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alexgriva [62]
2 years ago
15

How to solve y=-x+3 y=2x-1

Mathematics
2 answers:
svlad2 [7]2 years ago
7 0

Solve for x: 2x - y = 3 x + y = 3?
? can someone explain this for me plz
8 Answers • Mathematics

Best Answer (Chosen by Voter)
Hi,

For this question, we are given the following system of equations:

2x - y = 3
x + y = 3

Let's make the first equation in the form of y = mx + b as shown below:

y = 2x - 3
x + y = 3

Now, substitute the first equation into the second to get:

x + 2x - 3 = 3

Combine similar terms to get:

3x = 6

x = 2

We now know that x = 2 and can plug this value into one of the given equations to get:

2 + y = 3

y = 1

FINAL ANSWER: x = 2 ; y = 1

I hope that helps you out!
scZoUnD [109]2 years ago
4 0
\left \{ {{y=-x+3} \atop {y=2x-1}} \right. \\\\ equal\ both\ equations\\\\
-x+3=2x-1\ \ \ | subtract\ 2x\\\\
-3x+3=-1\ \ \ | subtract\ 3\\\\
-3x=-4\ \ \ | divide\ by\ -3\\\\
x=\frac{4}{3}\\\\
y=-x+3=-\frac{4}{3}+3=-1\frac{1}{3}+3=1\frac{2}{3}
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Answer:

f(x) = 3 \cdot 0.2^x

Step-by-step explanation:

f(x) = 3 \cdot 0.2^x


Domain of f: "What values of x can we plug into this equation?" This makes sense for all real numbers so the domain is \mathbb{R}

Range of f: "What values of f(x) can we get out of the function?" From the graph we see we can get any real number greater than 0 out of the function by choosing a suitable x-value in the domain. The range is therefore (0, \infty).

Continuity: Since the graph is one, unbroken curve (i.e. a curve that can be drawn in one movement without taking your pen off the paper). We see that "roughly speaking" the function is continuous.

Increasing or decreasing behaviour: For all x in the domain, as x increases, f(x) decreases. This means the function exhibits decreasing behaviour.

Symmetry: It is clear to see the graph of f(x) has no symmetry.

Boundedness: Looking at the graph we see it is unbounded above as when we choose negative values, the graph of f(x) explodes upwards exponentially. Choose a value of x, plug it in, next choose (x-1), plug this in and we observe f(x-1) > f(x) for all x in the domain.

The function is however bounded below by 0: no value of x in the domain exists which satisfies f(x) < 0.

Extrema: As far as I can tell, there are no turning points on the curve. (Is this what you mean by extrema?)

Asymptotes: Contrary to the curve's appearance, there are no vertical asymtotes for this curve. The negative-x portion of the curve is just growing so quickly it appears to look like an asymptote. There is a value of f(x) for all x<0. There is however a horizontal asymtote: y=0.

End behaviour: As x \rightarrow \infty, f(x) \rightarrow 0. As x \rightarrow -\infty, f(x) \rightarrow +\infty

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3 years ago
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