Question

X^2 - x - 42 = 0, ii. Y^2 - 17y + 72 = 0 to solve both the equations to find the values of x and y?

Answer 1

Answer:

First one:(x+6)(x-7) Second one:(y-8)(y-9)

Step-by-step explanation:

You do so by factorizing:

x^2-x-42=0

x^2-7x+6x-42=0

x(x-7)+6(x-7)=0

(x+6)(x-7)

y^2-17y+72=0

y^2-9y-8y+72=0

y(y-9)-8(y-9)

(y-8)(y-9)

Hope this helped.

Answer 2

PROCESS

1)

$x {}^{2} - x - 42 = 0 \\ x {}^{2} - 7x + 6x - 42 = 0 \\ x(x - 7) + 6(x - 7) = 0 \\ (x - 7)(x + 6) = 0 \\ therefore \: either \: \\ (x - 7) = 0 \\ x = 7 \\ or \\ (x + 6) = 0 \\ x = - 6$

2)

$y {}^{2} - 17y + 72 = 0 \\ y {}^{2} - 9y - 8y + 72 = 0 \\ y(y - 9) - 8(y - 9) = 0 \\ (y - 9)(y - 8) = 0 \\ either \: \: \\ (y - 9) = 0 \\ y = 9 \\ or \\ (y - 8) = 0 \\ y = 8$