Solve the system: y=14x2+2x−6 y=−14x2−3x+6?

x = -3/4, y = 3/8

x = 4/7, y = -2/7

Subtract the two eq'ns to eliminate 'y' 

Hence 

0 = 28x^2 + 5x -12 

Apply the Quadratic eq'n 

x = {- 5 +/- sqrt[(5)^2 - 4(28)(-12)]} / 2(28)

x = { - 5 +/- sqrt[25 + 1344]} / 56

x = { - 5 +/- sqrt[1369]} / 56 

x = { - 5 +/- 37}/56 

x = -42/56 = - 3/4(a) & 32/56 = 4/7(c)

Substitute for 'y' 

y = 14((-3/4)^2 + 2(-3/4) - 6 

y = 7 7/8 - 3/2 - 6 

y = 3/8 (b) 

&

y = 14(4/7)^2 - 2(4/7) - 6 

y = 4 4/7 - 8/7 - 6 

y = -2 4/7 (d) 

Hence a +b+c+d = -3/4 +3/8 +4/7  -2 4/7 = -2 3/8 

y=14x^2+2x-6-------(1)

y=-14x^2-3x+6-------(2)

(1)+(2)

=>

y=-x/2------(3)

combining (1) & (3), get

-x/2=14x^2+2x-6

=>

28x^2+5x-12=0

=>

x=-0.75~-1 (a)

y=0.375~0 correspondingly (b)

or

x=0.5714286~1 (c)

y=-0.2857143~0 (d)

=>

a+b+c+d=-1+0+1+0=0

approximately.

y = 14x² + 2x - 6

y = - 14x² - 3x + 6

y = y

14x² + 2x - 6 = - 14x² - 3x + 6

28x² + 5x = 12

x² + (5/28).x = 12/28

x² + (5/28).x + (5/56)² = (12/28) + (5/56)²

[x + (5/56)]² = 37²/56²

x + (5/56) = ± 37/56

x = - (5/56) ± (37/56)

x = (- 5 ± 37)/56

First case: x = (- 5 + 37)/56 = 32/56 = 4/7

y = 14x² + 2x - 6

y = 14.(4/7)² + 2.(4/7) - 6

y = (32/7) + (8/7) - (42/7)

y = - 2/7

→ First point (4/7 ; - 2/7)

Second case: x = (- 5 - 37)/56 = - 42/56 = - 3/4

y = 14x² + 2x - 6

y = 14.(- 3/4)² + 2.(- 3/4) - 6

y = (63/8) - (12/8) - (48/8)

y = 3/8

→ Second point (- 3/4 ; 3/8)

14x^2  + 2x  -6  = -14x^2 -3x + 6 

28x^2  + 5x -12 =  0  

use quadratic formula

x=  ( -5  +/- sqrt( 25 - 4*(-12)*(28) ) ) / 56  

x = (-5 +/- sqrt( 1369)  ) / 56  

x = (-5 +/- 37 ) /56  

x  =  (-42/56) or   (32/56)   

simplifying  

42/14  =3 ,  56/14 = 4

32/8 =4 ,    56/8  = 7

x =  -3/4 or    4/7      

y= 14*(-3/4)^2 +  2*(-3/4) - 6 = 14*(9/16) -6/4  -6  

y = 7*9/8 -6/4  -6   =  (63/8) - 12/8  - 48/8 =    3/8  

so one point is 

(-3/4,   3/8)  

2nd point  

y = 14*(4/7)^2   + 2*(4/7)  -6  =   14*(16/49)  +8/7-  42/7 

y =  (2*16)/7 +8/7 -42/7  =   40/7  -42/7 = -2/7   

2nd point 

(4/7,   -2/7)

a+b + c + d = (-3/4) + (3/8)  + 4/7-2/7 = 

a+ b + c + d =  -6/8 + 3/8    + 2/7  =   -3/8  + 2/7 

a+ b + c + d  =    -21/56   +    16/56   =   -5/56 

a + b + c + d =  -5/56  

approx.  - -0.089285714

y = 14x^2 + 2x − 6 

y = −14x2 − 3x + 6

14x^2 + 2x − 6 = −14x2 − 3x + 6

28x^2 + 5x - 12 = 0

(4 x + 3) (7 x - 4) = 0

Solutions: 

x = -3/4, y = 3/8

x = 4/7, y = -2/7

a + b + c + d = -3/4 + 3/8 + 4/7 - 2/7 = -5/56

Presuming these are:

y = 14x² + 2x - 6 and y = -14x² - 3x + 6

You have a system of two equations with two unknowns.  We can solve this with substistution.

Both expressions are equal to y so both expressions are equal to each other:

14x² + 2x - 6 = -14x² - 3x + 6

Now we can solve for x:

28x² + 5x - 12 = 0

I don't like trying to factor quadratics with such a high leading coefficient, so I'll use the quadratic equation:

x = [ -b ± √(b² - 4ac)] / (2a)

x = [ -5 ± √(5² - 4(28)(-12))] / (2 * 28)

x = [ -5 ± √(25 + 1344)] / 56

x = (-5 ± √1369) / 56

x = (-5 ± 37) / 56

x = -42/56 and 32/56

x = -3/4 and 4/7

These aren't integers, but I'll finish this, anyway.

We have two values of x so we can find two values of y:

y = 14x² + 2x - 6

y = 14(-3/4)² + 2(-3/4) - 6 and y = 14(4/7)² + 2(4/7) - 6

y = 14(9/16) - 3/2 - 6 and y = 14(16/49) + 8/7 - 6

y = 126/16 - 3/2 - 6 and y = 224/49 + 8/7 - 6

y = 63/8 - 3/2 - 6 and y = 224/49 + 8/7 - 6

y = 63/8 - 12/8 - 48/8 and y = 224/49 + 56/49 - 294/49

y = 3/8 and y = -14/49

y = 3/8 and y = -2/7

The solutions are:

(-3/4, 3/8) and (4/7, -2/7)

The sum of the four fractions is:

-3/4 + 3/8 + 4/7 - 2/7

-6/8 + 3/8 + 4/7 - 2/7

-3/8 + 2/7

-21/56 + 16/56

-5/56