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3 years ago

How many solutions does the equation below have. The answer choices are below

Mathematics

2 answers:

Schach [20]3 years ago

8 0

Answer:

The answer is <u>one solution</u>

Step-by-step explanation:

1)2(b+3)-17 = 3b-7+b

2)2b+6-17 = 4b-7

3)2b-11 = 4b-7

4)-11+7 = 4b-2b

5)-4 = 2b

6)2b = -4

7)b = -4/2

Final your answer is b = -2

Natali [406]3 years ago

5 0

<h3>Answer: D) One solution</h3>

==========================================

Work Shown:

2(b+3)-17 = 3b-7+b

2b+6-17 = 4b-7

2b-11 = 4b-7

-11+7 = 4b-2b

-4 = 2b

2b = -4

b = -4/2

b = -2

There is one solution and it is b = -2

A quick way to tell we have one solution is to note that both sides are a linear expression, and that the slopes of each linear expression are different values. If you had two lines with the same slope, then you'd have either parallel lines or coinciding lines (leading to no solutions and infinitely many solutions respectively). When you have two lines with two different slopes, then they are guaranteed to intersect only once. That intersection point is the solution.

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Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

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$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

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$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

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3 years ago

How many solutions does the equation below have. *The answer choices are below*

How many solutions does the equation below have. The answer choices are below