Current location - Training Enrollment Network - Mathematics courses - The solution of the math problem in the second volume of the seventh grade
The solution of the math problem in the second volume of the seventh grade
1, solution: Because the solution of equation group ① ② is also the solution of ③, ③ is solved first.

Because 2x+y=28, y=2x, put y=2x into ③, so 2x+2x=28, so 4x=28, so x=7, because y=2x, so y=2* (multiply) 7= 14. Because equations ① ② and ③ ④ have the same solution, their x and y are also equal. Substitute x=7 and y= 14 into ① to get 7+ 14=5k, so k=2 1/5, and substitute k=2 1/5 into (2) to get.

2. Solution: Because 3x+5x = m+2 1x5x+3y = 3m=2, and X and Y are reciprocal, so x+y=0, so ① is 8x=m+2, ② is 2x+3x+3y=m, so 3x+3y=0, so 2x =

Variant 1: because a=b and x=y, ③ is 4x+3y=7x=6-k, ④ is -x=k, so 7x=-7k, ③ is 7x=6-k, so-7k = 6-k.

Variant 2: Solve ⑤ and ⑦ first, because the solutions of the equations are the same, so 3x-y=5, 2x-3y+4=0. Because 3x-y=5, y=3x-5 is substituted into 2x-3y+4=0, so 2x-3 (3x-5)+. So y=22/7. Substitute x and y into ⑧ and ⑧, and you get a= 19/ 14, and b = 211(in detail, you should be able to solve it, right? )

3. Solution: No equation meets the meaning of the question, because it can be seen from ① that x=2 and y=2 are common solutions, so x=y, while ② and ③ contradict ①, so no equation meets the meaning of the question!