Search 75,306 tutors
FIND TUTORS
Ask a question
0 0

simultaneous linear and quadratic equations

Tutors, please sign in to answer this question.

2 Answers

The first problem has no real solution.  I'll show you how to solve the second one.

2. xy + y2 = 3  and  2x + y = 1

You can solve the linear equation for y and substitute that into the quadratic equation...
y = 1-2x, so...

x(1-2x) + (1-2x)2 = 3
x-2x2+1-4x+4x2 = 3
Combine terms...
4x2-2x2+x-4x+1-3 = 0
2x2-3x-2 = 0
Factor this using AC factoring...
2x2+x-4x-2 = 0
x(2x+1)-2(2x+1) = 0    (Watch the signs!)
(x-2)(2x+1) = 0

x-2 = 0,      x = 2
2x+1 = 0,    x = -1/2

Now plug these values for x into the linear equation and solve for y...  (y = 1-2x)

x = 2, y = 1-4 = -3    (2, -3) is in the solution set
x = -1/2,  y = 1 + 1 = 2   (-1/2, 2) is in the solution set.

Solution set = {(2,-3),(-1/2,2)}

You can try this procedure on the first problem.  The quadratic after substitution has no real roots.  Rebecca's suggestion should show you this immediately because the lines will not intersect.

I'll let you work through the other two.

Hope this helps.

The easiest way to solve this would be to graph the two lines and find out where they intersect for the solutions.