Create a system of linear equations that has one solution. Solve the system using substitution to prove your

Mathematics

1 answer:

bezimeni [28]3 years ago

3 0

Answer:

y = -2x + 3

y = -1/2x + 5

Step-by-step explanation:

So the best way to solve is to come up with two slopes, I'd prefer 2 perpendicular slopes because perpendicular lines only cross once hence one solution guaranteed
so say we choose a slopeA = 2 and slopeB = -1/2, the slopes are perpendicular only if when you multiply them you get -1
then we use the slope intercept form of an equation y = mx + b
we make up two equations y = -2x + b and y = -1/2x + b , now we can just make up number for b so the two equations are
y = -2x + 3
y = -1/2x + 5
Then we solve them by substituting for y so
-1/2x + 5 = -2x + 3 and x = -4/3
and then we put this to get y in either equation
y = -2(-4/3) + 3
y = 17/3
So the equations work

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What are the domain and range of the function f(x) = 3x 5?.

Bingel [31]

The domain and range of the given function both are the real numbers.

<h2>Given that:</h2>

The domain and range of function f(x) = 3x + 5 has to be found.

<h2>Explanation for domain and range:</h2>

Considering working in real numbers, domain is $\mathbb{R}$ and range is $\mathbb{R}$ too.

Domain can be complex numbers too, but for the sake of daily life cases, we consider working in real numbers.

Thinking of it as equation of straight line can help.

The given function is monotonically increasing and continuous.

Thus Range can be calculated as interval ( f(min value of domain), f(max value of domain) ) which gives us $\mathbb{R}$ as range.

Below is the plot of f(x) = 3x + 5 in real number plane.

In fact, any linear equation of the form

$y = mx + c ; \: \: c \in R \: and \: m \in {R-\{0\}}$