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finlep [7]
3 years ago
7

Find three consecutive integers such that three times the sum of the first two is eight more than 5 times the third.

Mathematics
1 answer:
klasskru [66]3 years ago
7 0
Consecutive integers are numbers one after the other on the number one (ie. 1,2,3,4,5 or 203,204,205).

times= multiplication
sum= addition
more than= addition
is= equal sign

x= first integer
x+2= second integer
x+3= third integer

3(x + (x + 2))= 5(x + 3) + 8
add inside parentheses

3(2x + 2)= 5(x + 3) + 8
multiply 3 and 5 by their parentheses

(3*2x) + (3*2)= (5*x) + (5*3) + 8
multiply in parentheses

6x + 6= 5x + 15 + 8

6x + 6= 5x + 23
subtract 5x from both sides

x + 6= 23
subtract 6 from both sides

x= 17 first integer

x + 1= 17 + 1= 18 second integer

x + 2= 17 + 2= 19 third integer


CHECK:
3(x + (x + 2))= 5(x + 3) + 8
3(17 + (17 + 2))= 5(17 + 3) + 8
3(17 + 19)= 5(20) + 8
3(36)= 100 + 8
108= 108


ANSWER: The three consecutive integers are 17, 18 and 19.

Hope this helps! :)
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  • The Midpoint of AB is (1,0).

Given that:

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To find:

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So, according the question

We know that,

The midpoint M of a line segment AB with endpoints A (x₁, y₁) and B (x₂, y₂) has the coordinates M ( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} ).

Now from question,

We know that the the coordinates of A is (3,1) and coordinates of B is (-1,-1) of line AB.

So, we can say that

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∵ The coordinates of midpoint M (X,Y)

          X = \frac{x_1 + x_2}{2}

              = \frac{3-1}{2}

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And

          Y = \frac{y_1 + y_2}{2}

              = \frac{1-1}{2}

              = 0/2

          Y = 0.

So, the midpoint of line AB is M (1,0)

To learn more about Midpoint of line, please click on the link;

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