Answer:
[tex]\boxed{x = -1; \, x = 5}[/tex]
Step-by-step explanation:
(a) Set the two functions equal to each other
[tex]\begin{array}{rcl}y & = & 2x - 3\\y & = & x^{2} - 2x - 8\\2x - 3 & = & x^{2} - 2x - 8\\2x & = & x^{2} -2x - 5\\x^{2} - 4x - 5 & = & 0\\\end{array}[/tex]
(b) Factor the quadratic equation
Find two numbers that multiply to give -5 and add to give ₄9.
Possible pairs are 1, -5; -1, 5;
By trial and error, you will find that 1 and -5 work:
1 × (-5) = -5 and 1 - 5= -4
x² - 4x - 8 = (x + 1)(x - 5)
(c) Solve the quadratic
[tex]\begin{array}{rlcrl}x+ 1 & =0 & \qquad & x - 5 & =0\\x & = -1 & \qquad & x & = 5\\\end{array}\\\text{The solution to the system of equations is }\boxed{\mathbf{x = -1; x = 5}}[/tex]
