solve the pairs of simultaneous equation to 3 decimal places for 2 question?

Question

(a) 2x+3y+4=0 , x^2+6xy+7=0
(b) x+5y=3 , x/2+3/y=4

Pls show step by step.Tq

pls show me the step.. coz i cant solve it.. tat y i will ask it at here.. tq for helping

i ask it at here is coz i duno how to solve it.. TAT y.. if nt dis website is for wad..zz tq for those helping.. and for those who ask me do it myself.. tq for reply too.. =.=|||

Answer 1

(a)
2x + 3y + 4 = 0
x^2 + 6xy + 7 = 0
Easiest looks like getting rid of y:
3y = -2x - 4
y = (-2x - 4)/3
x^2 + 6x(-2x - 4)/3 + 7 = 0
x^2 + 2x(-2x - 4) + 7 = 0
x^2 - 4x^2 - 8x + 7 = 0
- 3x^2 - 8x + 7 = 0
3x^2 + 8x - 7 = 0
x = (- 8 ± √(64 + 84))/6
x = (- 4 + √(16 + 21))/3, (- 4 - √(16 + 21))/3
x = (- 4 + √(37))/3, (- 4 - √(37))/3
x = 0.694, - 3.361
y = (-1.338 - 4)/3, (6.722 - 4)/3
y = -1.779, 1.361



(b)
x + 5y = 3
x/2 + 3/y = 4
xy/2 + 3 = 4y
x = 3 - 5y
(3 - 5y)y + 6 = 8y
3y - 5y^2 + 6 = 8y
5y^2 + 5y - 6 = 0
y^2 + y - 1.2 = 0
y = (- 1 ± √(1 + 4.8))/2
y = (- 1 ± √(5.8))/2
y = -1.704, 0.704
x = 11.521, -0.521

Answer 2

y = 2x + 3 , y = 3x² + 4x - 9 = 0 => 3x² + 4x - 9 = 2x + 3 => 3x² + 2x - 12 = 0 applying quadratic formula x = [ -2 +/a million squarert ( 4 + a hundred and forty four ) ] / 6 x = [ -2 +/- squarert ( 148 ) ] / 6 x = [ -2 +/- 12.sixteen ] / 6 => x = ( -2 + 12.sixteen ) / 6 or x = ( -2 - 12.sixteen ) / 6 => x = 10.sixteen / 6 or x = -14.sixteen / 6 => x = a million.sixty 9 or x = - 2.36 while x = a million.sixty 9 then from equation y = 2x + 3 y = 2 ( a million.sixty 9 ) + 3 = 6.38 while x = -2.36 then from equation y = 2x + 3 y = 2 ( -2.36 ) + 3 = - a million. seventy two thank you for declaring approximately ' y '

Answer 3

(a) In the second equation, there is only one term in y.
So it will be easier to find y in the first equation,
then substitute its value into the second equation.
I've left out intermediate steps which you should check.

2x + 3y + 4 = 0, therefore, y = -(4 + 2x) / 3

The second equation then becomes :
x^2 + 6x[-(4 + 2x) / 3] + 7 = 0
Therefore, 3x^2 + 8x - 7 = 0

By the quadratic formula :
x = {-8 +/- sqrt[8^2 - 4(3)(-7)]} / (2*3)
x = [-4 +/- sqrt(37)] / 3
x = 0.694 or -3.361

Substituting into first equation gives :
y = [-4 (-/+) 2*sqrt(37)] / 9
y = -1.796 or 0.907

Solutions (x, y) are : (0.694, -1.796) and (-3.361, 0.907)
-------------------------
(b) Multiplying second equation through by 2y gives :
xy + 6 = 8y

Find x in the first equation to be : x = 3 - 5y
and then substitute into the modified second equation.

(3 - 5y)y + 6 = 8y
Therefore, 5y^2 + 5y - 6 = 0

So, y = {-5 +/- sqrt[5^2 - 4(5)(-6)]} / (2*5)
y = [-5 +/- sqrt(145)] / 10
y = 0.704 or -1.704

Substituting into first equation gives :
x = [11 (-/+) sqrt(145)] / 2
x = -0.521 or 11.521

Solutions (x,y) are : (-0.521, 0.704) and (11.521, -1.704)

Answer 4

(b) 3-5y = 8-6/y (= x)

==> 3y-5y^2 = 8y - 6

==> 5y^2 +5y - 6 = 0

==> y^2 +y -6/5 = 0

==> (y + 1/2)^2 = 6/5 - 1/4 = 19/20

==> y = +/- sqrt(19/20) - 1/2 = .475 and -1.477

==> x = 3 - 5y = -.950 and 10.385

Answer 5

a) rearrange the equations and substitute
x = -2 - 1.5y

(-2 - 1.5y)^2 + 6(-2 - 1.5y)y + 7 = 0

now solve for x, and use that x value to solve for y

do the same thing for b
x = 3 - 5y

(3 - 5y)/2 + 3/y = 4
6/5 = y(1 - y)
...
after you get the y values, substitute both of them back in order to check which one is the right one

**actually get the answers by yourself...priceless**

Answer 6

you should do ur own homeowrk