Use the substitution method to solve the system of equations

Work out the size of angle X. Not dra x 164

beks73 [17]

This would be the correct answer

3 0

2 years ago

Can someone help me with these problems? I'm in k12 and this assignment was in classkick, so if anyone already completed this an

kherson [118]

 SOLVING QUESTIONS FROM 1ST PAGE

Given the point B(-5, 0)

y = -3x - 5

Putting x = -5, and y = 0 in y = -3x - 5

0 = -3(-5) - 5

0 = 15 - 5

0 = 10 ∵Putting B(-5, 0) in y = -3x - 5 does not equate the equation.

As L.H.S ≠ R.H.S

So,

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

Given the point C(-2, 1)

y = -3x - 5

Putting x = -2, and y = 1 in y = -3x - 5

1 = -3(-2) - 5

1 = 6 - 5

1 = 1 ∵Putting C(-2, 1) in y = -3x - 5 rightly equates the equation.

As L.H.S = R.H.S

So,

Does this check?

Answer: yes ∵ L.H.S = R.H.S

Given the point D(-1, -2)

y = -3x - 5

Putting x = -1, and y = -2 in y = -3x - 5

-2 = -3(-1) - 5

-2 = 3 - 5

-2 = -2 ∵Putting D(-1, -2) in y = -3x - 5 rightly equates the equation.

As L.H.S = R.H.S

So,

Does this check?

Answer: yes ∵ L.H.S = R.H.S

 SOLVING QUESTIONS FROM 2ND PAGE

Given the point B(6, 18)

y = -2x + 18

Putting x = 6, and y = 18 in y = -2x + 18

18 = -2(6) + 18

18 = -12 + 18

18 = 6 ∵Putting B(6, 18) in y = -2x + 18 does not equate the equation.

As L.H.S ≠ R.H.S

So,

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

Given the point C(9, 24)

y = 2x + 6

Putting x = 9, and y = 24 in y = 2x + 6

24 = 2(9) + 6

24 = 18 + 6

24 = 24 ∵Putting C(9, 24) in y = 2x + 6 rightly equates the equation.

As L.H.S = R.H.S

So,

Does this check?

Answer: yes ∵ L.H.S = R.H.S

 SOLVING QUESTIONS FROM 3RD PAGE

Given the point D(2, 7)

y = 3x + 4

Putting x = 2, and y = 7 in y = 3x + 4

7 = 3(2) + 4

7 = 6 + 4

7 = 10 ∵Putting D(2, 7) in y = 3x + 4 does not equate the equation.

As L.H.S ≠ R.H.S

So,

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

 SOLVING QUESTIONS FROM 4th PAGE

b. Which function could have produced the values in the table.

A. y = 3x + 4 

B. y = -2x + 18

C. y = 2x + 6

D. y = x + 9

The Table:

x y

3 12

6 18

9 24

Checking A) y = 3x + 4

Putting (3, 12), (6, 18) and (9, 24) in y = 3x + 4

For (3, 12)

y = 3x + 4

12 = 3(3) + 4

12 = 13 ∵ L.H.S ≠ R.H.S

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

For (6, 18)

18 = 3(6) + 4

18 = 18 + 4

18 = 22 ∵ L.H.S ≠ R.H.S

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

For (9, 24)

24 = 3(9) + 4

24 = 27 + 4

24 = 31 ∵ L.H.S ≠ R.H.S

Does this check?

Answer: no ∵ L.H.S ≠ R.H.S

So, the equation y = 3x + 4 could have not produced the all values in the table as the the ordered pairs in table do not satisfy(equate) the equation.

Checking B) y = -2x + 18

Putting (3, 12), (6, 18) and (9, 24) in y = -2x + 18

For (3, 12) ⇒ y = -2x + 18 ⇒ 12 = -2(3) + 18 ⇒ 12 = 12 ⇒ L.H.S = R.HS
For (6, 18) ⇒ y = -2x + 18 ⇒ 18 = -2(6) + 18 ⇒ 18 = 6 ⇒ L.H.S ≠ R.HS
For (9, 24) ⇒ y = -2x + 18 ⇒ 24 = -2(9) + 18 ⇒ 18 = 0 ⇒ L.H.S ≠ R.HS

So, y = -2x + 18 could have not produced all the values in the table, as (6, 18) does not equate the equation.

Checking C) y = 2x + 6

Putting (3, 12), (6, 18) and (9, 24) in y = 2x + 6

For (3, 12) ⇒ y = 2x + 6 ⇒ 12 = 2(3) + 6 ⇒ 12 = 12 ⇒ L.H.S = R.HS
For (6, 18) ⇒ y = 2x + 6 ⇒ 18 = 2(6) + 6 ⇒ 18 = 18 ⇒ L.H.S = R.HS
For (9, 24) ⇒ y = 2x + 6 ⇒ 24 = 2(9) + 6 ⇒ 24 = 24 ⇒ L.H.S = R.HS

So, y = 2x + 6 could have produced the values of in the table as all the orders pairs in the table satisfy/equate the equation.

Checking D) y = x + 9

Putting (3, 12), (6, 18) and (9, 24) in y = x + 9

For (3, 12) ⇒ y = x + 9 ⇒ 12 = 3 + 9 ⇒ 12 = 12 ⇒ L.H.S = R.HS
For (6, 18) ⇒ y = x + 9 ⇒ 18 = 6 + 9 ⇒ 18 = 15 ⇒ L.H.S ≠ R.HS
For (9, 24) ⇒ y = x + 9 ⇒ 24 = 9 + 9 ⇒ 24 = 18 ⇒ L.H.S ≠ R.HS

So, y = x + 9 could have also not produced all the values in the table, as (6, 18) and (9, 24) do not satisfy/equate the equation.

So, from all the verification we conclude that:

HENCE, ONLY y = 2x + 6 COULD HAVE PRODUCED THE VALUES IN THE TABLE AS ALL THE ORDERED PAIRS OF THE TABLE SATISFY/EQUATE THE EQUATION.

 SOLVING QUESTIONS FROM 5th PAGE

What is the domain and range of the relation?

{(-3, 7), (6, 2), (5, 1), (-9, -6)}

Domain: Domain is the set of all the x-coordinates of the ordered pairs of the relation, meaning the all first elements of the ordered pairs in a relation include in the domain of the relation.
Range: Range is the set of all the y-coordinate of the ordered pairs of the relation, meaning the all second elements of the ordered pairs in a relation include in the range of the relation.

As the given relation is {(-3, 7), (6, 2), (5, 1), (-9, -6)}

The domain is: {-3, 6, 5, -9}

Note: generally, we write the numbers in ascending order for both the domain and range.

The domain could also be written in order as: {-9, -3, 5, 6}

The range is: {7, 2, 1, -6}

The range could also be written in order as: {-6, 1, 2, 7}

Keywords: equation, point, ordered pair, domain, range

Learn more about points and equation from brainly.com/question/12597810

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8 0

3 years ago

I need help on this equation 6s-10=s

AysviL [449]

The answer is 2
One Solution

6s-10= 1s
-6s -6s
_________
-10=-5s
Divide -10 and -5
You'll get 2
X=1

8 0

3 years ago

Can you find the midpoint of a line?

natali 33 [55]

Answer:

sure.

Step-by-step explanation:

brainliest?

8 0

3 years ago

Read 2 more answers

A square has side length 6x+2. A rectangle with width 3x+1 has the same area as a square. What is the length of the rectangle

o-na [289]

Answer:

12x+4

Step-by-step explanation:

(6x+2)^2=36x^2+24x+4

(36x^2+24x+4)/3x+1=2(6x+2)=12x+4

6 0

3 years ago

Read 2 more answers