How many real solutions does this system of equations have? X^2 + 3 = y

3x-y+1 = 0

Answer choices:

A. 2

B. 1

C. 0

D. 3

Question

Mathematics

Macie Gillespie

17 February, 18:25

How many real solutions does this system of equations have? X^2 + 3 = y

3x-y+1 = 0

Answer choices:

A. 2

B. 1

C. 0

D. 3

+5

Answers (1)

Know the Answer?

Angeline · Answer 1

A

Step-by-step explanation:

x^2+3=y

3x-y+1=0

3x+1=y

so x²+3=3x+1

x²-3x+2=0

x²-x-2x+2=0

x (x-1) - 2 (x-1) = 0

(x-1) (x-2) = 0

x=1,2

two solutions

How many real solutions does this system of equations have? X^2 + 3 = y3x-y+1 = 0Answer choices:A. 2B. 1C. 0D. 3

How many real solutions does this system of equations have? X^2 + 3 = y

3x-y+1 = 0

Answer choices:

A. 2

B. 1

C. 0

D. 3