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3 years ago

Solve the system by substitution. y = 2x 3x + 2y = 21

2 answers:

soldier1979 [14.2K]3 years ago

6 0

$\left \{ {{y=2x} \atop {3x+2y=21}} \right \longrightarrow \left \{ {{y=2x} \atop {3x+2\cdot2x=21}} \right. \longrightarrow \left \{ {{y=2x} \atop {7x=21}} \right.\longrightarrow \left \{ {{y=2x} \atop \boxed{{x=3}}} \right. \\\\y=2x\longrightarrow y=2\cdot3\longrightarrow \boxed{y=6}$

Alexandra [31]3 years ago

6 0

Y=2x;3x+2y=21

Solve y=2x for y

Substitute (2x) for y in 3x+2y=21

3x+2y=21

3x+22x=21

7x=21 (Simplify)

7x / 7 = 21 / 7(Divide both sides by 7)

x=3

Substitute (3) for x in y=2x:

y=2x

y=(2)(3)

y=6 (Simplified)

y=6 and x=3

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Reorder the numbers and combine like terms to simplify. 4.5 + 2 1/2 + 6.0 - 5 1/2

DanielleElmas [232]

Answer: B

Step-by-step explanation:

4.5+6=10.5

2 1/2 - 5 1/2 = 2.5 - 5.5 = -3

10.5 - 3 = 7.5

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3 years ago

Cole paid $24 for 2 calculators and a pack of pens. Kai paid $45 for 3 calculators and 2 packs of pens. They both paid $6 for ea

s344n2d4d5 [400]

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4 years ago

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The population of Sunville increases by

UkoKoshka [18]

Answer: I think its 30000

Step-by-step explanation:

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3 years ago

Question 1,2, and 3 how do i factor those? Can you show the work and explain how?

DiKsa [7]

1: $3n^{2}+9n+6$

notice that each part is divisible by 3

$3n^{2}$ ÷ 3 = $n^{2}$

9n ÷ 3 = 3n

6 ÷ 3 = 2

so it becomes $3(n^{2} +3n+2)$

3n can be rewritten as 2n+n

-you want to rewrite it into two numbers that multiply to the number that's alone (in this case 2)

which would get you

$3(n^{2} +2n+n+2)$

Now that it's rewritten, you can factor out n + 2 from the equation.

the answer is

3(n+2)(n+1)

And you can check that by multiplying (n+2)(n+1) which is $n^{2} +2n+n+2$ and then each of those by 3, which is $3n^{2} +6n+3n+6$ or $3n^{2}+9n+6$ , our origional equation

2: $28+x^{2} -11x$

So I rewrote this as $x^{2} -11x+28$ (it's the same thing, just reordered using the commutative property)

now -11x can be rewritten as -4x-7x

(remember, the two numbers should multiply to equal 28, which is our constant.)

$x^{2} -4x-7x+28$

now we can factor out x from the first expression and -7 from the second

$x(x-4)-7(x-4)$

and lastly you factor out x-4,

which would give you

(x-4)(x-7)

Make sure to check your work and make sure it multiplies to $x^{2} -11x+28$

3: $9x^{2} -12x+4$

The first thing I notice when looking at this problem is that both 9 and 4 are perfect squares. Not only that, but they are the squares of 2 and 3, which are products of -12

So if you rewrite 9 as $3^{2}$ and 4 as $2^{2}$ , the equation becomes

$3^{2} x^{2} -12x+2^{2}$

now that $3^{2} x^{2}$ is ugly so it can be turned into $(3x)^{2}$

and -12x can be rewritten as $-2*3x*2$

so our equation now looks like $(3x)^2-2*3x*2+2^{2}$

There's a rule that says $a^{2} -2ab+b^{2} = (a-b)^{2}$

In our case, a=3x and b=2

so the final answer is

$(3x-2)^2$

5 0

3 years ago

A rule for the nth term of 22,9,-4,-17,-30

fenix001 [56]

Answer: In Exercises 21 and 22, describe and correct the error in writing a rule for the nth term of the arithmetic sequence 22, 9, −4, −17, −30, . . ..

3. 1, −1, −3, −5, −7, . . . 4.

5. 5, 8, 13, 20, 29, . . . 6.

7. 36, 18, 9, —9, —9, . . . 8. 24

9. —1, —3, 1, —5, —3, . . . 10. 24 42

12, 6, 0, −6, −12, . . .

3, 5, 9, 15, 23, . . .

81, 27, 9, 3, 1, . . .

—1 , —1 , —5 , —7 , —3 , . . . 62662

21.

22. ✗

11. WRITING EQUATIONS Write a rule for the arithmetic sequence with the given description.

a. The first term is −3 and each term is 6 less than the previous term.

b. The first term is 7 and each term is 5 more than the previous term.

12. WRITING Compare the terms of an arithmetic sequence when d > 0 to when d < 0.

In Exercises 13–20, write a rule for the nth term of the sequence. Then find a20. (See Example 2.)

The first term is 22 and the common difference is −13.

an =−13+(n−1)(22) an =−35+22n

13. 12, 20, 28, 36, . . . 14. 15. 51, 48, 45, 42, . . . 16.

7, 12, 17, 22, . . .

27. a =−5,d=−—1

17 2 21 2

86, 79, 72, 65, . . . 11 511

29. USING EQUATIONS One term of an arithmetic sequence is a8 = −13. The common difference

is −8. What is a rule for the nth term of the sequence?

○A an = 51 + 8n ○B an = 35 + 8n ○C an=51−8n ○D an=35−8n

17. −1, −—3, —3, 1, . . . 18. 19. 2.3, 1.5, 0.7, −0.1, . . . 20.

−2, −—4, −—2, —4, . . . 11.7, 10.8, 9.9, 9, . . .

422 Chapter 8 Sequences and Series

✗

Use a1 = 22 and d = −13. a =a +nd

In Exercises 23–28, write a rule for the nth term of th

Step-by-step explanation:

4 0

3 years ago