solve the simultaneous equation 9y²+8x=12 and 2x+3y=4?

Answer 1

Solve the second equation for x:

2x = 4 - 3y
x = (4 - 3y)/2

Plug that into the first equation:

9y^2 + 8(4 - 3y)/2 = 12
9y^2 + 16 - 12y = 12
9y^2 - 12y + 4 = 0

Use the quadratic equation to solve for y:
y = (-b +/- sqrt(b^2 - 4ac)) / 2a
y = (12 +/- sqrt(144 - 144)) / 18
y = 12 / 18
y = 2/3

x = 1

Answer 2

2x + 3y = 0 x - 3y = 9. Use removing approach. upload the two equation to a minimum of one yet another. 3y and - 3y cancel so which you will fairly 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 edge of the equation. 3y = - 6. Divide the two components by ability of three. y = - 2. solutions: x = 3, y = - 2 or as ordered pairs (3, - 2)

Answer 3

9y²+8x=12 and 2x+3y=4
There is an x¹ term in each equation so I'd isolate those first:
9y²+8x=12   ⇒  8x = 12 - 9y²
2x+3y=4  ⇒  2x = 4 - 3y  ⇒  8x = 16 - 12y

By transitive rule,
12 - 9y² = 16 - 12y
Rearrange to 9y² - 12y + 4 = 0

Solve for y, either by factoring:
9y² - 12y + 4 = (3y-2)²
or using quadratic formula.

Then plug value of y into one of the equations and solve for x.

Answer 4

9y²+8x=12 and 2x+3y=4

2x + 3y = 4
2x = 4 - 3y
x = 2 - 3/2y

9y² + 8(2 - 3/2y) = 12
9y² + 16 - 12y = 12
9y² - 12y = 12 -16
9y² - 12y = -4
9y² - 12y + 4 = 0
(3y - 2)²

3y - 2 = 0
3y = 2
y = 2/3

2x + 3y = 4
2x + 3(2/3) = 4
2x + 2 = 4
2x = 4 -2
2x = 2
x = 2

Therefore x = 1 and y = 2/3

Answer 5

9y²+8x=12 Let this be equation 1
2x+3y=4 Multiply with 4
8x+12y=16 Let this be equation 2

subract equation 1 from 2
9y²+8x=12
8x+12y=16
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9y²-12y=-4
9y²-12y+4=0
9y²-6y-6y+4=0
3y(3y-2)-2(3y-2)=0
(3y-2)(3y-2)=0
so 3y-2=0
3y=2
y=2/3

so when u substitute the value of y in either of the equation
2x+3y=4
2x+3(2/3)=4
2x+2=4
2x=4-2
2x=2
x=2/2
x=1

Answer 6

sub 2x = 4 - 3y

9y^2 + 4(4-3y) - 12 = 0
9y^2 - 12y +4 = 0
(3y-2)^2 = 0
y = 2/3
x = 1