If the solution of the system of equations 2x + 3Y = m, 2x + 5Y = m + 2 satisfies x + y = 12, the value of M is obtained

If the solution of the system of equations 2x + 3Y = m, 2x + 5Y = m + 2 satisfies x + y = 12, the value of M is obtained

By subtracting the two formulas, 2Y = 2
y=1
∴ 2x+3=m
x=(m-3)/2
∵x+y=12
(m-3)/2+1=12
∴m=25

5y-2y = 18 solution equation

3Y=18
Y=6

1.2y-y = 0.965y-3.7 = 8.57a + 3 × 4 = 40

1.2y-y=0.96
0.2y=0.96
y=0.96÷0.2
y=4.8
5y-3.7=8.5
5y=8.5+3.7
5y=12.2
y=12.2÷5
y=2.44
7a+3×4=40
7a+12=40
7a=40-12
7a=28
a=28÷7
a=4

1 / 3 (2y-1) & sup2; = 5Y (2y-1) solution equation
There's a third in front of me

4y²-4y+1=30y²-15y
26y²-11y-1=0
(13y+1)(2y-1)=0
13y + 1 = 0 or 2y-1 = 0
Y = - 1 / 13 or y = 1 / 2

The solution equation is y-4 = 3 / 5Y
2.4-（y-4/2.5）=3/5y

Multiply both sides by 5
12-5(y-4)/2.5=3y
12-2(y-4)=3y
12-2y+8=3y
transposition
3y+2y=12+8
5y=20
y=20÷5
y=4

Solve the equation: 3 / 5Y + 4 + 4 / Y-1 = 2-12 / 5y-5

Multiply both sides by 5
12-5(y-4)/2.5=3y
12-2(y-4)=3y
12-2y+8=3y
transposition
3y+2y=12+8
5y=20
y=20÷5
y=4

Solve the equations: 3x-y = 5,5y-1 = 3x + 5

To change the situation: 3x = 5 + y3x = 5y-1-5; 5 + y = 5y-6y = 11 / 4, we can get xx = 31 / 12

x:y:z=1：2：3 3x+5Y+7z=68

Let x = k, y = 2K, z = 3K; 3x + 5Y + 7z = 68,3k + 10K + 21k = 68,34k = 68, k = 2, x = 2, y = 4, z = 6

Solving equations 3x-y = 5, 5y-5 = 3x-1

The results are as follows: 3x = 5 + y; 3x = 5y-4; subtraction: 9-4y = 0; y = 9 / 4; substituting y = 9 / 4 into 3x = 5 + y; X = 29 / 12. Study hard!

The following equations {3x + 5Y = - 2 (1) 3x-y = 4 (2) are solved by the method of addition and subtraction

① - 2: 6y = - 6
y=-1
Substituting y = - 1 into 2, we get the following result
3x+1=4
3x=3
x=1
So x = 1, y = - 1