y = x²

Equate the ys:

x + 2 = x²

x² - x - 2 = 0

(x + 1)(x - 2) = 0

x = -1 or x = 2

When x = -1, y = 1

When x = 2, y = 4

So the points of intersection of the line and the parabola are
(-1, 1) and (2, 4).

y = 4x²

Equate the ys:

x + 3 = 4x²

4x² - x - 3 = 0

(4x + 3)(x - 1) = 0

x - y + 1 = 0

Substitute for y in Eqn 2:

x - (x² - 1) +1 = 0

x² - x - 2 = 0

(x - 2)(x + 1) = 0

x = 2 or x = -1

When x = 2, y = 3.

When x = -1, y = 0

So the points of intersection of the line and the parabola are

(2, 3) and (-1, 0).

y = x + x² - 10

y = 3x² - 2x - 19

Equate the ys:

x + x² -10 = 3x² - 2x -19

2x² - 3x - 9 = 0

(2x + 3)(x - 3) = 0

y = 3x² + 37x + 104

Equate the ys:

4x²+41x + 104 =
3x² + 37x + 104

x² + 4x = 0

x(x + 4) = 0

So x = 0 or x = -4

When x = 0, y = 104.

When x = -4, y = 4

So the points of intersection of the two parabolas are

(0, 104) and (-4, 4).

y = 2x² + 6x + 91

y = (x + 8)²

Equate the ys and expand the LHS:

2x² + 6x + 91 =
x² + 16x + 64

x² - 10x + 25 =0

(x - 5)² = 0

So x = 5 (only one solution)

When x = 5, y = 169

So the two parabolas
just touch each other at
(5, 169).

x + y = -1

Change subject to y in Eqn 2.

y = -1 - x

Substitute into Eqn 1:

x² + (-1-x)² = 5

x² + 1 + 2x + x² = 5

2x² + 2x -4 = 0

x² + x - 2 = 0

(x + 2)(x - 1) = 0

So x = -2 or 1

When x = -2, y = 1

When x = 1, y = -2

So the line meets the circle at (-2, 1) and (1, -2).

x² + y² = 4

Change subject to x in Eqn 1:

y = 3x

Substitute the y value in Eqn 2
into Eqn 1 and expand the brackets:

x² + 2x +1 + 9x² - 12x + 4 = 25

10(x² - x - 2) = 0

(x-2)(x+1) = 0

So x = 2 or x = -1

When x = 2, y = 6

When x = -1, y = -3.

So the line intersects with the circle at

(2, 6) and at (-1, -3).

2x - y = -1

Change subject to y in Eqn 2.

y = 2x + 1

2x² + (2x + 1)² = 11

2x² + 4x² + 4x + 1 = 11

6x² + 4x - 10 = 0

2(3x + 5)(x - 1) = 0

xy = 8

y = x + 7

Just to do somethng different because the 2nd equation is written as y = ...
substitute Eqn 2 into Eqn 1:

x(x + 7) = 8

x² + 7x - 8 = 0

(x + 8)(x - 1) = 0

So x = -8 or x = 1

When x = -8, y = -1.

When x = 1, y = 8

So the line meets the hyperbola at
(-8, -1) and (1, 8).

Equate the ys:

So the line meets the hyperbola at
(-6, -1) and at (2, 3).

x - 2y = 3

Change the subject of Eqn 2 to x:

x = 2y + 3

y(2y + 3) = 2

2y² + 3y - 2 = 0

(2y-1)(y+2) = 0

3x - 2y = 12

Change the subject of Eqn 1 to y:

So the line touches the hyperbola at just the point (2, -3).

x - y = 1

Change the subject of Eqn 2 to x:

x = 1 + y

1 + 2y + y² - 2y² = 5

y² - 2y +4 = 0

This equation cannot be factorised and so there is no point of intersection between the line and the hyperbola.

x² + y² = 13

xy = 8

x + 2y - z = -5 (Eqn 1)

2x - 3y + 4z = 28 (Eqn 2)

4x + 5y - 3z = -10 (Eqn 3)

I will not eliminate x as the hint suggested but leave that to you. I will remove the z instead.

Multiply Eqn 1 by 4:

4x + 8y - 4z = -20 (Eqn 4)

Add Eqn 2 to the Eqn 4:

6x + 5y = 8 (Eqn 5)

Multiply Eqn 1 by 3:

3x + 6y - 3z = -15 (Eqn 6)

Subtract Eqn 6 from Eqn 3:

x - y = 5 (Eqn 7)

Multiply this eqn by 5 and add
to the other new equation:

11x = 33

x = 3

So y = 3 - 5 = -2

3 - 4 - z = -5

z = 4.

So x = 3, y = -2 and z = 4.

YOU substitute these values into one of the three equations to verify these solutions.

e - f + g = 10 (Eqn 1)

4e + 2f - 3g = 8 (Eqn 2)

3e - 5f + 2g = 34 (Eqn 3)

Eqn 1 × 3
3e - 3f + 3g = 30 (Eqn 4)

Add Eqn 2 and Eqn 4:

7e - f = 38 (Eqn 5)

Eqn 1 × 2

2e - 2f + 2g = 20 (Eqn 6)

Subtract Eqn 6 from Eqn 3

e - 3f = 14 (Eqn 7)

Now substitute for e in Eqn 5
using Eqn 7:

7(3f + 14) - f = 38

20f = -60

f = -3

So e = 14 + 3(-3) = 5

Substitute for e and f in Eqn 1:

5 -(-3) +g = 10

g = 2

So e = 5, f = -3 and g = 2.

6x + 4y - 2z = 0 (Eqn 1)

3x - 2y + 4z = 3 (Eqn 2)

5x - 2y + 6z = 3 (Eqn 3)

(Eqn 3 - Eqn 2)

2x + 2z = 0 (Eqn 4)

(Eqn 2 ×2)

6x - 4y + 8z = 6 (Eqn 5)

Eqn 1 plus Eqn 5

12x + 6z = 6 (Eqn 6)

Divide by 6:

2x + z = 1 (Eqn 7)

Eqn 4 - Eqn 7

z = -1

So x = 1

Substitute into Eqn 1:

6(1) + 4y -2(-1) = 0

4y = -8

y = -2

So x = 1, y = -2 and z = -1.

2a - b - c = 9 (Eqn 1)

5a + 2c = -3 (Eqn 2)

7a - 2b = 1 (Eqn 3)

Eqn 3 only has a and b terms so we could eliminate c from the other 2 eqns:

Eqn 1 × 2
4a - 2b - 2c = 18 (Eqn 4)

Add Eqn 2 and Eqn 4:

9a - 2b = 15 (Eqn 5)

Subtract Eqn 5 from Eqn 3

2a = 14

a = 7

Now substitute for a in Eqn 3:

b = 24

Now substitute for a and b in Eqn 1:

2(7) - 24 - c = 9

c = -19

So a = 7, b = 24 and c = -19.

3p - q - 2r = -15 (Eqn 1)

20p - 3q - 5r = -15 (Eqn 2)

5p + 2q + 3r = -16 (Eqn 3)

q is the variable which has the lowest coefficients and so it is esiest to eliminate.

Eqn 1 × 2
6p - 2q - 4r = -30 (Eqn 4)

Add Eqn 3 and Eqn 4:

11p - r = -46 (Eqn 5)

Eqn 1 × 3

9p - 3q - 6r = -45 (Eqn 6)

Subtract Eqn 6 from Eqn 2

11p + r = 30 (Eqn 7)

Add Eqn 5 to Eqn 7:

22p = -16

Using Eqn 7:

Using Eqn 1:

x - 2y = 3 (Eqn 1)

4y - 3z = 4 (Eqn 2)

x + 3z = 2 (Eqn 3)

Subtract Eqn 1 from Eqn 3
2y + 3z = -1 (Eqn 4)

Add Eqn 2 and Eqn 4:

6y = 3

y = 0.5

Substitute into Eqn 1

x - 2(0.5) = 3

x = 4

Substitute into Eqn 3

4 + 3z = 2