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Morgarella [4.7K]
2 years ago
11

Solve this system of equations by graphing y=5/3x+3 y=1/3x-3

Mathematics
1 answer:
Thepotemich [5.8K]2 years ago
4 0

The solution to the system of equation is x = 9 /2 and y = - 3 / 2

<h3>Simultaneous equation</h3>

y = 5 / 3 x + 3

y = 1 / 3 x - 3

Therefore, let subtract equation(i) from equation (ii) to eliminate y

5 / 3 x - 1 / 3 x = 3 + 3

5x - x / 3 = 6

4x / 3 = 6

cross multiply

18 = 4x

x = 18 / 4

x = 9 / 2

Therefore,

y = 1 / 3(9 / 2) - 3

y = 9 / 6 - 3

y = 9 - 18 / 6

y = - 9 / 6

y = - 3 / 2

learn more on system of equation here: brainly.com/question/3060334?referrer=searchResults

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sergeinik [125]

Answer:

48km

5miles = 8km

×6 = ×6

30miles= 48km

6 0
3 years ago
Find the y intercept of the straight line with the following equations y=7+3x <br><br>3y=6x+12
neonofarm [45]

y = 7 + 3x

y = 3x + 7

y = mx + b, therefore the y-intercept is 7.


3y =  6x + 12

y = 2x + 4

y = mx + b, therefore the y-intercept is 4.

8 0
3 years ago
Dionne builds a pattern that is represented by the equation b = 2f + 6, where f is the figure number of the pattern and b is the
GenaCL600 [577]

I don't know i just want some points

Step-by-step explanation:

well thank you anyways hope someone else can help you i am sorry.

7 0
3 years ago
Write the sentence as an equation.<br> 143 more than the product of g and 369 is equal to 141.
DiKsa [7]

Answer: 143>(369g=141)

Step-by-step explanation:

143>(369g=141)

3 0
3 years ago
For each function below, determine whether or not the function is injective and whether or not the function is surjective. (You
Harlamova29_29 [7]

Answer: First, injective means that for every y, there is only one x such f(x) = y, and surjective means that if f is a function that goes from the set {x} to the set {y}, for every element y in the codomain there is at least one element x in the domain such f(x) = y

(a) f:R----> + R given by f(x) = x2 (i guess you written x^{2})

The function is not injective, because f(-2) = f(2), and is surjective, because the codomain is +R, and x^{2} is always a positive real number.

(b) f:N----> + N given by f(n) = n2  (i guess you written n^{2})

As the naturals have no negative numbers, in this case the function is injective, but isn't surjective, because there is no number that when squared is equal to 5, for example.

(c) f: Zx Z → Z given by f(n, k) = n +k:

here the domain is of the form (n,k) and the function is n +k, so the numbers (z,w) and (w,z) return the same value when evaluated in f, then f is not injective. And is easy to see that f is surjective, because in the sum you can reach al the integers on the codomain.

3 0
3 years ago
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