A number is six greater than 12 times another number the sum of the two numbers is 214 find both numbers

1 answer:

6 0

Let x and y represent the two numbers.
.. x = 6 +12y . . . . . one number is 6 greater than 12 times another
.. x + y = 214 . . . . their total is 214

Substitute for x in the second equation using the first equation.
.. (6 +12y) +y = 214
.. 13y = 208 . . . . . . . . simplify and subtract 6
.. y = 16 . . . . . . . . . . . divide by 13
.. x = 214 -y = 198

The numbers are 198 and 16.

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These are solved easily any of several ways on a graphing calculator. You should learn to use yours for that purpose.

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7 0

3 years ago

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Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S

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Recall that a relation is an equivalence relation if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

Reflexive: We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that $A=J^{-1}AJ$ . Thus, A↔A.

Symmetric: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that $A=P^{-1}BP$ . In this equality we can perform a right multiplication by $P^{-1}$ and obtain $AP^{-1} =P^{-1}B$ . Then, in the obtained equality we perform a left multiplication by P and get $PAP^{-1} =B$ . If we write $Q=P^{-1}$ and $Q^{-1} = P$ we have $B = Q^{-1}AQ$ . Thus, B↔A.

Transitive: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have $A=P^{-1}BP$ and from B↔C we have $B=Q^{-1}CQ$ . Now, if we substitute the last equality into the first one we get