Question

What are the solutions of this system of equations?

Answer 1

Answer:

Option C is correct.

Step-by-step explanation:

y = x^2-x-3     eq(1)

y = -3x + 5      eq(2)

We can solve by substituting the value of y in eq(2) in the eq(1)

-3x+5 = x^2-x-3

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

x^2+2x-8=0

Now factorizing the above equation

x^2+4x-2x-8=0

x(x+4)-2(x+4)=0

(x-2)(x+4)=0

(x-2)=0 and (x+4)=0

x=2 and x=-4

Now finding the value of y by placing value of x in the above eq(2)

put x =2

y = -3x + 5

y = -3(2) + 5

y = -6+5

y = -1

Now, put x = -4

y = -3x + 5

y = -3(-4) + 5

y = 12+5

y =17

so, when x=2, y =-1 and x=-4 y=17

(2,-1) and (-4,17) is the solution.

So, Option C is correct.

Answer 2

Answer: Third Option

(2, -1) and (-4, 17)

Step-by-step explanation:

We have the following system of equations:

$y = x^2 - x-3$

$y=-3x + 5$

We have the first three steps to solve the system.

$x^2- x-3 = -3x +5$   equal both equations

$0 = x^2 + 2x - 8$      Simplify and equalize to zero

$0 = (x-2)(x+4)$     Factorize

Then note that the equation is equal to zero when $x = 2$ or $x = -4$

Now substitute the values of x in either of the two situations to obtain the value of the variable y.

$y=-3(2) + 5$

$y=-6 + 5$

$y=-1$

First solution: (2, -1)

$y=-3(-4) + 5$

$y=12 + 5$

$y=17$

Second solution: (-4, 17)

The answer is the third option