Solve the equation system {2x-4y3y = 1 {4 (X-Y) - 3 (2x + y) = 17

Solve the equation system {2x-4y3y = 1 {4 (X-Y) - 3 (2x + y) = 17

{2x of 3-4y = 1 of 2
8x-9y=6 (1)
{4(x-y)-3(2x+y)=17
-2x-7y=17
-8x-28y=68 (2)
(1)+(2)
-37y=74
y=-2
Y = - 2 into (1)
8x+18=6
x=-3/2
The solution of the equations is x = - 3 / 2, y = - 2

Solving equations Y+1 4＝x+2 Three 2x−3y＝1 ．

The equations are as follows: 1
4x−3y＝−5①
2x−3y＝1② ，
① - 2: 2x = - 6,
That is, x = - 3,
Substituting x = - 3 into ①, we get: y = - 7
3，
Then the solution of the equations is
x＝−3
Y
＝−7
3 ．

To solve the equations 2x−y＝3(1) 3x+y＝7(2)

Solution 1: (1) + (2), get 5x = 10,
Ψ x = 2, (3 points)
Substituting x = 2 into (1) yields 4-y = 3,
ν y = 1, (2 points)
The solution of the system of equations is
x＝2
Y = 1. (1 point)
Solution 2: from (1), y = 2x-3, ③ (1 point)
Substituting ③ into (2) gives 3x + 2x-3 = 7,
Ψ x = 2, (2 points)
Substituting x = 2 into ③ gives y = 1, (2 points)
The solution of the system of equations is
x＝2
Y = 1. (1 point)

The solution of the equation system 3x + y = 30, y-2x = 5 is? Given the equation system 4x + 3Y = 15,2x-y = 17, then the 2012 power of (x + 2Y) =?

y-2x=5
y=5+2x
3x+5+2x=30
5x=25
X=5
y=5+2x5=15
Question two
4x+3y-(2x-y)=15-17
4x+3y-2x+y=-2
2x+4y=-2
x+2y=-1
The 2012 power of (x + 2Y) = (- 1) to the 2012 power = 1

Solving equations: 2x + 3y-z = 30, (x-4): (Y-3): (Z-2) = 1:2:3 I have limited IQ.

(X-4):(Y-3)=1:2
Y-3=2(X-4)
Y=2X-5.(1)
（X-4）:(Z-2)=1:3
Z-2=3(X-4)
Z=3X-10.(2)
Substitute (1) (2) into 2x + 3y-z = 30
2X+3(2X-5)-(3X-10)=30
5X=30+15-10=35
X=7
Y=2*7-5=9
Z=3*7-10=11

The solution of the equation y + 3 x + 3 x + 3 x = 2 x + 3 x = 1 component of the equation

2 / x + 3 / y = 1, multiply 4 and subtract 2x + 3Y = - 1 to get y = - 3 x = 4
25% x + 40% y = 6 × 30%, multiply 8 and subtract 2x + y = 6 to get y = 42 / 11 x = 12 / 11

If x and y are opposite numbers in the solution of the system of equations about X and Y 4x-3y = 7K + 1 / 2x-y = K-3, find the value of K Please give the detailed process thank you

①4x-3y=7
②k+1/2x-y=k-3;
③x=-y;
By substituting formula 3 into Formula 1, we can get: x = 1, y = - 1. Then we can get k = 5 by substituting X and Y into formula ②

If equations 4x−3y＝k If x and y are equal in 2x + 3Y = 5, then K is equal to () A. 1 or - 1 B. 1 C. 5 D. -5

According to the meaning of the title:
4x−3y＝k(1)
2x+3y＝5(2)
x＝y(3) ，
Substituting (3) into (2) gives x = y = 1,
Substituting (1) gives k = 1
Therefore, B

If equations 4x−3y＝k The solution of 2x + 3Y = 5 satisfies x + y ≤ K

4x−3y＝k(1)
2x+3y＝5(2)
① + 2 gives 6x = 5 + K and x = 5 + K
6，
Let x = 5 + K
Substituting 6 into ①, y = 10 − K is obtained
9，
∵ x + y ≤ K, i.e. 5 + K
6+10−k
9≤k，
∴k≥35
17．

If equations 4x−3y＝k If x and y are equal in 2x + 3Y = 5, then K is equal to () A. 1 or - 1 B. 1 C. 5 D. -5

According to the meaning of the title:
4x−3y＝k(1)
2x+3y＝5(2)
x＝y(3) ，
Substituting (3) into (2) gives x = y = 1,
Substituting (1) gives k = 1
Therefore, B