Solve the system of equations 5 x 6 y 8,1 x

5/x+6/y=8（1）

1/x-3/y=-11（2）

1), (2) multiply the two formulas by xy.

5y+6x=8xy(3)

y-3x=-11xy(4)

4) Multiply by 2 to get 2y-6x=-22xy(5).

5) + (3) gives 7y = -14xy

So x= 1 2(6).

Bring (6) into (4).

y=1/3

Summary. Solve the system of equations x: y = 5:6, (x 2): y = 7: 9x 5y 6 substituting 5y 6-2 7y 915y 18-14y 18 2y 36x 5 6 36 30

Solve the system of equations x:y=5:6,(x2):

y=7:9 to solve the equation without jujube base group x:y=5:

6, (x 2): y=7: 9x 5y 6 substitution 5y 6-2 7y 915y 18-14y 18 2y 36x 5 6 36 30 Kujin Iwano

The first time a rope is used 1 4, the second time it is used 15 meters, and the ratio of the number of meters used to the remaining meters is 2:3 How many meters the rope is originally long.

Let the original length be x meters (x 4+15) :(x-x 4-15) 2: 33x 4+45 3x 2-30

15＝3x/4

75＝3x/4

The length of the rope is 100 meters.

15 and 75 are always equal to 3x 4

Sorry, the one above was written incorrectly.

The bottom was changed immediately.

75=3x/4

2(x-y)/3

x+y/4=-1

1)6(x+y)-4(2x-y)=16

2) Solution: 1) Multiply both sides of the equation by 12.

8(x-y)-3(x+y)=-12

Mergers of the same kind.

5x-11y+12=0

3) (2) Consolidation of similar items.

10y-2x-16=0

4) (3) * 2 + (4) * 5.

28y-56=0

So y=2 substituting y=2 into (1) gives x=2

x+y=-1 1）

x2+4y2=8 2）

1) Quarrel with the spine stupidity, get y=-1-x

Substitution 2), get.

x^2+4(-1-x)^2=8

5x^2+8x-4=0

5x-2)(x+2)=0

x1=2/5,x2=-2

y1=-7/5,y2=1

x -y =5(x+y) (1) x +xy+y =43 (2) solution: obtained from (1).

x+y)(x-y-5)=0

x=-y x=y-5

Substituting x=-y into (2) to get it.

y²=43y=√43 y=-√43

x=- 43 x= 43y= 43 y=- 43 substitute x=y-5 into (2).

y-5)²+y(y-5)+y²=43

3y²-15y-18=0

y²-5y-6=0

y-6)(y+1)=0

y=6 y=-1

x=1 x=-6

y=6 y=-1

x=-√43 x=√43 x=1 x=-6

y=√43 y=-√43 y=6 y=-1

x^2-5x=y^2+5y

x-5/2)^2=(y+5/2)^2

x-5 2=y+5 2 or x-5 2=-y-5 2

x=y+5 or x=-y

y 2+10y+25+y 2+5y+y 2-43=0 or y 2-y 2+y 2=43

3y^2+15y-18=0

y^2+5y-6=0

y=1 or y=-6 or y=root43 or - root43

x=6 or x=-1 x=-root number 43 or root number 43

So x=6 x=-1 x=-root43 x=root43

y=1 y=-6 y=root43 y=- root43

x -y =5(x+y) (x-y)(x+y)-5(x+y)=0 (x+y)(x-y-5)=0 x=-y or x=y+5

x²+xy+y²=43

x+y)²-xy=43

Substituting x=-y into it, we get:

0+y²=43 y=±√43

x = 43 or x = 5 43

Substituting x=y+5 yields:

2y+5)²-y²-5y=43

3y²+15y-18=0

y²+5y-6=0

y+6)(y-1)=0

y=-6 or y=1

x=6 or x=-1

(x+y)(x-y)=5(x+y)

x-y=5 into the second equation.

Got it.

5/9*x+4/9*y=50

3/5*x+2/5*y=

1) Multiply both sides of the equation by 9, 5x+4y=450

2) Multiply both sides of the equation by 5, 3x+2y=243

3)-(4)×2,-x=-36

x=36 substituting x=36 into (3) to get y=

The solution of the system of equations is x=36,y=

The question should be: 6x x -1 +5x x-1 = x+4 x+1 ?

4x²+8x+4=0

x1=x2=﹣1

When x = 1, x+1 x-1 = 0

x1=x2=1 is the added root of the original fractional equation, and the original equation has no solution.