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guapka [62]
2 years ago
10

What is the solution to the system of equations below?

Mathematics
1 answer:
liubo4ka [24]2 years ago
7 0

y=2x+1
3x + y=-9


3x + y=-9 is the same as y=-3x-9

y=2x+1
y=-3x-9

2x+1 = -3x-9

—-add 3x on both sides

5x+1=-9

—-subtract 1 from both sides

5x=-10

—-divide 5 on both sides

5x/5 = -10/5

x = -2

——now plug in x in one of the equations to find y

y=2x+1

y=2(-2)+1

y=-4+1

y=-3


The solution is (-2,-3).
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In an arithmetic series, if the fifth term is 74 and the twelfth term is 116, find the sum of the first 30 terms.
Assoli18 [71]
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(116 - 74)/(12 - 5) = 6.
now we can find the first term. the fourth term is 74 - 6, the third term is 74 - 2(6), the second term is 74 - 3(6) and the first term is 74 - 4(6) = 50. 
now using the first term, we can find the last term. the last term is the 30th term and to get that number, some might think it is:
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but take a look at the second term, it is
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basically the nth term is
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so the 30th term is:
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now we can compute what we want:
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now you may be wondering why the formula is the way it is. there is actually a story associated with this. there was a famous mathematician named carl friedrich gauss. legend has it that when he was a kid, once when he was in school, the teacher wanted to slack off, so he (or she...idk which one) told all his (her) students to add the numbers from 1 to 100. everyone instantly went 1+2 = 3, 3 + 3 = 6, 6 + 4 = 10, etc. carl, on the other hand, applied a neat trick that got him the answer really quickly. he wrote the integers from 1 to 100 then wrote that same sequence below it but in reverse order. so:
1,      2,   3,   4,    5, ...,   95, 96, 97, 98, 100
100, 99, 98, 97, 96, 95...5,    4   3,   2,   1
now add the first term of the top sequence to the first term of the bottom sequence (1 + 100 = 101), and do the same thing for the 2nd term of top and bottom sequence, and do it for all 100 terms. you'll find that each term adds to 101. so the total sum of the 2 identical sequences is (101)(100). the 101 is the first term+last term while the 100 is the number of terms (starting to look like the formula i proposed in the beginning!). now remember this is the sum of 2 sequences but we only want the sum of 1. since the sum of the 2 sequences are the same (because they have the same number but written in different orders) we can just take (101)(100) and divide that by 2 to get the sum of the first 100 positive integers. this can be extended to any arithmetic sequence like so:
a,   a+d,    a+2d...    a+(n-2)d, a+(n-1)d
a+(n-1)d, a+(n-2)d, ...a+d,       a
a is the first term of the sequence and d is the distance and the sequence has n terms. each of the n pairs sum to a + (a+(n-1)d) and we can do the same thing as carl did to get the formula i used to solve this problem.

let me know if you have any questions!!! 
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