Question

solve the following pairs of equation by cross multiplication method 3x+y=5 and 5x+3y=3 ​

Answer 1

Method of cross-multiplication

When the simultaneous equations are expressed as :

$\sf\qquad a_1x+b_1y+c_1=0$

$\sf and,\quad a_2x+b_2y+c_2=0$

$\small\sf Then,\qquad\qquad x = \dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1} \ \ and \ y = \dfrac{c_1a_2-c_2a_1}{a_1b_2 - a_2b_1}$

Here, we can use this method.

Standard form of the equation in cross Multiplication method is :

ax + by + c = 0

_____________________________

3x + y = 5

Then,

3x + y - 5 = 0

Similarly,

5x + 3y = 3

Then,

5x + 3y - 3 = 0

Now we have our equations as,

______________________________

3x + y - 5 = 0

5x + 3y - 3 = 0

______________________________

Comparing

$\sf \qquad a_1x+b_1y+c_1=0 \: with \ 3x + y-5=0$

$\sf and \quad a_2x+b_2y+c_2=0 \ with \ 5x+3y-3=0, \ we \ get \ :$

• $\sf a_1 = 3$
• $\sf b_1 = 1$
• $\sf c_1 = -5$
• $\sf a_2 = 5$
• $\sf b_2 = 3$
• $\sf c_2 = -3$

Now we can apply the formula

$\sf x = \dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}$

$\sf \qquad x = \dfrac{(1)(-3) - (3)(-5)}{(3)(3) - (5)(1)} = \dfrac{-3 - (-15)}{9-5} = \dfrac{-3 + 15}{4}$

$\sf \implies x = \dfrac{12}{4} = 3$

$\qquad\qquad \bigstar\boxed{\bf x = 3}\bigstar$

$\sf y = \dfrac{c_1a_2-c_2a_1}{a_1b_2 - a_2b_1}$

$\sf \qquad y = \dfrac{(-5)(5) - (-3)(3)}{(3)(3) - (5)(1)} = \dfrac{-25 - (-9)}{9-5} = \dfrac{-25 + 9}{4}= \dfrac{-16}{4}$

$\qquad\qquad \bigstar\boxed{\bf y= -4}\bigstar$