Question

Solve the simultaneous equation..


`x - 2y = 1 \\ y {}^(2) - 3xy + 8 = 0`
​

Answer 1

x-2y = 1 -----> equation (1)

y²-3xy + 8 = 0 -----> equation (2)


so equation (1) is x = 1 + 2y

now we will put equation (1) in equation (2)

y² - 3xy + 8 = 0

y² - 3 (1+2y) y + 8 = 0

y² - 3y - 6y² + 8 = 0

-5y² - 3y + 8 = 0

5y² + 3y - 8 = 0

(5y+8) (y-1) = 0

5y + 8 = 0
5y = -8
y = -⁸/₅


y-1 = 0

y = 1


so y = -⁸/₅ and y = 1

for y₁
x = 1 + 2y

x = 1 + 2(-⁸/₅)

x = 1 - ¹⁶/₅

x = -¹¹/₅

for y = 1

x = 1 + 2y

x = 1 + 2(1)

x = 1 + 2

x = 3




so (x₁,y₁) = (-¹¹/₅ , -⁸/₅)

and (x₂,y₂) = (3 , 1)