how do i solve the simultaneous equation x+2y=1 and 4x^2+3y^2=52?

Answer 1

Let's solve the first for x:
x = -2y + 1

Then substitute into the second:
4x² + 3y² = 52
4(-2y + 1)² + 3y² = 52

Expand the square using FOIL:
4[ 4y² - 4y + 1] + 3y² = 52

Multiply the 4 through:
16y² - 16y + 4 + 3y² = 52

Add the y² terms:
19y² - 16y + 4 = 52

Subtract 52 from both sides:
19y² - 16y - 48 = 0

At this point you would use the quadratic formula to solve.

y = [ -b ± sqrt( b² - 4ac ) ] / 2a
y = [ 16 ± sqrt( 256 - 4(19)(-48) ] / 38
y = [ 16 ± sqrt( 256 + 3648 ] / 38
y = [ 16 ± sqrt( 3904 ) ] / 38
y = [ 16 ± 8 sqrt( 61 ) ] / 38
y = [ 8 ± 4 sqrt(61) ] / 19

So values of y are:
y = [ 8 + 4 sqrt(61) ] / 19
or
y = [ 8 - 4 sqrt(61) ] / 19

Then substitute these back in to get the values of x:
For y = [ 8 + 4 sqrt(61) ] / 19
x = 1 - 2y
x = 1 - 2[ 8 + 4 sqrt(61) ] / 19
x = [ 19 - 16 - 8 sqrt(61) ] / 19
x = [ 3 - 8 sqrt(61) ] / 19

For y = [ 8 - 4 sqrt(61) ] / 19
x = 1 - 2y
x = 1 - 2[ 8 - 4 sqrt(61) ] / 19
x = [ 19 - 16 + 8 sqrt(61) ] / 19
x = [ 3 + 8 sqrt(61) ] / 19

So your answers are:
( [ 3 - 8 sqrt(61) ] / 19, [ 8 + 4 sqrt(61) ] / 19)
and
( [ 3 + 8 sqrt(61) ] / 19, [ 8 - 4 sqrt(61) ] / 19)

In decimal approximations the answers are:
( -3.1306, 2.0653 )
and
( 3.4464, -1.2232 )

The attached graph shows the two equations and the intersection points, if that helps confirm the answers.

Answer 2

2x + 3y = 0 x - 3y = 9. Use removal technique. upload the two equation to a minimum of one yet another. 3y and - 3y cancel so which you would be able to relatively remedy for x. 3x = 9. x = 3. Now replace x = 3 in the two unique equation and remedy for y. 2x + 3y = 0. 2(3) + 3y = 0. 6 + 3y = 0. Subtract 6 from the two area of the equation. 3y = - 6. Divide the two facets via 3. y = - 2. solutions: x = 3, y = - 2 or as ordered pairs (3, - 2)

Answer 3

Substitution

Solve for x: x=1-2y then substitute for x in other 4(1-2y)^2 + 3y^2 =52 then expand 4(1-4y+4y^2) +3y^2= 52 then collect like terms 19y^2 - 16y +4=52 subtract so 19y^2-16y-48 =0 then factor or use quadratic formula so y=(8+/- 4sqrt(61)) / 19 and x=1-2y

Answer 4

x+2y=1
1 = -1
equall:x+2y-1

4x^2+3y^2=52
-52=-52
equal: 4x^2+3y^2-50

Answer 5

Use substitution.

x + 2y = 1
x = 1 - 2y

4x^2 + 3y^2 = 52
4(1 - 2y)^2 + 3y^2 = 52
4(1 - 4y + 4y^2) + 3y^2 = 52
4 - 16y + 16y^2 + 3y^2 = 52
19y^2 - 16y - 48 = 0

y = {16 ± √[16^2 - 4(19)(-48)]}/[(2)(19)
y = {16 ± √[256 + 3648]}/38
y = {16 ± √3904}/38
y = {16 ± 8√61}/38
y = {8 ± 4√61}/19
y = -1.223, 2.065

Answer 6

1.Get both to equal 0 (algebraically)
2. Solve either one for x or y
3. Sub the answer from step 3 into the other equation to get an answer for x or y (whichever one you didn't find in step 2)
4. Sub the answer from step 3 into either equation to find the other unknown value.
By looking at the two equations, your solution set should consist of two answers, since one is a conic and the other is a line. Good luck!

Answer 7

x + 2y = 1
4x^2 + 3y^2 = 52

x + 2y = 1
x = -2y + 1

4(1 - 2y)^2 + 3y^2 = 52
4((1 - 2y)(1 - 2y)) + 3y^2 = 52
4(1 - 2y - 2y + 4y^2) + 3y^2 = 52
4(1 - 4y + 4y^2) + 3y^2 = 52
4 - 16y + 16y^2 + 3y^2 = 52
19y^2 - 16y + 4 = 52
19y^2 - 16y - 48 = 0

y = (-(-16) ± sqrt((-16)^2 - 4(19)(-48)))/(2(19))
y = (16 ± sqrt(256 + 3648))/38
y = (16 ± sqrt(3904))/38
y = (16 ± sqrt(64 * 61))/38
y = (16 ± 8sqrt(61))/38
y = (1/19)(8 ± 4sqrt(61))

x = ((-2/19)(8 ± 4sqrt(61)) + 1

x = (-16/19) ± (-8/19)sqrt(61) + 1
x = (1/19)(3 ± 8sqrt(61))

ANS :
x = (4/19)(2 + sqrt(61))
y = (1/19)(3 - 8sqrt(61))
or
x = (4/19)(2 - sqrt(61))
y = (1/19)(3 + 8sqrt(61))

Answer 8

x = 1 - 2y

replacing x in the 2nd equation, we get
4(1-2y)^2 + 3y^2 = 52
4(1-4y+4y^2)+3y^2=52
4 - 16y + 4y^2 + 3y^2 = 52
7y^2 - 16y - 48 = 0
y^2 - 16/7 y - 48/7 = 0
y - 16/7 y + (8/7)^2 - (8/7)^2 - 48/7 = 0
(y - 8/7)^2 - (8/7)^2 - 48/7 = 0
....solve for y then replace in x=1-2y and get x

Answer 9

x+2y=1 .............(1)
4x^2+3y^2=52...(2)

from(1),y=(1-x)/2
substitute this value into (2)
4x^2+3((1-x)/2)^2=52
multiply through by 4
16x^2+3(1-x)^2=208
16x^2+3-6x+3x^2=208
19x^2-6x-205=0
using the quadratic formula
x={6+or-sqrt(36+15580)}/38
={6 +or-sqrt(64*244)}/2
={6+or-16sqrt(61)}/38
={3+or-8sqrt(61)/19
=3.446420916 or -3.130631422

when x=(3+8sqrt(61)/19
=3.446420916,
y=(1-(3+8sqrt(61))/19)/2
=(16-8sqrt(61))/38
=(8-4sqrt(61))/19
= -1.223210458
when x=(3-8sqrt(61)/19
= -3.130631422,
y=(1-(3-8sqrt(61))/19/2
=(16+8sqrt(61))/38
=(8+4sqrt(61))/19
= 2.065315721

i hope that this helps

Answer 10

(x+2y=1
(4x²+3y²=52

x = 1 - 2y
Substitute:
4(1 - 2y) + 3y² = 52
4 - 8y + 3y² = 52
3y² - 8y + 4 - 52 = 0
3y² - 8y - 48 = 0

delta = -8² - 4.3.-48
delta = 64 + 576 = 640
y = (8 +/-\/640) : 2.3
y' = (8 + 25,3) : 6 = 5,55
y" = (8 - 25,3) : 6 = -2,8333...

x = 1 - 2y
x = 1 - 2(5,55)
x' = 1 - 11,1 = -10,1 or
x" = 1 - 2(-2,833)
x" = 1 + 5,66 = 6,66
Solution: {x, y E R | x' = -10,1; x' = 6,66 and y' = 5,55; y = -2,833...}
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