What is the solution of the linear-quadratic system of equations? will give brainliest and extra !

d. (1, 3) and (-3, -1)

Step-by-step explanation:

Equating the expressions for y, we have ...

... x² +3x -1 = y = x +2

Subtracting x+2 gives ...

... x² +2x -3 = 0

... (x +3)(x -1) = 0 . . . . . factored form

... x = -3, 1 . . . . . . . . . . .values that make the factors zero

The second equation tells us, y = x+2, so

... For x = -3, y = -3 +2 = -1. The solution is (-3, -1)

... For x = 1, y = 1 +2 = 3. The solution is (1, 3)

(- 3, - 1 ) or (1, 3 )

Since both equations express y in terms of x, equate both sides

x² + 3x - 1 = x + 2 ( subtract x + 2 from both sides )

x² + 2x - 3 = 0

(x + 3 )(x - 1 ) = 0

equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x - 1 = 0 ⇒ x = 1

Substitute these values into either of the 2 equations for y

y = - 3 : y = - 3 + 2 = - 1 ( using y = x + 2 )

x = 1 : y = 1 + 2 = 3

solutions are (- 3, - 1 ) or (1, 3 )

x² + 3x + 7 = 0       [ax² + bx + c]

a = 1

b = 3

c = 7

quadratic formula:

    (+/- is this ±)

plug in the numbers into the variables

your answer is a

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5