Question

1. Solve the following systems of equations using cross multiplication method.

Answer 1

equation is given : (3x/2) - (5y/3) = -2 , (x/3) + (y/2) = 13/6

3x/2 - 5y/3 =-2
⇒(3 × 3x - 2 × 5y)/6 = -2
⇒9x - 10y = -12
⇒9x - 10y + 12 = 0 ------------(1)

x/3 + y/2 = 13/6
⇒(2 × x + 3 × y)/6 = 13/6
⇒2x + 3y = 13
⇒2x + 3y - 13 = 0 -------(2)

Now, apply cross multiplication rule ,
see the attachment , here it is clearly shown
∴ x/{(-10) × (-13) - 12 × 3} = -y/{9 × (-13) - 12 × 2} = 1/(9 × 3- (-10) × 2}
x/(130-36) = y/(24 -(-117)) = 1/(27-(-20))
x/(94) = y/(24+117) = 1/(27+20)
x/94 = x/141 = 1/47
x/2 = y/3 = 1
Hence, x = 2 and y = 3

Answer 2

QUESTION :
Solve the following systems of equations using cross multiplication method.
(i) 3 x + 4 y = 24, 20 x - 11 y = 47 (ii) 0.5 x + 0.8 y = 0.44 , 0.8 x + 0.6 y = 0.5
(iii) (3x/2) - (5y/3) = -2 , (x/3) + (y/2) = 13/6

SOLUTION IS IN THE ATTACHMENT.

CROSS - MULTIPLICATION METHOD:
The general form of a pair of linear equations
a1x + b1y + c1 = 0 , & a2x + b2y + c2 = 0.
When a1 / a2 ≠ b1 / b2, the pair of linear equations will have a unique solution.



       x                    y            1
----------- = ----------------- =---------
b1       c1      c1         a1       a1   b1
b2        c2     c2         a2       a2    b2


To solve this pair of equations for x and y using cross-multiplication, we’ll arrange the variables x and y and their coefficients a1, a2, b1 and b2, and the constants c1 and c2 as shown below
⇒ x = b1 c2 - b2 c1 / a1 b2 - a2 b1
⇒ y = c1 a2 - c2 a1 / a1 b2 - a2 b1

The above equation is generally written as :
x/ b1 c2 - b2 c1 = y/ c1 a2 - c2 a1 = 1/a1 b2 - a2 b1

HOPE THIS WILL HELP YOU...