Login/Sign Up
Home » Solve each of the following systems of equations by the method of cross-multiplication:
Solve each of the following systems of equations by the method of cross-multiplication:
CBSE | Class 10 | Exercise 3.4 | Guides | Maths | Maths | RD Sharma
Solve each of the following systems of equations by the method of cross-multiplication:

 

    \[\mathbf{x}\text{ }+\text{ }\mathbf{2y}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}\]

    \[\mathbf{2x}\text{ }\text{ }\mathbf{3y}\text{ }\text{ }\mathbf{12}\text{ }=\text{ }\mathbf{0}\]

 

    \[\mathbf{3x}\text{ }+\text{ }\mathbf{2y}\text{ }+\text{ }\mathbf{25}\text{ }=\text{ }\mathbf{0}\]

    \[\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }+\text{ }\mathbf{10}\text{ }=\text{ }\mathbf{0}\]

Solution: 

Given

    \[\mathbf{x}\text{ }+\text{ }\mathbf{2y}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}\]

    \[\mathbf{2x}\text{ }\text{ }\mathbf{3y}\text{ }\text{ }\mathbf{12}\text{ }=\text{ }\mathbf{0}\]

Cross multiplication Method

R D Sharma Solutions For Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables ex 3.4 - 1

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

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

    \[{{a}_{2}}=2,{{b}_{2}}=3,{{c}_{2}}=12\]

Calculation,

\frac{x}{-24+3}=\frac{y}{-12-2}=\frac{1}{-3-4}

    \[\frac{x}{-21}=\frac{-y}{-14}=\frac{1}{-7}\]

\frac{x}{-21}=\frac{1}{-7}

= x=3

And,

\frac{y}{-14}=\frac{1}{-7}

= y=2

x=3 and y=2

Solution:

Given

    \[\mathbf{3x}\text{ }+\text{ }\mathbf{2y}\text{ }+\text{ }\mathbf{25}\text{ }=\text{ }\mathbf{0}\]

                 

    \[\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }+\text{ }\mathbf{10}\text{ }=\text{ }\mathbf{0}\]

By cross multiplication method

Comparing the two equations

    \[{{a}_{1}}=3,{{b}_{1}}=2,{{c}_{1}}=25\]

    \[{{a}_{2}}=2,{{b}_{2}}=3,{{c}_{2}}=10\]

Calculation

\frac{x}{20-25}=\frac{y}{30-50}=\frac{1}{3-4}

    \[\frac{x}{-5}=\frac{-y}{-20}=\frac{1}{-1}\]

\frac{x}{-5}=\frac{1}{-1}

= x=5

And,

\frac{y}{-20}=\frac{1}{-1}

= y=20

x=5 and y=20

i

More Material