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IrinaK [193]
3 years ago
9

3(x – 2) = 22 – x How many solutions exist for the equation?

Mathematics
2 answers:
Naily [24]3 years ago
7 0
3(x - 2) = 22 - x

First, we need to expand the problem (remove parentheses). The 3 outside will need to multiply into each term inside the parentheses. 
3x - 6 = 22 - x

Second, since our goal is to get the variable on one side by itself, we can start by adding 6 to each side. 
3x = 22 - x + 6

Third, now we need to simplify the right side of the problem. 22 - x + 6, don't look at the variable in between the 22 and 6, and just add those two numbers. This will allow the solution to 22 + 6 to be subtracted to x.
3x = 28 - x

Fourth, knowing that we have to get the variable (x) by itself on one side, we can continue by adding 'x' to both sides, which will flip the problem around.
3x + x = 28

Fifth, we can simplify 3x + x now. Basically, the variable by itself is considered a (1). So, now we can look at it as 3x + 1, which equals 4x.
4x = 28

Sixth, still focusing on the goal of getting the variable by itself, we can remove the 4 from the x by dividing both sides by 4. This will create a fraction.
x =  \frac{28}{4}

Seventh, ask yourself, what times 4 equals 28? Another way to look at 28/4 is (28 \div 4 , anyway you put it, your answer should be 7. Now we know that there is only one solution for this problem.
x = 7

Answer: \fbox {x = 7}
matrenka [14]3 years ago
3 0

Answer:

infinitely many

Step-by-step explanation:

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A manufacturer determines that the number of toys it can sell is given by the formula
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Answer:

The price where the manufacture sells the maximum number of toys is $20

Step-by-step explanation:

The given equation for that represents the number of toys the manufacturer can sell is given as follows;

T = -4·p² + 160·p - 305

Where;

p = The price of the toys in dollars

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Therefore, at the maximum point;

The slope = 0 = dT/dp = d(-4·p² + 160·p - 305)/dp = -8·p + 160

∴ -8·p + 160 = 0

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Answer:

Follows are the solution to the given point:

Step-by-step explanation:

In point a:

¬∃y∃xP (x, y)  

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In point b:

¬∀x∃yP (x, y)

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In point d:

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In point e:

¬∃y(∀x∃zT (x, y, z) ∨ ∃x∀zU (x, y, z))

∀y(∃x ∀z)>T(x,y,z) \wedge ∀x ∃z> V (x,y,z))  

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