Question

N

Answer 1

Answer:

C

Step-by-step explanation:

Given: $y=x^2-x-12\,,\,y-x-3=0$

To find: solution set for this linear-quadratic system of equations

Solution:

A linear equation is an equation whose degree is 1 and degree of quadratic equation is 2.

$y-x-3=0\\y=x+3$

Put $y=x+3$ in $y=x^2-x-12$

$x+3=x^2-x-12\\x^2-2x-15=0\\x^2-5x+3x-15=0\\x(x-5)+3(x-5)=0\\(x+3)(x-5)=0\\x=-3\,,\,x=5$

For x = -3, y = -3+3=0

For x = 5, y = 5 + 3 = 8

So solutions are $\left ( -3,0 \right )\,,\,\left ( 5,8 \right )$

Option C. is correct