Solve the system of equations y=3x+2 y=x^2+2
1 answer:
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consecutive integers are 1 apart
16 of them can be represented as
x, x+1, x+2, x+3, x+4, x+5, x+6, x+7, x+8, x+9, x+10, x+11, x+12, x+13, x+14, x+15
the average is
there might be some trick to get the average of the first 8 integers only
the average of the first 8 integers would be
how can we get from
first, let's match the denomenators (make them both 1 for ease)
multiply the first eqution by 8 to get
x+x+1+x+2+x+3+x+4+x+5+x+6+x+7=8?
multiply the 2nd equation by 16 to get
x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488
notice that we can try and force the 1st equation into the 2nd equation by adding some numbers
we can already do this:
x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488
[x+x+1+x+2+x+3+x+4+x+5+x+6+x+7]+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488
[8?]+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488
we can try to force another 8? into it
[8?]+x+7+1+x+7+2+x+7+3+x+7+4+x+7+5+x+7+6+x+7+7+x+7+8=488
[8?]+(x)+7+(1+x)+7+(2+x)+7+(3+x)+7+(4+x)+7+(5+x)+7+(6+x)+7+(7+x)+7+8=488
[8?]+(x)+7+(x+1)+7+(x+2)+7+(x+3)+7+(x+4)+7+(x+5)+7+(x+6)+7+(x+7)+7+8=488
group
[8?]+[x+x+1+x+2+x+3+x+4+x+5+x+6+x+7]+7+7+7+7+7+7+7+7+8=488
[8?]+[8?]+8(7)+8=488
16?+64=488
minus 64 from both sides
16?=424
divide both sides by 16
?=26.5
the average of the first 8 integers is 26.5
I'm not sure if there is a simpler way to do it or not