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: Third Option
(2, -1) and (-4, 17)
Step-by-step explanation:
We have the following system of equations:
[tex]y = x^2 - x-3[/tex]
[tex]y=-3x + 5[/tex]
We have the first three steps to solve the system.
[tex]x^2- x-3 = -3x +5[/tex] equal both equations
[tex]0 = x^2 + 2x - 8[/tex] Simplify and equalize to zero
[tex]0 = (x-2)(x+4)[/tex] Factorize
Then note that the equation is equal to zero when [tex]x = 2[/tex] or [tex]x = -4[/tex]
Now substitute the values of x in either of the two situations to obtain the value of the variable y.
[tex]y=-3(2) + 5[/tex]
[tex]y=-6 + 5[/tex]
[tex]y=-1[/tex]
First solution: (2, -1)
[tex]y=-3(-4) + 5[/tex]
[tex]y=12 + 5[/tex]
[tex]y=17[/tex]
Second solution: (-4, 17)
The answer is the third option
