Solve the following systems of equations. Show all work.

Question

1). 4x² + 8x + 5 - y = 1

   3x² - x + 3 = y + x + 8

2). 4x² – 2y² = 4

   x²  + y² = 10

Answer 1

4x² + 8x + 5 - y = 1->y=4x^2+8x+4->4x^2+4x+4x+4->4x(x+1)+4(x+1)->(4x+4)(x+1)

3x² - x + 3 = y + x + 8->y=3x^2-2x-5 3x^2+3x-5x-5->(3x-5)(x+1)

4x^2-2y^2=4->4x^2=4+2y^2->4x^2-4=2y^2->2x^2-2=y^2->y=sqrt(2x^2-2)->sqrt(2(x^2-1))->sqrt(2(x+1)(x-1)

x^2+y^2=10->y^2=10-x^2->y=sqrt(10-x) or y=-sqrt(10-x)

Answer 2

1) Subtract the second equation from the first:

x²+9x+2=-x-7

x²+10x+9=0

(x+1)(x+9)=0 --> x = -1, -9

x = -1, y=0

x=-9, y=256

2) Add the two equations to eliminate y²

6x²=24

x= ±√4 = ±2

4(2²)-2y²=4

y²=6

y = ±√6

x=±2

y=±√6

Answer 3

1) As there is a single y term, we express all other terms in relation to y

4x^2 + 8x + 4 = y (first equation) - to make the math easy when we substitute, note that this is 4(x+1)^2 = y

3x^2 - 2x - 5 = y (second equation)

Subtracting,

x^2 + 10x + 9 = 0

(x + 9)(x + 1) = 0

x = -9, -1

Substituting, we get solutions for (x, y) of

(-9, 256) and (-1, 0)

2) Let u = x^2 and v = y^2

4u - 2v = 4

u + v = 10 (Multiply by 4 to eliminate top equation by subtracting it).

4u + 4v = 40

6v = 36

v = 6

u = 10 - v = 4

(u, v) = (4, 6)

(x, y) = (±2, ±√6); we may also write the terms separately as (-2, -√6), (-2, √6), (2, -√6), (2, √6)