Help me solve a system of binary equations, a system of binary equations, who can help me solve it?

3(x-1)=y+5 simplified: y=-5+3x-3=3x-8 (one) 5(y-1)=3(x+5) simplified: 5y-5=3x+15 simplified: 5y=3x+20 (two).

1 formula) * 5 = (2 formulas) so get: 5 (3x-8) = 3x + 20 15x-40 = 3x + 20 12x = 60 from this x=5

Combine x = 5 generations to get y=15-8=7

So x=5 y=7

3x-3=y+5

3x+15=5y-5

The synthesis yields 4y=28

Obtain y=7 and substitute x=5

The addition of 12 formulas simplifies y, and substituting x

x=6，y=4

5x+3y=42 ①

x+y=10 ②

From x=10 y

Substitute x=10 y

Get 5 (10 y) 3y = 42

y=4 substitute y=4 into x=10 y

x=6x=6, y=4

2x+2y=80

xy=750

x+y=40

x-y) Defeat cover inspection trouble=(Yu Da x+y) -4xy=40 -4 750=-1400 0 No solution.

a=150,b=

Subtract the two formulas to give b=; Substitution yields a=150

The one on the first floor seems to have miscalculated?!! b=2/3，a=700/3

Solution: a+b=-3 2

a*b=-7/2

a=-3/2-b

b(-3/2-b)=-7/2

b2+3/2b-7/2=0

2b2+3b-7=0

a=(-3+root65)4,b=(-3-root:65)4,a=(-3-root:65)4,b=(-3+root:65)4,

a=(-3+root65)4,b=(-3-root:65)4,a=(-3-root:65)4,b=(-3+root:65)4,

Misunderstanding in the range of real numbers.

The above system of equations can be deformed as: 2a 2-3a+7=0 because (-3) 2-4*2*7=-47<0, so there is no solution in the range of real numbers.

However, there is a solution in the imaginary range.

It should be provided for i 2 = -1;1) = i can be used to find the root formula.

The first equation touches x=()y

Then substitute equation 2 to solve one.

Then substitute the first formula and pull it.

This equation is generational and generational. I don't count the data, I didn't learn math well in primary school...

The addition of the two formulas yields 60 y=3

Therefore, y=30 can be solved, then x=25

From the formula (1) to get a = b + 3, and then substitute this formula into (2) to get: 16 (b + 3) + 4b - 3 = 5, 20b = 56, so b = then a =

Simplifying 2 first, we get 4a+b=2 b=2-4a.Substituting b=2-4a into equation 1 gives a-2+4a-3=0, a=1, substituting a=1 into equation 1, there is 1-b-3=0, and b=-2