Answer:
The points of intersections are: (1.5 , 1.25) and (4 , 0)
Step-by-step explanation:
Given:
y = x² - 6x + 8 ⇒ (1)
2y + x = 4 ⇒ (2)
And required the solution of the system of equations.
By graphing the system of equations, the points of intersections are:
(1.5 , 1.25) and (4 , 0)
See the attached figure.
Another solution:
By substitution of y from the second equation at the first equation.
From (1) ⇒ y = 0.5 (4-x)
At (2): and solve for x
0.5 ( 4 - x ) = x² - 6x + 8 ⇒ multiply both sides by 2
4 - x = 2x² - 12x + 16
2x² - 12x + 16 + x - 4 = 0
2x² - 11x + 12 = 0
The general solution of the quadratic equation:
[tex]x = \frac{-b \pm \sqrt{b^2-4ac} }{2a}[/tex]
so, a = 2 , b = -11 and c = 12
∴[tex]x=\frac{-(-11) \pm \sqrt{(-11)^2-4*2*12} }{2*2}=\frac{11 \pm \sqrt{25} }{4} =\frac{11 \pm 5}{4} \\x = \frac{11+5}{4} = \frac{16}{4}=4\\ OR \ x = \frac{11-5}{4}=\frac{6}{4}=1.5\\[/tex]
∴ at x = 4 ⇒ y = 0.5 (4-x) = 0.5 * 0 = 0
And at x = 1.5 ⇒ y = 0.5 (4-x) = 0.5 * ( 4 - 1.5 ) = 0.5 * 2.5 = 1.25
So, the solution are the points (4,0) and (1.5 , 1.25)
