Wtite down Three of the first six square numbers are two more than a prime number
1 answer:
You might be interested in
First equation is x-3y+2z=11
We coundt x and we've got
x=11+3y-2z
Third equation is 2x-2y-4z=2
We count also x and we've got:
2x=2+2y+4z /:2
x=1+y+2z
so:
11+3y-2z=1+y+2z
2y-4z=-10 /:2
y-2z=-5
y=2z-5
so
x=1+y+2z
x=1+2z-5+2z
x=4z-4
Now substitute this to second equation:
-x+4y+3z=5
-(4z-4) + 4(2z-5) +3z=5
-4z+4+8z-20+3z=5
7z-16=5
7z=21 /:7
z=3
And count rest:
x=4z-4 = 4*3-4=8
y=2z-5=2*3-5=1
So the solution is:
We coundt x and we've got
x=11+3y-2z
Third equation is 2x-2y-4z=2
We count also x and we've got:
2x=2+2y+4z /:2
x=1+y+2z
so:
11+3y-2z=1+y+2z
2y-4z=-10 /:2
y-2z=-5
y=2z-5
so
x=1+y+2z
x=1+2z-5+2z
x=4z-4
Now substitute this to second equation:
-x+4y+3z=5
-(4z-4) + 4(2z-5) +3z=5
-4z+4+8z-20+3z=5
7z-16=5
7z=21 /:7
z=3
And count rest:
x=4z-4 = 4*3-4=8
y=2z-5=2*3-5=1
So the solution is: