Question

Some the system of equations

Y=2x^2 - 3
Y=3x - 1

How many real number solutions are there to the equation 0=3x^2 + x - 4?

What are the solutions of the system
Y=x + 5
Y=x^2 - x + 2

Answer 1

y = 2x² - 3
y = 3x - 1

2x² - 3 = 3x - 1
+ 1 + 1
2x² - 2 = 3x
2x² - 3x - 2 = 0
2x² - 4x + x - 2 = 0
2x(x) - 2x(2) + 1(x) - 1(2) = 0
2x(x - 2) + 1(x - 2) = 0
(2x + 1)(x - 2) = 0
2x + 1 = 0 or x - 2 = 0
- 1 - 1 + 2 + 2
2x = -1 or x = 2
2 2
x = -0.5

y = 3x - 1 or y = 3x - 1
y = 3(-0.5) - 1 or y = 3(2) - 1
y = -1.5 - 1 or y = 6 - 1
y = -2.5 or y = 5
(x, y) = (-0.5, -2.5) or (x, y) = (2, 5)
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3x² + x - 4 = 0
3x² - 6x + 2x - 4 = 0
3x(x) - 3x(2) + 2(x) - 2(2) = 0
3x(x - 2) + 2(x - 2) = 0
(3x + 2)(x - 2) = 0
3x + 2 = 0 or x - 2 = 0
- 2 - 2 + 2 + 2
3x = -2 or x = 2
3 3
x = ⁻²/₃
There are two real number solutions in the equation.
————————————————————————————————
y = x + 5
y = x² - x + 2

x + 5 = x² - x + 2
+ x + x
2x + 5 = x² + 2
- 2 - 2
2x + 3 = x²
0 = x² - 2x - 3
0 = x² - 3x + x - 3
0 = x(x) - x(3) + 1(x) - 1(3)
0 = x(x - 3) + 1(x - 3)
0 = (x + 1)(x - 3)
0 = x + 1 or 0 = x - 3
- 1 - 1 + 3 + 3
-1 = x or x = 3

y = x + 5 or y = x + 5
y = -1 + 5 or y = 3 + 5
y = 4 or y = 8
(x, y) = (-1, 4) or (x, y) = (3, 8)