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icang [17]
4 years ago
14

Solve the system algebraically. Check your work.

Mathematics
1 answer:
Luden [163]4 years ago
8 0

answer.

Answer:

x=2 and y=0 is the required result.

Step-by-step explanation:

We have been given system of equations:

5x+2y=105x+2y=10     (1)

And 3x+2y=63x+2y=6      (2)

We will use elimination method:

Multiply 1st equation by 3 and 2nd equation by 5 we get:

15x+6y=3015x+6y=30      (3)

15x+10y=3015x+10y=30      (4)

Now subtract  (4) from (3) we get:

-4y=0−4y=0

y=0y=0

Now, put y=0 in (1) equation:

5x+2(0)=105x+2(0)=10

5x=105x=10

x=2x=2

Hence, x=2 and y=0

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<h3>Answer: Arithmetic, common difference = -4</h3>

-------------------

Explanation:

Pick any term of the sequence, and subtract off the previous term to find that,

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Problem 2

<h3>Answer: Arithmetic, common difference = 5</h3>

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Similar to problem 1, this sequence is also arithmetic because we add on 5 to each term to get the next one

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==========================================================

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<h3>Answer: Not arithmetic</h3>

-------------------

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Unlike the previous two problems, this sequence is not arithmetic.

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This sequence is instead geometric because

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Each quotient is 4, showing the common ratio is 4. To find the next term, we multiply the current term by 4. So the next term after 192 would be 4*192 = 768, then 4*768 = 3072 is next, and so on.

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