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3 years ago

Which ordered pair is the solution to the system of equations?

Mathematics

2 answers:

Tcecarenko [31]3 years ago

6 0

The best way to answer this item is to use the Substitution method. Substitute the value of y from the first equation to the y of the second equation such that the second equation becomes,

-4x + 3(x - 4) = -3

Simplifying the equation,
-x = 9

Dividing both sides by -1 gives an answer of,
<em>x = -9</em>

Then, substitute the value of x in the first equation.
y = -9 - 4
<em> y = -13</em>

The answer to this item is letter B.

lys-0071 [83]3 years ago

4 0

Hope this helped...

MARK ME BRAINIEST PLEASE

Answer:

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Answer:

The statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ is a contingency.

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Step-by-step explanation:

A tautology is a proposition that is always true.

A contradiction is a proposition that is always false.

A contingency is a proposition that is neither a tautology nor a contradiction.

a) To classify the statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ , you need to use the logic laws as follows:

$(A\rightarrow \lnot B)\land (B\rightarrow A) \equiv$

$\equiv (\lnot A \lor\lnot B)\land(\lnot B \lor A)$ by the logical equivalence involving conditional statement.

$\equiv (\lnot B\lor \lnot A )\land(\lnot B \lor A)$ by the Commutative law.

$\equiv \lnot B \lor (\lnot A \land A)$ by Distributive law.

$\equiv \lnot B \lor (A \land \lnot A)$ by the Commutative law.

$\equiv \lnot B \lor F$ by the Negation law.

Therefore the statement $(A\rightarrow \lnot B)\land (B\rightarrow A)$ is a contingency.

b) To classify the statement $(P\rightarrow \lnot P)\land P$ , you need to use the logic laws as follows:

$(P\rightarrow \lnot P)\land P \equiv$

$\equiv (\lnot P \lor \lnot P)\land P$ by the logical equivalence involving conditional statement.

$\equiv P \land (\lnot P \lor \lnot P)$ by the Commutative law.

$\equiv (P \land \lnot P) \lor (P \land \lnot P)$ by Distributive law.

$\equiv F \lor F \equiv F$ by the Negation law.

Therefore the statement $(P\rightarrow \lnot P)\land P$ is a contradiction.

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