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

Answer:

Choice A is correct answer.

Step-by-step 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,


${x}^{2} = 9$

$x = \pm \sqrt{9}$

$x = \pm 3$

$x = 3 \: or \: x = - 3$

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).