Can anybody solve this simultaneous equation and tell me how you did it? 2x + y = 4, xsquared + 2ysquared = 17

Answer 1

2x + y = 4
x^2 + 2y^2 = 17

y = -2x + 4

x^2 + 2(-2x + 4)^2 = 17
x^2 + 2((-2x + 4)(-2x + 4)) = 17
x^2 + 2(4x^2 - 8x - 8x + 16) = 17
x^2 + 2(4x^2 - 16x + 16) = 17
x^2 + 8x^2 - 32x + 32 = 17
9x^2 - 32x + 32 = 17
9x^2 - 32x + 15 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-32) ± sqrt((-32)^2 - 4(9)(15)))/(2(9))
x = (32 ± sqrt(1024 - 540))/18
x = (32 ± sqrt(484))/18
x = (32 ± 22)/18
x = (64/18) or (10/18)
x = 3 or (5/9)

y = -2x + 4
y = -2(3) + 4 = -6 + 4 = -2
y = -2(5/9) + 4 = (-10/9) + 4 = (-10/9) + (36/9) = (-10 + 36)/9 = (26/9)

ANS :
(3,-2) and ((5/9),(26/9))

Answer 2

2x + y = 4
x^2 + 2 y ^2 = 17

Rearrange the first equation to get the value of one of the variables, x or y, in terms of the other, for example:
y = 4 - 2x

Now substitute this into the second equation
x^2 + 2 (4-2x)^2 = 17

The term
(4-2x)^2
= (4-2x)(4-2x)
= 16 - 8x - 8x + 4(x^2) (by multiplying out each term)
= 4(x^2) - 16x + 16

Multiply this by 2 and put it back into the equation:
x^2 + 8(x^2) - 32x + 32 = 17

And adding like terms (and subtracting 17 from both sides) gives
9(x^2) - 32 x + 15 = 0

To solve this, you need the equation:
(-b +- sqrrt(b^2 - 4ac))/2a
where a is the number in front of the x^2, b the number in front of the x and c the remaining number.

so possible solutions for x are either
(32+sqrrt(32^2 - 4*9*15))/18; or
(32-sqrrt(32^2 - 4*9*15))/18

sqrrt(1024-540) = sqrrt(484) = 22

so solutions for x are
(32+22)/18; or
(32-22)/18

so x = 54/18 = 3 or x = 10/18 = 5/9

Now 2x+y=4, so

If x = 3...
2*3 + y = 4
6 + y = 4
y = -2

and if x = 5/9...
2*(5/9) + y = 4
10/9 + y = 4
y = 36/9 - 10/9 (where 4=36/9)
= 26/9
= 2 8/9

There should always be two answers if the equation goes up to the power of 2. If you write your answers in the form (x,y) you get the two answers (3,-2) and (5/9,2 8/9)

Hope that helps

Answer 3

x+y=4, x^2 + y^2 = 40 Take the 1st equation and isolate y: y=4-x because of the fact that's y, plug it into the 2nd equation for y: x^2 + (4-x)^2 = 40 x^2 + 4^2 -8x + x^2 = 40 x^2 + sixteen - 8x + x^2 = 40 2x^2 - 8x + sixteen = 40 2 (x^2 - 4x + 8) = 40 x^2 - 4x + 8 = 20 x^2 - 4x - 12 = 0 (x-6)(x+2) = 0 So x is 6 or -2. (you ought to apply the quadratic formulation for that, too.) we've 2 solutions for x, so plug the two in for x interior the unique equation: one million) 6 + y = 4; y = -2 2) -2 + y = 4, y = 6 Do they make experience? 6^2 + -2^2 = 36 + 4 = 40 -6^2 + 2^2 = 36 + 4 = 40 Yup! Wolfram Alpha could supply you the respond yet no longer tutor you the stairs in this actual case, in spite of if that's in simple terms placing apart one variable and plugging it into the different equation. once you are trying this, although, in simple terms be careful which you do no longer DIVIDE the two components with the aid of an expression that ought to equivalent 0. in case you are trying this, you will wander into undefined outcomes, that's the place you get those bizarre "one million=2" proofs you notice on the internet each now and then.

Answer 4

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>2 2
square the first equation 2x+y=4 >>>> 4x + y = 16

so that both equations have like terms; then add them together...
2 2
4x + y = 16
2 2
+ x + 2y = 17
---------------------
2 2
5x + 3y = 33 >>>>>>>>>>>> then solve for X in terms of Y...
2 2
...Isolate X... 5x = 33 - 3y >>>> divide ea. side by 5 to get X...
2 2 _____________
x = 33/5 - 3y /5 >>>>>>>>x = / 2
\/ 33/5 - 3y /5

>>>then plug the value in for X in the second equation>>>

2 2
33/5 - 3y /5 + 2y = 17 >>multiply both sides by 5>>>>>>
2 2
33 - 3y + 10y = 85 >>>>>>>>>>>> then solve>>>>>>>
2 2 _____
7y = 52 >>>>>>>>>> y = 52/7 >>>>>>> y = \/52/7 >>>

>>>plug y's value into the equation for x >>>
_____
2x + y = 4 >>>>> 2x = 4 - y >>> x = 2 - (\/ 52/7)/2 >>>>>
2 2
>>> x = 4 - (52/7)/4 >>> x = 16/4 - (52/7)/4 >>>
2
x = (112/7)/4 - (52/7)/4
2
x = (60/7)/4
2
x = 60/28
2
x = 15/7
________ _________
x = \/ 15/7 and y = \/ 52/7

Answer 5

2x + y = 4_______(1)

x^2 + 2*y^2 = 17_____(2)

Substitute 'y' from (1) in (2). Then you get

x^2 + 2*(4 - 2x)^2 = 17

x^2 + 2*[4^2 - 2*4*(2x) + (2x)^2] = 17

x^2 + 32 - 32x + 8*x^2 = 17

9*x^2 - 32*x + 32 - 17 = 0

9*x^2 - 32*x + 15 = 0

9*x^2 - 27*x - 5*x + 15 = 0

9x(x - 3) - 5(x - 3) = 0

(9x - 5)(x - 3) = 0

So either x = 3 or x = 5/9

From (1) the corresponding values of 'y' are -2 and 26/9.

Solution: x = 3, y = -2 and x = 5/9, y = 26/9

Answer 6

2x + y = 4 => y = 4 - 2x

x^2 + 2y^2 = 17
x^2 + 2(4 - 2x)^2 = 17
x^2 + 2(16 - 16x + 4x^2) = 17
x^2 + 32 - 32x + 8x^2 - 17 = 0
9x^2 - 32x + 15 = 0
(x - 3)(9x - 5) = 0
x = 3 or 5/9

Substitute x=3 and 5/9 into y = 4 - 2x,
y = 4 - 2(3) = -2

and

y = 4 - 2(5/9) = 4 - 1 1/9 = 2 8/9

Answer 7

Method (I'm not working it out):
1) Rearrange the first equation to express one variable in terms of the other, eg y=(4-2x).
2) Substitute this expression wherever that variable appears in the second equation. eg xsquared + 2(4-2x)squared = 17.
3) Multiply out and simplify to get a quadratic equation in one variable.
4) Solve the quadratic the usual way.
5) Substitute the solution in the first equation to solve for the other variable.

Answer 8

rearrange first one: y = 4 - 2x
then replace 'y' in the second with '4-2x' so:

x^2 + 2(4-2x)^2 = 17
x^2 + 2(16 - 16x +4x^2) = 17
9x^2 - 32x +32 = 17
9x^2 - 32x + 15 = 0
(9x - 5) (x-3) = 0
so x = 5/9 or x = 3

so, using first equation, y = 26/9 or y = -2.

now give me best answer. by the way, 'x^2' means 'xsquared'

Answer 9

4=4-2x

Substitute into x^2 + 2 y^2 = 17

x^2 + 2(4-2x)^2 =17
x^2 + (32 - 32x + 8x^2) = 17
9x^2 - 32x + 15 =0

And you can work out the rest im sure

Answer 10

y= 4- 2x
substitute y in the second equation by 4-2x we have
x^2 +(4-2x)^2 =17
simplify it we have
5x^2- 16^x +16=17
x=3.2
x=-0.06