Question

Y = x2 + 3
y = x + 5

Answer 1

(2,7)

Explanation:

SOLVE FOR X:

x * 2 + 3 = x + 5

2x + 3 = x +5

-x -x

-3 -3

2x - x = 5 - 3

x = 5 - 3

x = 2

SOLVE FOR Y

y = x + 5

subsitute 2 for x

y = 2 + 5

y = 7

(x,y)

(2,7)

Answer 2

For this case we must solve the following system of two equations with two unknowns.

`y = x ^ 2 + 3\\y = x + 5`

We match:

`x ^ 2 + 3 = x + 5\\x ^ 2-x + 3-5 = 0\\x ^ 2-x-2 = 0`

Where:

`a = 1\\b = -1\\c = -2`

The roots will be given by:

`x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}`

We replace:

`x = \frac {- (- 1) \pm \sqrt {(- 1) ^ 2-4 (1) (- 2)}} {2 (1)}\\x = \frac {1 \pm \sqrt {1 + 8}} {2}\\x = \frac {1 \pm \sqrt {9}} {2}\\x = \frac {1 \pm3} {2}`

So, we have two roots:

`x_ {1} = \frac {1 + 3} {2} = \frac {4} {2} = 2\\x_ {2} = \frac {1-3} {2} = \frac {-2} {2} = - 1`

Answer:

`x_ {1} = 2 \ then \ y_ {1} = 2 + 5 = 7\\x_ {2} = - 1 \ then \ y_ {2} = - 1 + 5 = 4`