help!!!what is the answer to this simultaneous equation?y=x^2 + x - 6y=2x - 1?

Question

please help!! i just need the answer and the workings please. everything i've done just doesnt seem to work, where am i going wrong!!!

ok i messed up...written the question wrong.
y = x^2 + x - 6 and y = 2x - 1

Answer 1

Substitute 2x-1 for y in the middle part, you get:
x^2 + x -6(2x-1) = 2x-1
This you can reduce to a quadratic equation for x.

Answer 2

y = x^2 + x - 6 and y = 2x - 1

Substitute linear into quadratic to get:
x^2 + x - 6 = 2x - 1
Rearrange to get:
x^2 - x - 5 = 0
Complete the square:
(x-1/2)^2 - 21/4 = 0
x-1/2 = sqrt(21/4)
x = 1/2 + sqrt(21/4) and x = 1/2 - sqrt(21/4)

Answer 3

Equate one to the other:

2x-1=x^2+x-6

Make it equalto zero:

x^2-x-5=0

No easy factorisation use quadratic formula:

a=1, b=-1, c=-5

(1 +/- SQROOT((1- (-20))))/2

=(1+/- SQR(21))/2

The answer would be a surd. If you are newto quadratics I would guess you have written the question wrong. If you are at GCSE/A Level standard then the answer would be a surd.

Answer 4

y=x^2 + x - 6y
=> x^2 + x = 7y ---(1)

x^2 + x - 6y = 2x - 1
=> x^2 + x - 6y - 2x + 1 = 0
=> x^2 - x +1 = 6y ---(2)

(1)x6, 6x^2 + 6x = 42y ---(3)
(2)x7, 7x^2 - 7x + 7 = 42y ---(4)
(4)-(3), x^2 - 13x + 7 = 0

x = [13 +- SQRT(169-4(1)(7))] / (2)(1)
x = [13 +- SQRT(141)] / 2
x = 12.4 or 0.563 (to 3 sig fig)

Substitute these values of x into any of the equations to find y.

Answer 5

y=x^2 + x - 6y=2x - 1
for the moment, forgrt the y= part
x^2+x-6y=2x+1 subtract 2x-1
x^2-x+1-6y=0 since y=2x-1
x^2-x+1-6(2x-1)=0
x^2-13x +7=0
x=(13+/-√(169-28))/2=(13+/-√(141))/2
substitute this in y=2x-1 to find y

Answer 6

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

Equation (1) - equation (2)

0 = x^2 - x - 5

So x = ½(1 ± √41) by the quadratic formula

When x = ½(1 + √41)
y = √41
and when x = ½(1 - √41)
y = - √41

Answer 7

Y = X^2 - x - 6y + 1 = 0
Y + 6Y = X^2 - x + 1
SO Y = X^2 - x + 1 all over 7.

Answer 8

y=x^2+x-6......(1)
y=2x-1
......x=(y+1)/2.......(2)
substitute for x into (1)
y=((y+1)/2)^2+(y+1)/2-6
4y=y^2+2y+1+2y+2-24
y^2=21
>>>>>y=+or- sqrt21
from(2),
when y= +sqrt21
x=(sqrt21+(1))/2
when y= -sqrt21
x=(1-sqrt21)/2

1 hope that this helps

Answer 9

You sure you have the question write? Simultaneous equations should have two distinct equations

Answer 10

x^2 -x+1=7y
(x^2-x+1)/7=y
((1/7)+- sqr (1/7)^2 -4*(1/7)*(1/7))7 solve that for your answer