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Solve each of the following systems of equations by the method of cross-multiplication:
CBSE | Class 10 | Exercise 3.4 | Guides | Maths | Pair of Linear Equations in Two Variables | RD Sharma
Solve each of the following systems of equations by the method of cross-multiplication:

    \[\mathbf{13}.\]

    \[~\mathbf{x}/\mathbf{a}\text{ }+\text{ }\mathbf{y}/\mathbf{b}\text{ }=\text{ }\mathbf{a}\text{ }+\text{ }\mathbf{b}\]

    \[\mathbf{x}/{{\mathbf{a}}^{\mathbf{2}~}}+\text{ }\mathbf{y}/{{\mathbf{b}}^{\mathbf{2}~}}=\text{ }\mathbf{2}\]

    \[\mathbf{14}.\]

    \[\mathbf{x}/\mathbf{a}\text{ }=\text{ }\mathbf{y}/\mathbf{b}\]

    \[\mathbf{ax}\text{ }+\text{ }\mathbf{by}\text{ }=\text{ }{{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}}\]

Solution:

Given

    \[x/a\text{ }+\text{ }y/b\text{ }\text{ }\left( a\text{ }+\text{ }b \right)~=\text{ }\mathbf{0}\]

    \[x/{{a}^{2~}}+\text{ }y/{{b}^{2~}}\text{ }2~=\text{ }\mathbf{0}\]

Cross multiplication Method

    \[\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{-y}{{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]

Comparing the two equations

    \[{{a}_{1}}=\frac{1}{a},\]

Let

    \[{{b}_{1}}=\frac{1}{b},\]

Let

    \[{{c}_{1}}=\left( a+b \right)\]

    \[{{a}_{2}}=\frac{1}{{{a}^{2}}},{{b}_{2}}=\frac{1}{{{b}^{2}}},{{c}_{2}}=-2\]

\frac{x}{\frac{-2}{b}+\frac{a}{{{b}^{2}}}+\frac{1}{b}}=\frac{-y}{\frac{-2}{a}+\frac{1}{a}+\frac{b}{{{a}^{2}}}}=\frac{1}{\frac{-1}{a{{b}^{2}}}-\frac{-1}{a2b}}

    \[\frac{x}{\frac{a-b}{{{b}^{2}}}}=\frac{-y}{\frac{-a-b}{{{a}^{2}}}+\frac{1}{a}+\frac{b}{{{a}^{2}}}}=\frac{1}{\frac{-1}{a{{b}^{2}}}-\frac{-1}{a2b}}\]

    \[\frac{x}{\frac{a-b}{{{b}^{2}}}}=\frac{1}{\frac{-1}{a{{b}^{2}}}-\frac{-1}{a2b}}\]

x={{a}^{2}}

    \[\frac{-y}{\frac{-a-b}{{{a}^{2}}}+\frac{1}{a}+\frac{b}{{{a}^{2}}}}=\frac{1}{\frac{-1}{a{{b}^{2}}}-\frac{-1}{a2b}}\]

y={{b}^{2}}

x={{a}^{2}}and y={{b}^{2}}

Solution:

Given

    \[x/a\text{ }\text{ }y/b~=\text{ }\mathbf{0}\]

    \[ax\text{ }+\text{ }by\text{ }\text{ }\left( {{a}^{2}}~+\text{ }{{b}^{2}} \right)~=\text{ }\mathbf{0}\]

Cross multiplication Method

    \[\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{-y}{{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]

Comparing the two equations

    \[{{a}_{1}}=\frac{1}{a},\]

Let

    \[{{b}_{1}}=\frac{1}{b},\]

    \[{{c}_{1}}=0\]

    \[{{a}_{1}}=a,{{b}_{1}}=b\]

, Let

    \[{{c}_{1}}=\left( {{a}^{2}}+{{b}^{2}} \right)\]


\frac{x}{\frac{{{a}^{2}}+{{b}^{2}}}{b}}=\frac{-y}{\frac{{{a}^{2}}+{{b}^{2}}}{b}}=\frac{1}{\frac{a}{b}+\frac{b}{a}}

\frac{x}{\frac{{{a}^{2}}+{{b}^{2}}}{b}}=\frac{1}{\frac{a}{b}+\frac{b}{a}}

=x=a

And,

\frac{-y}{\frac{{{a}^{2}}+{{b}^{2}}}{b}}=\frac{1}{\frac{a}{b}+\frac{b}{a}}

=y=b

x=a and y=b

i

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