Can you help me with this simultaneous equation question?

Question

Basically the equations are:

3x + 4y = 5
x^2 + 5xy + y^2 = -5

Answer 1

From the first equation solve for x in terms of y:

x = (5 - 4y) / 3

Then substitute this value of x in the second equation:

((5 - 4y) / 3) ^ 2 + 5((5 - 4y) / 3)y + y ^2 = -5 and then solve for y.

Answer 2

Ok, the solution follows:

Rearranging the linear equation we have: y = (5 - 3x)/4

Substituting this expression for y into the quadratic, we have:

x^2 + 5x[(5 - 3x)/4] + [(5 - 3x)/4]^2 = -5

To get rid of the fractions we can multiply through by 16 to obtain:

16x^2 + 20x(5 - 3x) + (5 - 3x)^2 = -80

Expanding we get:

16x^2 + 100x - 60x^2 + 25 - 30x + 9x^2 = -80

Simplfying, this becomes:

-35x^2 + 70x + 105 = 0

Dividing through by -35 we get:

x^2 - 2x - 3 = 0

This factorises to:

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

Therefore x = 3 or x = -1

Now we use our rearranged linear equation, y=(5 - 3x)/4 to find the corresponding values of y;

If x = 3 then y = (5 - 9)/4

therefore y = -1


If x = -1 then y = (5 + 3)/4

therefore y = 2


So the solutions to this pair of simultaneous equations are:

x = 3, y = -1

and x = -1, y = 2


Hope that helps!

Answer 3

3x + 4y = 5................(1)
x^2 + 5xy + y^2 = -5.. (2)

here's another way

square(1) and multiply by 45

45x^2+120xy+80y^2=125....(3)
multiply (2) by 24
24x^2+120xy+24y^2= -120..(4)
subtract (3) from(4)
21x^2 +56y^2=245
divide through by 7
3x^2 8y^2=35..(5)
from(1),x=(5-4y)/3
substitute into (5)
3((5-4y)/3)^2 +8y^2 =35
((5-4y)^2)/3 +8y^2 =35
multiply through by 3
(5-4y)^2 +24y^2 =105
40y^2-40y-80 =0
divide through by 40
y^2-y-2 =0
(y-2)(y+1) =0
>>>y =2 or -1

substitute back into (1)

when y=2,x= -1
wheny= -1,x=3

there are a lot of ways to solve this problem
but basically they are all the same

i hope that this helps

Answer 4

Ok... first, you can solve the first equation for x:
3x = 5 - 4y
x = (5 - 4y)/3

Then, you can plug in this value of x into the other equation. That equation will now only have y's in it, and you can solve for y. Then, using your answer(s) for y, you can find all possible values for x by plugging your y values into the first (or second) equation again. Hope this helps!