Question

N

What is the solution set for this linear-quadratic system of equations?

y= x2 - x-12

y-x-3 = 0


A. {(-3,0), (0,3)}

B. {(-3,0), (4, 0);

C. {(-3,0), (5,8)}

D. {(4,0), (0,3)}

Answer 1

C

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