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

    \[\mathbf{9}.\]

    \[\mathbf{5}/\left( \mathbf{x}\text{ }+\text{ }\mathbf{y} \right)\text{ }\text{ }\mathbf{2}/\left( \mathbf{x}\text{ }-\mathbf{y} \right)\text{ }=\text{ }-\mathbf{1}\]

    \[\mathbf{15}/\left( \mathbf{x}\text{ }+\text{ }\mathbf{y} \right)\text{ }+\text{ }\mathbf{7}/\left( \mathbf{x}\text{ }\text{ }\mathbf{y} \right)\text{ }=\text{ }\mathbf{10}\]

    \[\mathbf{10}.\]

    \[\mathbf{2}/\mathbf{x}\text{ }+\text{ }\mathbf{3}/\mathbf{y}\text{ }=\text{ }\mathbf{13}\]

    \[\mathbf{5}/\mathbf{x}\text{ }\text{ }\mathbf{4}/\mathbf{y}\text{ }=\text{ }-\mathbf{2}\]

Solution:

Let substitute :

    \[1/\left( x\text{ }+\text{ }y \right)\text{ }=\text{ }u\]

And

    \[1/\left( x\text{ }\text{ }y \right)\text{ }=\text{ }v\]

Equations

    \[5u\text{ }\text{ }2v\text{ }=\text{ }-1\]

    \[15u\text{ }+\text{ }7v\text{ }=\text{ }10\]

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

    \[{{a}_{1}}=5,{{b}_{1}}=-2,{{c}_{1}}=1\]

    \[{{a}_{2}}=15,{{b}_{2}}=7+,{{c}_{2}}=-10\]

\frac{u}{20-7}=\frac{-v}{-50-15}=\frac{1}{35+30}

\frac{u}{13}=\frac{-v}{-65}=\frac{1}{65}

\frac{u}{13}=\frac{1}{65}

=u=\frac{1}{5}

\frac{1}{u}=x+y

x+y=5.........\left( i \right)

And,

\frac{1}{v}=x-y

x-y=1.........\left( ii \right)

    \[\left( i \right)+\left( ii \right)\]

2x=6

x=3

Substitute the value of x in equation (i)

3+y=5

y=2

 

    \[x\text{ }=\text{ }3\]

    \[y\text{ }=\text{ }2\]

    \[\mathbf{10}.\]

    \[\mathbf{2}/\mathbf{x}\text{ }+\text{ }\mathbf{3}/\mathbf{y}\text{ }=\text{ }\mathbf{13}\]

    \[\mathbf{5}/\mathbf{x}\text{ }\text{ }\mathbf{4}/\mathbf{y}\text{ }=\text{ }-\mathbf{2}\]

Solution:

Let 1/x = u and

    \[1/y\text{ }=\text{ }v\]

    \[2u\text{ }+\text{ }3y\text{ }=\text{ }13\]

    \[\Rightarrow 2u\text{ }+\text{ }3y\text{ }\text{ }13\text{ }=\text{ }0\]

    \[5u\text{ }\text{ }4y\text{ }=\text{ }-2\]

    \[\Rightarrow 5u\text{ }\text{ }4y\text{ }+\text{ }2\text{ }=\text{ }0\]

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

    \[{{a}_{1}}=2,{{b}_{1}}=3,{{c}_{1}}=-13\]

    \[{{a}_{2}}=5,{{b}_{2}}=-4,{{c}_{2}}=2\]

\frac{u}{6-52}=\frac{-v}{4+65}=\frac{1}{-8-15}

\frac{u}{-46}=\frac{-v}{-69}=\frac{1}{-23}

\frac{u}{-46}=\frac{1}{-23}

=u=2

\frac{1}{u}=x

x=\frac{1}{2}

And,

\frac{-v}{69}=\frac{1}{-23}

v=3

\frac{1}{v}=y

=y=\frac{1}{3}

x=\frac{1}{2}, y=\frac{1}{3}

i

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