What is 1x^2+x-12=0,y^2-5y+6=0 ?

The solution to 1x^2+x-12=0,y^2-5y+6=0 is

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1x^2+x-12=0,y^2-5y+6=0

1 x^{2} + x - 12 = 0, y^{2} - 5 y + 6 = 0

x = 3, x = 3, x = - 4, x = - 4, y = 3 y = 2 y = 3 y = 2

Solution steps

[1 \cdot x^{2} + x - 12 = 0 y^{2} - 5 y + 6 = 0]

Solve

1 x^{2} + x - 12 = 0 : x = 3, x = - 4

For

y^{2} - 5 y + 6 = 0

, subsitute

x

with

3 : y = 3, y = 2

For

y^{2} - 5 y + 6 = 0

, subsitute

x

with

- 4 : y = 3, y = 2

Solve

y^{2} - 5 y + 6 = 0 : y = 2, y = 3

For

1 x^{2} + x - 12 = 0

, subsitute

y

with

2 : x = 3, x = - 4

For

1 x^{2} + x - 12 = 0

, subsitute

y

with

3 : x = 3, x = - 4

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

1 x^{2} + x - 12 = 0, y^{2} - 5 y + 6 = 0

are

x = 3, x = 3, x = - 4, x = - 4, y = 3 y = 2 y = 3 y = 2

4 x y - 2 x^{2} = - 6, 6 x + 3 y = - 6

f (x) = x^{2} - 5 x + 3, f (x) = - 3

y = - x^{2} + 2, y = x

2 x - 5 y = - 3, 2 x - 5 y = 3

x - y = 9, x \cdot y = 14