Solution of binary linear equation: 1, addition and subtraction and elimination: solution of equation with the same coefficient, such as 2x+y= 1 and 2x-3y=4: 2x+y= 1①, 2x-3y = 4②-②, and the solution is y =-3. 2x-5y= 10② is obtained from ①, x=8-y, ③ substitutes ③ into ②, and obtains 2(8-y)-5y= 10, and y=6/7④ substitutes ④ into ①. X=50/7 (the method of addition, subtraction and elimination is limited to equations with unknown coefficient of 1) 3. Substitution method, such as: (x+5)+(y-4)=8 (x+5)-(y-4)=4 ling X+5 = m+n=8 m-n=4 = N. 0? Equation with 5=3 (equation with b equal to 0) (Note: When solving the equation by formula method, whether δ is ≥0 should be calculated, otherwise there is no solution) 2. Formula method (all quadratic equations with one variable can be solved in junior high school) 3. Matching method: the form is x 2+2x-3 = 0. For example, solve the equation: x 2+2x-3 = 0. Solution: Shift the constant term to: x 2+2x = 3. Add 1 to both sides of the equation. 2 = 4 solution: x 1=-3, x2= 1 (all equations can be solved in junior high school) 4. Factorization method: shape: 2x 2+3x = 0 Solution: 2x 2+3x = 0x (2x+3) = 0 (using common factor method, (2) Solution: 6x2+5x-50 = 0 (2x-5) (3x+10) = 0 (in cross multiplication) 0? 5-4(3x+5)+3=0 Solution: Let y=3x+5, then y? 0? 5-4y+3=0 factorization, (y-1) (y-3) = 0 ∴ y1=1,y2=3 ∴3x+5= 1 or 3 ∴.