Can anyone solve this simultaneous equation?

Question

2x + y = 6
x² + y² = 20

Answer 1

First, square both sides of the first equation.

(2x+y)^2=6^2
4x^2+4xy+y^2=36.

Then use the elimination method and subtract the second equation from the new first equation.

4x^2+4xy+y^2=36
-(x^2 +y^2) =-20
3x^2+4xy =16

Then factor out x.

x(3x+4y) =16

Since 3x+4y can be rewritten as 4(2x+y) - 5x, substitute 2x+y into 4(2x+y) - 5x.

x(4 times 6 -5x)=16.

Simplify the equation.

-5x^2+24x-16=0
(-5x+4)(x-4)=0
x=4/5 or x=4

Then plug-in the two values of x back into the original first equation.

2 times 4/5 +y=6
2 times 4 + y =6

Simplify the equations.

8/5+y=6 or 8+y=6
y=22/5 or y= -2
x=4/5 or x= 4

(x,y)= (4/5,22/5) or (x,y)= (4, -2)

Answer 2

Its referred to as algebraic technique because of the fact which you could desire to make certain what the variables (or letters) values are. Trial and mistake can paintings, even though it heavily isn't plenty use to you for greater durable equations. first element to do is to objective to do away with 1 variable via subtracting or including the equations jointly. you could no longer try this subsequently because of the fact which you will finally end up with the two variables nonetheless. So make the two the numbers till now the x's or the y's an identical like this: call the equations (a million) 5x + 3y = -a million and (2) 3x + 4y =6 Take (a million) and multiply it via 4 so which you have (3) 20x +12y = -4. Take (2) and multiply it via 3 to get (4) 9x +12y = 18. Now your numbers till now the y's are an identical so as which you will get rid of that variable. Take equation (3) and subtract (4) which aspects 11x = -22 so x = -2. placed this lower back into the two of the unique equations to paintings out the cost of y. In (a million) which skill (-2)*5 + 3y = -a million; -10 +3y = -a million. So 3y = 9 and y = 3. finally verify those are the main suitable solutions via putting them into equation (2): (3*-2)+(4*3) = (-6) + 12 = 6

Answer 3

2x + y = 6
x² + y² = 20
rearrange the top equation
y=6-2x
subsitute into the other equation
x^2 + (6-2x)^2 = 20
x^2 + (6-2x) (6-2x) = 20
x^2 + 36 -24x +4x^2 =20
5x^2 -24x +36=20
5x^2 -24x +16=0
(5x - 4) (x - 4)
x = 4/5, 4
y=6-2x
y=6-2(4/5)
y=6-1 3/5 = 4 2/5
y=6-2x
y=6-2(4) = -2
this gives co-ordinates of
(4/5, 4 2/5)
(4,-2)

Answer 4

Use substitution.

y = 6 - 2x

x^2 + (6-2x)^2 =20

x^2 + 36-12x-12x+4x^2 = 20

5x^2 -24x -16 =0

Now get the roots. Sub the roots into the other equation for 'y' values.

Answer 5

y = 6 - 2x

Substitute for y in equation 2

x² + (6 - 2x)² = 20

x² + 36 - 24x + 4x² = 20

5x² - 24x + 16 = 0

(5x - 4) (x - 4) = 0

x = 4/5 or x = 4

Plugging into equation 1

y = 22/5 or y = -2

So x = 4/5 and y = 22/5 or

x = 4 and y = -2

I hope this helps.

Answer 6

y = 6 - 2x
x² + (6 - 2x)² = 20
x² + 36 - 24x + 4x² = 20
5x² - 24x + 16 = 0
(5x - 4)(x - 4) = 0
x = 4/5 , x = 4

8/5 + y = 6
y = 22/5

8 + y = 6
y = - 2

Solution is ( 4/5 , 22/5 ) ; ( 4 , - 2 )

Answer 7

2x +y =6, so
y = 6 - 2x
xx + (6 - 2x)(6-2x) = 20
xx + 36 - 12x - 12x + 4xx = 20
5xx -24x +36 = 20
5xx - 24x + 16 = 0
(5x - 4)(x - 4) = 0

x = 4/5 & 4
y= 6 - 2x, so
y = 6 -2(4/5) = 4 2/5 and
y = 6-2(4) = -2

so, x= 4/5 and 4
y=4 2/5 and -2

Answer 8

y = 6 - 2x
x^2 + (6 - 2x)^2 = 20
x^2 + 36 - 24x + 4x^2 = 20
5x^2 - 24x + 16 = 0
(5x - 4)(x - 4) = 0
x = 4/5, 4

y = 6 - 2*(4/5)
y = 22/5

y = 6 - 2*(4)
y = -2

(x, y) = (4/5, 22/5) and (4, -2)

Answer 9

2 x + y = 6 =
x + y = square root of 20

2x+y=6 -
x+y=square root of 20

= x = 1.34

so y = 3.32

these are only approximates because of square rooting 20 leaves a long messy number...

Hopes this helps (but i'm not sure if it's 100% right...)

Answer 10

I think: x=square root of + or - '-16/5'

Then sub this into the first equation to get y.