldots\text{(i)}
\displaystyle ax-by-2ab=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{(-b^2)(-2ab)-(-b)[-ab(a+b)]}=\frac{-y}{a^2(-2ab)-a[-ab(a+b)]}
\displaystyle =\frac{1}{a^2(-b)-a(-b^2)}
\displaystyle \Rightarrow \frac{x}{-ab^2(a-b)}=\frac{-y}{-a^2b(a-b)}=\frac{1}{-ab(a-b)}
\displaystyle \frac{x}{-ab^2(a-b)}=\frac{1}{-ab(a-b)}\Rightarrow x=b
\displaystyle \frac{-y}{-a^2b(a-b)}=\frac{1}{-ab(a-b)}\Rightarrow -y=a
\displaystyle \Rightarrow y=-a
\displaystyle \therefore x=b,\qquad y=-a.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle \frac{x}{a}+\frac{y}{b}=a+b,\qquad\frac{x}{a^2}+\frac{y}{b^2}=2 
\displaystyle \text{Answer:}
\displaystyle \frac{x}{a}+\frac{y}{b}=a+b
\displaystyle \Rightarrow bx+ay-ab(a+b)=0\qquad\ldots\text{(i)}
\displaystyle \frac{x}{a^2}+\frac{y}{b^2}=2
\displaystyle \Rightarrow b^2x+a^2y-2a^2b^2=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{a(-2a^2b^2)-a^2[-ab(a+b)]}=\frac{-y}{b(-2a^2b^2)-b^2[-ab(a+b)]}
\displaystyle =\frac{1}{b(a^2)-b^2(a)}
\displaystyle \Rightarrow \frac{x}{a^3b(a-b)}=\frac{-y}{-ab^3(a-b)}=\frac{1}{ab(a-b)}
\displaystyle \frac{x}{a^3b(a-b)}=\frac{1}{ab(a-b)}\Rightarrow x=a^2
\displaystyle \frac{-y}{-ab^3(a-b)}=\frac{1}{ab(a-b)}\Rightarrow y=b^2
\displaystyle \therefore x=a^2,\qquad y=b^2.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle \frac{x}{a}=\frac{y}{b},\qquad ax+by=a^2+b^2
\displaystyle \text{Answer:}
\displaystyle \frac{x}{a}=\frac{y}{b}
\displaystyle \Rightarrow bx-ay=0\qquad\ldots\text{(i)}
\displaystyle ax+by-(a^2+b^2)=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{(-a)[-(a^2+b^2)]-b(0)}=\frac{-y}{b[-(a^2+b^2)]-a(0)}=\frac{1}{b(b)-a(-a)}
\displaystyle \Rightarrow \frac{x}{a(a^2+b^2)}=\frac{-y}{-b(a^2+b^2)}=\frac{1}{a^2+b^2}
\displaystyle \frac{x}{a(a^2+b^2)}=\frac{1}{a^2+b^2}\Rightarrow x=a
\displaystyle \frac{-y}{-b(a^2+b^2)}=\frac{1}{a^2+b^2}\Rightarrow y=b
\displaystyle \therefore x=a,\qquad y=b.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle \frac{a^2}{x}-\frac{b^2}{y}=0,\qquad\frac{a^2b}{x}+\frac{b^2a}{y}=a+b,\quad x,y\ne0\hfill\text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{The given equations become}
\displaystyle a^2u-b^2v=0\qquad\ldots\text{(i)}
\displaystyle a^2bu+ab^2v-(a+b)=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{u}{(-b^2)[-(a+b)]-ab^2(0)}=\frac{-v}{a^2[-(a+b)]-a^2b(0)}
\displaystyle =\frac{1}{a^2(ab^2)-a^2b(-b^2)}
\displaystyle \Rightarrow \frac{u}{b^2(a+b)}=\frac{-v}{-a^2(a+b)}=\frac{1}{a^2b^2(a+b)}
\displaystyle \frac{u}{b^2(a+b)}=\frac{1}{a^2b^2(a+b)}\Rightarrow u=\frac{1}{a^2}
\displaystyle \frac{-v}{-a^2(a+b)}=\frac{1}{a^2b^2(a+b)}\Rightarrow v=\frac{1}{b^2}
\displaystyle \therefore \frac{1}{x}=\frac{1}{a^2}\Rightarrow x=a^2
\displaystyle \text{and }\frac{1}{y}=\frac{1}{b^2}\Rightarrow y=b^2.
\displaystyle \therefore x=a^2,\qquad y=b^2.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle ax+by=\frac{a+b}{2},\qquad3x+5y=4
\displaystyle \text{Answer:}
\displaystyle ax+by-\frac{a+b}{2}=0\qquad\ldots\text{(i)}
\displaystyle 3x+5y-4=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{b(-4)-5\left(-\frac{a+b}{2}\right)}=\frac{-y}{a(-4)-3\left(-\frac{a+b}{2}\right)}=\frac{1}{5a-3b}
\displaystyle \Rightarrow \frac{x}{\frac{5a-3b}{2}}=\frac{-y}{-\frac{5a-3b}{2}}=\frac{1}{5a-3b}
\displaystyle \frac{x}{\frac{5a-3b}{2}}=\frac{1}{5a-3b}\Rightarrow x=\frac{1}{2}
\displaystyle \frac{-y}{-\frac{5a-3b}{2}}=\frac{1}{5a-3b}\Rightarrow y=\frac{1}{2}
\displaystyle \therefore x=\frac{1}{2},\qquad y=\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle 2(ax-by)+a+4b=0,\qquad2(bx+ay)+b-4a=0\hfill\text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle 2ax-2by+a+4b=0\qquad\ldots\text{(i)}
\displaystyle 2bx+2ay+b-4a=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{(-2b)(b-4a)-(2a)(a+4b)}=\frac{-y}{(2a)(b-4a)-(2b)(a+4b)}
\displaystyle =\frac{1}{(2a)(2a)-(2b)(-2b)}
\displaystyle \Rightarrow \frac{x}{-2(a^2+b^2)}=\frac{-y}{-8(a^2+b^2)}=\frac{1}{4(a^2+b^2)}
\displaystyle \frac{x}{-2(a^2+b^2)}=\frac{1}{4(a^2+b^2)}\Rightarrow x=-\frac{1}{2}
\displaystyle \frac{-y}{-8(a^2+b^2)}=\frac{1}{4(a^2+b^2)}\Rightarrow y=2
\displaystyle \therefore x=-\frac{1}{2},\qquad y=2.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle (a+2b)x+(2a-b)y=2,\qquad(a-2b)x+(2a+b)y=3
\displaystyle \text{Answer:}
\displaystyle (a+2b)x+(2a-b)y-2=0\qquad\ldots\text{(i)}
\displaystyle (a-2b)x+(2a+b)y-3=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{(2a-b)(-3)-(2a+b)(-2)}=\frac{-y}{(a+2b)(-3)-(a-2b)(-2)}
\displaystyle =\frac{1}{(a+2b)(2a+b)-(a-2b)(2a-b)}
\displaystyle \Rightarrow \frac{x}{-2a+5b}=\frac{-y}{-a-10b}=\frac{1}{10ab}
\displaystyle \frac{x}{-2a+5b}=\frac{1}{10ab}\Rightarrow x=\frac{5b-2a}{10ab}
\displaystyle \frac{-y}{-a-10b}=\frac{1}{10ab}\Rightarrow y=\frac{a+10b}{10ab}
\displaystyle \therefore x=\frac{5b-2a}{10ab},\qquad y=\frac{a+10b}{10ab}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle (a-b)x+(a+b)y=2a^2-2b^2,\qquad(a+b)(x+y)=4ab
\displaystyle \text{Answer:}
\displaystyle (a-b)x+(a+b)y-2(a^2-b^2)=0\qquad\ldots\text{(i)}
\displaystyle (a+b)x+(a+b)y-4ab=0\qquad\ldots\text{(ii)}
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{(a+b)(-4ab)-(a+b)[-2(a^2-b^2)]}
\displaystyle =\frac{-y}{(a-b)(-4ab)-(a+b)[-2(a^2-b^2)]}
\displaystyle =\frac{1}{(a-b)(a+b)-(a+b)^2}
\displaystyle \Rightarrow \frac{x}{2(a+b)(a^2-2ab-b^2)}
\displaystyle =\frac{-y}{2(a-b)(a^2+b^2)}=\frac{1}{-2b(a+b)}
\displaystyle \frac{x}{2(a+b)(a^2-2ab-b^2)}=\frac{1}{-2b(a+b)}
\displaystyle \Rightarrow x=-\frac{a^2-2ab-b^2}{b}
\displaystyle \Rightarrow x=\frac{-a^2+2ab+b^2}{b}
\displaystyle \frac{-y}{2(a-b)(a^2+b^2)}=\frac{1}{-2b(a+b)}
\displaystyle \Rightarrow y=\frac{(a-b)(a^2+b^2)}{b(a+b)}
\displaystyle \therefore x=\frac{-a^2+2ab+b^2}{b},\qquad y=\frac{(a-b)(a^2+b^2)}{b(a+b)}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Solve the following system of equations by the method of cross-multiplication:}
\displaystyle bx+cy=a+b
\displaystyle ax\left(\frac{1}{a-b}-\frac{1}{a+b}\right)+cy\left(\frac{1}{b-a}-\frac{1}{b+a}\right)=\frac{2a}{a+b}
\displaystyle \text{Answer:}
\displaystyle bx+cy=a+b\qquad\ldots\text{(i)}
\displaystyle \frac{1}{a-b}-\frac{1}{a+b}=\frac{2b}{a^2-b^2}
\displaystyle \frac{1}{b-a}-\frac{1}{b+a}=-\frac{2a}{a^2-b^2}
\displaystyle \therefore ax\left(\frac{2b}{a^2-b^2}\right)-cy\left(\frac{2a}{a^2-b^2}\right)=\frac{2a}{a+b}
\displaystyle \Rightarrow \frac{2a(bx-cy)}{(a-b)(a+b)}=\frac{2a}{a+b}
\displaystyle \Rightarrow bx-cy=a-b\qquad\ldots\text{(ii)}
\displaystyle bx+cy-(a+b)=0
\displaystyle bx-cy-(a-b)=0
\displaystyle \text{Using cross-multiplication,}
\displaystyle \frac{x}{c[-(a-b)]-(-c)[-(a+b)]}
\displaystyle =\frac{-y}{b[-(a-b)]-b[-(a+b)]}=\frac{1}{b(-c)-b(c)}
\displaystyle \Rightarrow \frac{x}{-2bc}=\frac{-y}{2b^2}=\frac{1}{-2bc}
\displaystyle \frac{x}{-2bc}=\frac{1}{-2bc}\Rightarrow x=\frac{a}{b}
\displaystyle \frac{-y}{2ab}=\frac{1}{-2bc}\Rightarrow y=\frac{b}{c}
\displaystyle \therefore x=\frac{a}{b},\qquad y=\frac{b}{c}.
\displaystyle \\

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