"8x+6y=0" and "3x+4y=9" solve this?

Answer 1

frm eqtion (1)
8x+6y=0
8x= -6y
x = -6y/8
x=-3y/4..............(2)
put x= -3y/4 in eqtion
3x+4y=9
3*(-3y/4)+4y=9
-9y/4+4y=9
-9y+16y=36
7y=36
y= 36/7
put y=36/7 in eqtion (2)
x=-3y/4
x==-3/4*36/7
x= -27/7

hence(-27/7 ,36/7)

Answer 2

8x + 6y = 0 and 3x + 4y = 9
8x + 6y = 0
therefore 8x = – 6y
or 4x = – 3y
or 12 x = – 9 y

3x + 4y = 9 ------- eq(2)
multiply the eq. by 4
12 x + 16 y = 36 ----------eq(3)
putting the value of 12 x = – 9 y in the eq. 3
– 9 y + 16 y = 36
or 7 y = 36
or y = 36/7

Putting the value of y in the eq. 4x = – 3y
4 x = – 36 × 3/7
x = – 27/7
so x = – 27/7 and y = 36/7

Answer 3

8x+6y=0==> x= (-6y)/8 =-3/4 y

Substitute this in 3x+4y=9 ,

3(-3/4)y+4y=9 Multiply this equation by 4

=> -9y+16y=36
=> 7y=36
==> y= 36/7

x= -3/4 y => -3/4 * 36/7 => -27/7

(-27/7,36/7)

Answer 4

it incredibly is an Euler-Cauchy DE, so assume a polynomial answer as a results of fact right here: y = x^(n). . . . y' = n x^(n - a million). .. . . y'' = n(n - a million) x^(n - 2) Substitution: x^2 * n * (n - a million) * x^(n - 2) + 3x * n * x^(n - a million) + 5 * x^(n) = 8x n(n - a million) x^(n) + 3 * n * x^(n) + 5 x^(n) = 0 x^n * ( n * (n - a million) + 3n + 5 ) = 0 x^n = 0 ----> ln(x^n) = ln(0) ( no pass ) n^2 - n + 3n + 5 = 0 n^2 + 2n + 5 = 0 n^2 + 2n + a million + 4 = 0 (n + a million)^2 = -4 n + a million = +/- 2i n = -a million +/- 2i. . . a +/- b i standard answer would be as : y(x) = x^(a) * [ C1 * cos(b * ln(x)) + C2 * sin(b * ln(x)) ] y(x) = x^(-a million) * [ C1 * cos(2 * ln(x)) + C2 * sin(2 * ln(x)) ] particular answer: Yp(x) = Ax + B Yp ' (x) = A Yp '' (x) = 0 Subbing back into: x^2 * (0)+ 3x * ( A) + 5 * ( Ax + B ) = 8x 0 + 3xA + 5Ax + 5B = 8x 8Ax + 5B = 8x [ 8A = 8 ] x ----> A = a million B = 0 y(x) = x^(-a million) * [ C1 * cos(2 * ln(x)) + C2 * sin(2 * ln(x)) ] + a million * x y(x) = x^(-a million) * [ C1 * cos(2 * ln(x)) + C2 * sin(2 * ln(x)) ] + x ========= loose to digital mail if have a query

Answer 5

Given Equations,
8x+6y=0 ---- (1)
3x+4y=9 ---- (2)

Multply eq (1) with 4,
32x+24y=0 --- (3)

Multiply eq(2) with 6,
18x+24y=54 --- (4)

Subtract eq(4) from eq(3),
14x=-54 ---> x=-54/14=-27/7

From eq(1), y=-8x/6 = (-8/6)*(-27/7) = 36/7

Result:
x = -27/7
y = 36/7

Answer 6

8x+6y=0 x 3
3x+4y=9 x 8

24x+18y=0
24x+32y=56
On changing the signs,

-14y=-72
y=36/7.
Substituting the value of y in one of the equations, the value of x comes out to be -27/7
ANS:x=-27/7
y=36/7

Answer 7

8x+6y=0 ------ (1)
3x+4y=9 ------ (2)
(1)*2 => 16x+12y=0
(2)*3 => 9x+12y=27
(1)-(2) => 7x= -27
x= -27/7 ------- (3)
Substituting (3) in (2):
=> 3* -27/7 +4y=9
=> 4y=9+81/7=144/7
y=144/28=36/7

Answer 8

8x + 6y =0 => 12x + 9y=0 (1)
and 3x + 4y=9 => 12x + 16y=36 (2)
(2)-(1): 7y=36 => y=36/ 7= 5.14 and x= 3.86

Answer 9

ok.. this problem is not a pretty one for say, substitution, so we have to rely on "Linear combination".. code name for adding and subtracting the two lines..

8x +6y =0

3x +4y = 9

the quest here is to find where on earth do this two lines intersect, or "cross each other"..

we choose to eliminate y, by finding a common multiple. of 6, and 4.. or12....

mulitply top equation by 2.. and get

16x +12y=0
bottom equation by -3 to get

-9x-12y= -27

now combining like terms .. you see the y's will add to zero y...

and get. 7x=-27 or x=-27/7

to solve for y, substitute into origal equation above, say,
8(-27/7) + 6y=0..

solve to get...
-30 6/7 +6y=0.

6y=30 and 6/7dividing by 6 we obtain..

y=35/36

hopefully that helps..

Answer 10

8x+6y=0
3x+4y=9
multiply first eqn with 4 and second eqn with 6,we get
32x+24y=0
18x+24y=54
Therefore 42y will be cancelled and the resultant will be
14x=-54
x=-27/7
substituting the value of x in eqn 1,we get
8*-27/7+6y=0
after calculating further you will get the value of y.