The problem of solving the one variable linear equation which is a little difficult in the sixth grade Similar to x + 12 = x (4 * 360%), about 50 channels, In addition, I only need formula problems, exercises, not application problems

The problem of solving the one variable linear equation which is a little difficult in the sixth grade Similar to x + 12 = x (4 * 360%), about 50 channels, In addition, I only need formula problems, exercises, not application problems

2(x-2)-3(4x-1)=9(1-x) 11x+64-2x=100-9x 15-(8-5x)=7x+(4-3x) 3(x-7)-2[9-4(2-x)]=22 3/2[2/3(1/4x-1)-2]-x=2 2(x-2)+2=x+1 0.4(x-0.2)+1.5=0.7x-0.38 30x-10(10-x)=100 4(x+2)=5(x-2) 120-4(x+5)=25 15x+863-65x=5...

In the sixth grade, the problem of one variable linear equation,
If the root of the equation 6N + 4Y = 7y-3m about y is 1, then the relation between M and N is ()
A.m+2n=1 B.m+2n=-1 C.m-2n=1 D.3m+6n=11

A
6n+4y=3y-3m
4y-7y=-3m-6n
-3y=-3m-6n
y =m+2n
And because y = 1
So m + 2n = 1

One variable linear equation in Grade 6
1. A train starts from a certain station at the speed of 48 kilometers per hour, and then leaves a bus from the same station in the same direction 50 minutes later. It is known that the speed of the bus is one and one sixth of that of the train. How long does it take for the train to catch up with the train?

The distance between bus and coach is s = V, t = 48km / h * 5 / 6h = 40km
The speed of the bus relative to the train v = (7 / 6-1) v = 8km / h
The time for the bus to catch up with the train is 40km / (8km / h) = 5h
The initial distance between the train and the bus is 48 × 50 / 60 = 40 km
The speed of the bus is 48 × (1 + 1 / 6) = 48 × 7 / 6 = 56 km / h
The time to catch up with the train is: 40 △ 56 = 5 / 7 hours