In mathematics, solving linear equations is one of the important topics. We can say concept of linear equations is the base of advance algebra. Many students are scared of math and this anxiety should be the first thing that needs to be fixed as soon as possible. Many times students are afraid of asking any question related to it because they do not want others to think that they did not understand the concept. So we have designed cross multiplication method concept in such a way so that students can learn at their own pace and clear their doubts. Go through the below content and at the end you will be confident about the concept and able to solve problems at our own. Also we provide NCERT Solutions on cross multiplication method problems which are prepared by experts. We provide accurate and easy solutions for all questions covered in NCERT textbooks. Answers have been structured in a logical and easy language for quick revisions during examination or tests.

In this section, we will discuss about simultaneous linear equations by using cross-multiplication method.A method can be used to determine the value of a variable from any equation. Usually in elementary algebra and elementary arithmetic, rational expressions and equations involving fractions are solved by using cross-multiplication method.

General form of a linear equation in two unknown quantities: ax + by + c = 0, (a, b â‰ 0)

Assume two linear equation for x and y variables be

$a_1$ x + $b_1$y + $c_1$ = 0 ...(1)

$a_2$ x + $b_2$ y + $c_2$ = 0 ...(2)

The coefficients of x are: $a_1$ and $a_2$

The coefficients of y are: $b_1$ and $b_2$

The constant terms are: $c_1$ and $c_2$

Use method of elimination to solve both the equations:

Equation (1) is multiplied with $b_2$

$b_2$($a_1$ x + $b_1$y + $c_1$ = 0)

$a_1$$b_2$ x + $b_1$$b_2$y + $c_1$$b_2$ = 0 ...(3)

Equation (2) is multiplied with $b_1$

$b_1$($a_2$ x + $b_2$ y + $c_2$ = 0)

$a_2$$b_1$ x + $b_2$$b_1$ y + $c_2$$b_1$ = 0 ...(4)

Subtract equation 4 from equation 3, we have

($a_1$$b_2$ - $a_2$$b_1$) x + ($b_1$$b_2$ - $b_2$$b_1$)y + ($c_1$$b_2$ - $c_2$$b_1$) = 0

This implies

x = $\frac{ b_1\ c_2 -\ b_2\ c_1}{ b_2\ a_1 -\ a_2\ b_1}$; where ($a_1$$b_2$ - $a_2$$b_1$) $\neq$ 0

To obtain the value of y, substitute the value of x in equation (1),

y = $\frac{ c_1\ a_2 -\ c_2\ a_1}{a_1\ b_2\ -\ a_2\ b_1}$; where ($a_1$$b_2$ - $a_2$$b_1$) $\neq$ 0

From the value of x and y we can obtain the result as:

Example: Solve below system of linear equations using cross multiplication method

3x + y = 10 and x + 2y = 5

Solution:

3x + y - 10 = 0 ...equation(1)

x + 2y - 5 = 0 ...equation (2)

Here,

a$_1$ = 3, b$_1$ = 1, c$_1$ = -10

a$_2$ = 1, b$_2$ = 2, c$_2$ = -5

Use above derived formula, to find the value of x and y.

x = $\frac{ b_1 c_2- b_2 c_1}{a_1 b_2 - a_2 b_1}$

x = $\frac{ b_1 c_2- b_2 c_1}{a_1 b_2 - a_2 b_1}$

= $\frac{-5 \times 1\ - ( -10)\ \times 2}{3\ \times 2\ -\ 1\ \times 1}$

= $\frac{(-5) - (-20)}{6 -1)}$

= $\frac{(-5) - (-20)}{6 -1)}$

= $\frac{15}{5}$

= 3

Value of x is 3

Now,

y = $\frac{ c_1\ a_2 -\ c_2\ a_1}{a_1\ b_2\ -\ a_2\ b_1}$

Value of x is 3

Now,

y = $\frac{ c_1\ a_2 -\ c_2\ a_1}{a_1\ b_2\ -\ a_2\ b_1}$

= $\frac{(-10)(1)-(-5)(3)}{(3)(2) - (1)(1)}$

= $\frac{(-10) - (-15)}{6 - 1}$

= $\frac{5}{5}$

= 1

Therefore, solution is : x = 3 and y = 1

Therefore, solution is : x = 3 and y = 1