dawnmsinagra dawnmsinagra
  • 02-06-2020
  • Mathematics
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solve the system of equations y=4x+1, y=x2+2x-2

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turtlefishangel
turtlefishangel turtlefishangel
  • 02-06-2020
option D
(-1,-3) and (3,13)

Given in the question two equations
y = 4x + 1
y = x² + 2x - 2
Equate both functions
4x + 1 = x² + 2x - 2
rearrange the x term and constant
-x² + 4x - 2x + 2 + 1 = 0
-x² + 2x + 3 = 0
factors
-x * 3x = -3x²
-x + 3x = 2x
-x² -x + 3x + 3 = 0
-x(x+1) +3(x+1) = 0
solve
(x+1)(3-x) = 0
x = -1
and
x = 3
Plug these values in equation to find y
x = -1
y = 4(-1)+ 1
y = -3
x = 3
y = 4(3)+ 1
y = 13
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dhshshshsb dhshshshsb
  • 02-06-2020
Option d is the answer to your question
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