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andrezito [222]
2 years ago
5

Which of the following is a solution of the system 2x + y = 9 and 3x + 5y = 19.​

Mathematics
1 answer:
Nezavi [6.7K]2 years ago
4 0

2x+y=9

3x+5y=19

I will do this problem in 2 ways. I.)Substitution II.)Elimination

Solution I.) Substitution

We can subtract 2x from both sides in the first equation.

y=9-2x

Now we can substitute the y in the second equation with 9-2x

3x+5(9-2x)=19

-7x+45=19

-7x=-26

x=26/7

y=9-2(26/7)=11/7

Solution II.)Elimination

We can multiply both side of first equation by 5 to get a 5y in both equations.

10x+5y=45

Now because both are positive 5y we just need to do simple subtraction of the 2 equation, each side respectively.

(10x+5y)-(3x+5y)=45-19

7x=26

x=26/7

2*26/7+y=9

y=11/7

Ultimately you get the same answer, both are viable methods, some problems are faster with one method but I recommend mastering both since they are very useful.

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200 σ j=1 2j( j 3) describe the steps to evaluate the summation. what is the sum?
ziro4ka [17]

The sum of the equation is  = 5494000.

<h3>What does summation mean in math?</h3>

The outcome of adding numbers or quantities mathematically is a summation, often known as a sum. A summation always has an even number of terms in it. There may be just two terms, or there may be 100, 1000, or even a million. Some summations include an infinite number of terms.

<h3>Briefing:</h3>

Distribute 2j to (j+3).

Rewrite the summation as the sum of two individual summations.

Evaluate each summation using properties or formulas from the lesson.

The lower index is 1, so any properties can be used.

The sum is 5,494,000.

<h3>Calculation according to the statement:</h3>

\sum_{j=1}^{200} 2 j(j+3)

simplifying them we get:

\sum_{j=1}^{200} 2 j^{2}+6 j

Split the summation into smaller summations that fit the summation rules.

\sum_{j=1}^{200} 2 j^{2}+6 j=2 \sum_{j=1}^{200} j^{2}+6 \sum_{j=1}^{200} j

\text { Evaluate } 2 \sum_{j=1}^{200} j^{2}

The formula for the summation of a polynomial with degree 2

is:

\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}

Substitute the values into the formula and make sure to multiply by the front term.

(2)$$\left(\frac{200(200+1)(2 \cdot 200+1)}{6}\right)$$

we get: 5373400

Evaluating same as above : 6 \sum_{j=1}^{200} j

we get: 120600

Add the results of the summations.

5373400 + 120600

= 5494000

The sum of the equation is  = 5494000.

To know more about  summations visit:

brainly.com/question/16679150

#SPJ4

6 0
1 year ago
The product of two consecutive positive integers is 56. Find the two integers
vredina [299]

Answer:

Two consecutive integers are of the form

x and x + 1

Their product is equal to 56

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Solve and find the two numbers x and x + 1. The above equation may be written as follows

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Step-by-step explanation:

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3 years ago
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user100 [1]

Answer:

3.  9x+5+x-7=25

4.  10x-2=25

5.  10x=27

Step-by-step explanation:

3.  9x+5+x-7=25

4.  10x-2=25

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5.  10x=27

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x=2.7

7 0
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