Question

What are the solutions to the following system of equations? y = x2 + 3x − 7 3x − y = −2 (2 points) (3, 11) and (−3, −7) (11, 3) and (−3, −7) (3, 11) and (−7, −3) No real solutions

Answer 1

Choice A is correct answer.

Explanation:

We have given a system of equations.

y = x² + 3x − 7 eq(1)

3x − y = −2 eq(2)

We have to find the solution of given system of equations.

Adding -3x to both sides of eq(2), we have

-3x+3x-y = -3x-2

-y = -3x-2

y = 3x+2 eq(3)

Now, putting the value of y in eq(1), we have

3x+2 = x²+3x-7

Adding -3x to both sides of above equation, we have

-3x+3x+2 = -3x+x²+3x-7

2 = x²-7

Adding 7 to both sides of above equation, we have

2+7 = x²-7+7

9 = x²

Taking square root to both sides of above equation, we have

√9 = √x²

±3 = x

Putting the value of x in eq(3), we have

y = 3(±3)+2

y = 3(3)+2 and y = 3(-3)+2

y = 9+2 and y = -9+2

y = 11 and y = -7

Hence, the solutions of given system are :

(3,11) and (-3,-7)

Answer 2

ANSWER

(3, 11) and (−3, −7)


EXPLANATION

The given system of equations are:



`y = {x}^(2) + 3x - 7`

and


`3x - y = - 2`

or


`y = 3x + 2`


We equate the two equations to obtain,




`{x}^(2) + 3x - 7 = 3x + 2`

We rewrite in standard quadratic form to obtain,


`{x}^(2) + 3x - 3x - 7 - 2 = 0`


This simplifies to


`{x}^(2)- 9= 0`


We solve for x to obtain,



`{x}^(2) = 9`


`x = \pm √(9)`


`x = \pm 3`


`x = 3 \: or \: x = - 3`

When

`x = 3`


`y = 3(3) + 2 = 11`

When x=-3,


`y = 3( - 3) + 2 = - 7`


Therefore the solution for the system is (3, 11) and (−3, −7).