Question

AX-9=0 (A-5)X=A-5 (A+7)(A-1)X=A-1

Answer 1

Ответ:

Объяснение:

AX-9=0

X=9/A

(A-5)*X=A-5

X= A-5/A-5

X=1

(A+7)(A-1)*X=A-1

X=A-1/(A+7)(A-1)

X=1/A+7

Answer 2

To solve these equations, we need to isolate the variable "X" on one side of the equation.

Starting with the first equation:

AX - 9 = 0

Adding 9 to both sides, we get:

AX = 9

Dividing both sides by A, we get:

X = 9/A

Now, moving on to the second equation:

(A - 5)X = A - 5

Substituting X with 9/A, we get:

(A - 5)(9/A) = A - 5

Expanding the left side of the equation:

9 - 45/A = A - 5

Multiplying both sides by A, we get:

9A - 45 = A^2 - 5A

Moving all the terms to one side:

A^2 - 14A + 45 = 0

Factoring the equation:

(A - 9)(A - 5) = 0

Therefore, A = 9 or A = 5.

If A = 9, then X = 1.

If A = 5, then the equation (A - 5)X = A - 5 becomes 0X = 0, which is true for any value of X.

Finally, let's solve the third equation:

(A + 7)(A - 1)X = A - 1

Substituting X with 9/A, we get:

(A + 7)(A - 1)(9/A) = A - 1

Expanding the left side of the equation:

9(A^2 + 6A - 7) = A - 1

Multiplying both sides by A:

9A^2 + 54A - 63 = A - 1

Moving all the terms to one side:

9A^2 + 53A - 62 = 0

Factoring the equation:

(9A - 2)(A + 31) = 0

Therefore, A = 2/9 or A = -31.

If A = 2/9, then X = 81/2.

If A = -31, then the equation (A + 7)(A - 1)X = A - 1 becomes 0X = 0, which is true for any value of X.

So the solutions for the given system of equations are:

A = 9, X = 1 A = 5, X can be any value A = 2/9, X = 81/2 A = -31, X can be any value.