ok ok..so how would you solve this one? like im confused cuase its?

Question

1.find four consecutive integers such that 6 times the sum of the first and the fourth is 26 less than 10 times the 3rd.

and this one really got me

2.find 3 consecutive even integers such that the sum of the 1st and 2nd = the sum of the 3rd and -10

PS. this is my last 2 thank you to all that helped i really appreciate it. =]

Answer 1

1.
Let x=1st no.
x+1=2nd no.
x+2=3rd no
x+3=4th no.

so,

6[x+(x+3)]=10(x+2)-26

6(2x+3)=10x+20-26

12x+18=10x-6

2x=-24

x=-12

substituting...
integers are -12,-11,-10,-9...

2.
Let x=1st even no.
x+2=2nd even no.
x+4=3rd even no.

x+x+2=x+4-10

2x+2=x-6

x=-8

substituting...
integers are -8,-6,-4

Answer 2

As people stated in your other consecutive integers question, you write cons. integers by starting with x, then adding 1 more each time to get to the next number in order (consecutive).

So if you need 4 cons. integers, they would be:
x = 1st
x+1 = 2nd
x+2 = 3rd
x+3 = 4th

1.) 6[1st + 4th] = 10 (3rd) - 26
6[x + (x+3)] = 10 (x + 2) - 26

Add the x and x + 3 in the brackets first.

6[2x+3] = 10(x+2) -26

Then you need to distribute (multiply) the 6 and the 10 by the binomials inside the parenthesis.

12x+18 = 10x + 20 - 26
12x+18 = 10x-6

When you move all the x terms to one side and the numbers to the other to finish solving, you should find that x = -12

2.) Since this question asks for cons. EVEN integers, you only need every other number. If you think of any even number, you need to add 2 to it to get to the next even. So this time your 3 numbers will be:

x = 1st
x+2 = 2nd
x+4 = 3rd

1st + 2nd = 3rd + (-10)
x + x + 2 = x + 4 - 10
2x + 2 = x - 6

Here, x = -8.

Answer 3

1.find four consecutive integers such that 6 times the sum of the first and the fourth is 26 less than 10 times the 3rd.
integers are n, n+1, n+2, & n+3
6(n+n+3)=10(n+2)-26
12n+18=10n+20-26 subtract 10 n from each side
2n+18=-6 subtract 18 from each side
2n=-24 divide by 2
n=-12
the integers are -12, -11, -10 & -9
check 6(-12-9)=10(-10)-26
6(-21)=-100-26
-126=-126
They are negative integers but still integers

2.find 3 consecutive even integers such that the sum of the 1st and 2nd = the sum of the 3rd and -10
integers are n, n+2 & n+4
n+n+2=n+4-10
2n+2=n-6 subtract n from each side
n+2=-6 subtract 2 from each side
n=-8
the integers are -8, -6 & -4
check -8-6=-4-10
-14=-14

Answer 4

1. OK, so 4 consecutive integers. Let's call the first integer x. They are consecutive, so the next three are (x+1), (x+2), and (x+3). So, six times the sum of the first and the fourth is twenty-six less than ten times the third. Let's make this an equation.
So 6(x+(x+3))=10(x+2)-26.
6(2x+3)=10x+20-26
12x+18=10x-6
2x=-24
x=-12
So the integers are: -12, -11, -10, and -9.

2. 3 consecutive even integers. Let's call the first one y. That makes the next two (y+2) and (y+4). Remember, all of these are even numbers. OK, now the sum of the first and second is the sum of the third minus ten. So, formula:
y+(y+2)=(y+4)-10
2y+2=y-6
y=-8
So the consecutive even integers are -8, -6, and -4.