Let x = 1, y = - 1; X = 2, y = 2 of the binary linear equations ax + by = - 2, try to judge whether x = 3, y = 5 are the solutions of the equation

Let x = 1, y = - 1; X = 2, y = 2 of the binary linear equations ax + by = - 2, try to judge whether x = 3, y = 5 are the solutions of the equation

Substituting x = 1, y = - 1 into ax + by = - 2 to get A-B = - 2, substituting x = 2, y = 2 into 2A + 2B = 2, combining the two equations, the solution is a = - 1 / 2, B = 3 / 2, so the original equation becomes - 0.5x + 1.5y = - 2, at this time, substituting x = 3, y = 5 to see if the left and right phases of the equation are not equal

1. It is known that {x = 3, y + 4, {x + 1, y + 5 are the solutions of the equation AX + by = 22. Try to judge whether {x = 5, y = 3 is the solution of the equation
2. Given that the M + N-5 power of the equation 1 / 2x and the 2m-4n + 1 power of the equation 1 / 3Y = 1 are binary linear equations, we can find the values of M and n
be careful:

If x = 3, y = 4, x = 1, y = 5 are substituted into ax + by = 22, 3A + 4B = 22a + 5B = 22, a = 2, B = 4, then when x = 5, y = 3, ax + by = 2 * 5 + 4 * 3 = 22 means that it is the solution of the equation. 2. Binary linear equation, so the exponents of X and y are 1, so m + N-5 = 1 (1) 2m-4n + 1 = 1 (2) (1) * 4 + (2)

The square of square 4 of Y under the root sign of y = - 2 times the root sign of 5 to 5
Solve the equation. Y is the square of Y under the root sign + 4 = - 2 times the root sign 5 to 5?

y/√(y^2+4^2)=-2√5/5【＜0】
So, y < 0
The square of both sides is y ^ 2 / (y ^ 2 + 16) = - 2 √ 5 / 5) ^ 2 = 4 / 5
===> 5y^2=4(y^2+16)
===> 5y^2=4y^2+64
===> y^2=64
===> y=±8
And because y < 0
So, y = - 8

If 2 is a root of x ^ 2-C = 0, what is the constant C? Can you get other roots of this equation?

Substituting x = 2 into the equation, we get the following result:
4-c=0
c=4
The original equation becomes
x^2-4=0
(x-2)(x+2)=0
X = 2 or x = - 2
The other root is x = - 2