Ask for help to solve a simple system of binary equations

y÷3-(x+1)÷6=3

2×(x-y÷2)=3×(x+y÷18)

Multiply 1 by 6 and simplify it to.

2y-(x+1)=18---3）

2 formulas. 2x-y=3x+(y/6)

Weigh the item. 7y/6=x－－－4）

3)-(4) De.

2y-(x+1)-(7y/6)=18-x

2y-x-1+7y/6=18-x

2+(7/6)]y=18+1

19y/6=19

y=6 substituting y=6 back to (4) gives -7y 6=x, x=-7 and the solution of the equation system is x=-7, y=6

y/3+(x+1)/6=3

2(x-y/2)=3(x+y/18)

1)*6, we get 2y+x+1=18(3).

2)*6, to get 12x-6y=18x+y, simplify, to get 6x+7y=0(4).

3)*6-(4), we get 5y=102

So y= x=

Multiply 6 by the two sides of the equation at the same time to get 2y-x=19

Open the parentheses on both sides of the equation to get 2x-y=3x+y 6, and multiply 6 by 6 on both sides.

12x-6y=18x+y can be introduced to solve the equation 2y-x=19 with 6x+7y=0

6x+7y=0

We get x=-7 y=6

<> hello. I use the substitution method to evaluate, first turn the coefficient of an unknown number into 1, and then substitute it into another formula, from a binary equation to a one-dimensional equation, and solve the value of an unknown number into any formula to calculate it, you look at the solution process, if you don't understand, you can ask me, if it helps you, help like it, thank you.

51x-34y=-7 ①

92x-230y=7 ②51-92 gets: -11730y+3128y=357+644-8602y=1001

Solution: y= —1001 8002Substituting y=—1001 8002 or solution: x=—1848 8602

The solution of the original system of equations is: x = —1848 8602, y = —1001 8002

Set two kinds of color TVs to buy x,y sets,then。

x+y=73000x+1300y=15900

Find x=4y=3

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.