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# simultaneous linear and quadratic equations

find the truth set of the following pairs of equations.

1. 3x2 –y2=35 and 2x-y=1

2. Xy+y2=3 and 2x+y=1

3. X2-xy=14 and x+y=3

4. 3xy-x=0 and x+3y=2

### 2 Answers by Expert Tutors

Gene G. | You can do it! I'll show you how.You can do it! I'll show you how.
5.0 5.0 (257 lesson ratings) (257)
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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.