Question

Solve the following pair of equations using cross-multiplication method :
x - 3y - 7 = 0
3x - 5y - 15 = 0​

Answer 1

Therefore the solution of pair of equations are

$x=\frac {5}{2}$     and     $y=-\frac{3}{2}$

Step-by-step explanation:

Cross- multiplication method is a method for solving linear equation of two variables.

If two linear equations are

a₁x+b₁y+c₁=0

a₂x+b₂y+c₂=0

Then the solution is

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{a_2c_1-a_1c_2}=\frac{1}{a_1b_2-a_2b_1}$

Given pair of linear equation are

x-3y-7=0 ......(1)

3x - 5y -15 = 0 .....(2)

Here a₁ = 1, b₁=-3, c₁=-7,a₂=3,b₂=-5, c₂= -15

$\frac{x}{-3(-15)-(-7)(-5)}=\frac{y}{3.(-7)-1.(-15)}=\frac{1}{1.(-5)-3.(-3)}$

$\Rightarrow \frac{x}{10}=\frac{y}{-6}=\frac{1}{4}$

$\therefore x =\frac{10}{4}$    and   $y=\frac{-6}{4}$

$x=\frac {5}{2}$     and     $y=-\frac{3}{2}$

Therefore the solution of pair of equations are

$x=\frac {5}{2}$     and     $y=-\frac{3}{2}$

Answer 2

The values of x and y are  $\frac{5}{2}$ and $\frac{-3}{2}$  respectively

Step-by-step explanation:

Given equations are $x-3-7=0\hfill (1)$ and  

$3x-5y-15=0\hfill (2)$​

To solve the given equations by Cross Multiplication method :

Now solving the given two equations

For the equations $a_1x+b_1y+c_1=0$  and $a_2x+b_2y+c_2=0$

The formula is $\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

-3   -7      1     -3

-5   -15     3    -5

• Substitute the values in the formula we get
• $\frac{x}{(-3)(-15)-(-5)(-7)}=\frac{y}{(-7)(3)-(-15)(1)}=\frac{1}{(1)(-5)-(3)(-3)}$
• $\frac{x}{45-35}=\frac{y}{-21+15}=\frac{1}{-5+9}$
• $\frac{x}{10}=\frac{y}{-6}=\frac{1}{4}$
• Now equating $\frac{x}{10}=\frac{1}{4}$

• $x=\frac{1}{4}\times 10$
• $=\frac{5}{2}$
• $x=\frac{5}{2}$

Therefore the value of x is $\frac{5}{2}$

• Now equating $\frac{y}{-6}=\frac{1}{4}$
• $y=\frac{1}{4}\times (-6)$
• $=\frac{-3}{2}$
• $y=\frac{-3}{2}$

Therefore the value of y is $\frac{-3}{2}$

From the solved given equations we get the values of x and y are  $\frac{5}{2}$ and $\frac{-3}{2}$  respectively