Answer:
btfrt34598fc
Step-by-step explanation:
Answer:
Step-by-step explanation:
REcall the following definition of induced operation.
Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.
So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.
For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).
Case SL2(R):
Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So
.
So AB is also in SL2(R).
Case GL2(R):
Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So
.
So AB is also in GL2(R).
With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).
-3(x + 5) = -9
-3x + -15 = -9
-3x - 15 = -9
-3x - 15 + 15 = -9 + 15 (we added 15 to both sides of the equation because now, the 15 on the left side cancels out).
-3x = 6
-3x/-3 = 6/-3
X = -2
So, in conclusion, X is equal to -2.
5x = 6x-7
subtract 6x from each side
-x=-7
x=7
y = 4x - 3 would be it !!!!