Two unknowns in Arithmetic Progression I have a problem in my maths book which says 

Find the arithmetic sequence in which $T_8 = 11$ and $T_{10}$ is the additive inverse of $T_{17}$

I don't have a first term of common difference to solve it, so I managed to make two equations to find the first term and common difference from them. Here they are.
first equation $a + 7d = 11$
second equation since $T_{10} + T_{17} = 0$
therefore $a+9d + a+16d = 0 ~~\Rightarrow~~ 2a + 25d = 0$
So what I did is subtracting the first equation from the second one to form this equation with two unknowns $a – 18d = 11$
This is what I came up with and I can't solve the equations, any help?
 A: To solve linear equations with n variables, you need a minimum of n equations. Here, we have 2 variables and 2 equations, so it can be solved. 
There are many methods to solve the equations. But the most common method is elimination method.
In this method, you multiply an equation with a constant, such that the coefficient of the equation becomes equal to the coefficient of the other equation. Then you subtract them, and you are left with one variable. 
You have the equations:
$$ a + 7d = 11 $$
$$ 2a + 25d = 0 $$
Now multiply the first equation by 2, you get:
$$2a+14d=22$$
Subtract this with the second equation. You get:
$$ 2a+14d - 2a - 25d = 22$$
$$ - 11d = 22 $$
This gives $d= -2 $. Putting this value in first equation, we get:
$$a = 11 - 7.(-2) = 25 $$
So we have, $a=25, d=-2$
A: A: $T_8 = 11$ means $a + 7d = 11$
B: $T_{10}=$ additive inverse of $T_{17}$ means $(a+ 9d) + (a+ 16d) = 0$ or $2a + 25d = 0$.
Substract A) from B) and you get:
C:  $a + 18d = -11$  (you subtracted wrong).
Well, subtract A: from C: now.
D:$11d = -22$
So $d = -2$
Plug it into A:
and you get $a + 7(-2) = 11$
$a - 14 = 11$
$a = 25$
===
Note.  Well you has $2a + 25d = 0$, that's enough to make a relation betweend $a$ and $d$.
$2a = -25d$
$a = -\frac {25}2 d$
Plug that into $A$ and $-\frac {25}2d + 7d = 11$
$\frac {-25 +14}2d = 11$
$\frac {-11}2d = 11$
$-11d = 22$
$d = -2$
Plug that into $A$.
Or 
when you subtracted A: for B:  Instead of subtracting A; subtract $2$ times A:
$2a + 25d = 0$ and subtract $2a + 14d = 22$ to get $11d = -22$.
A: It's simple .
a+7d=11
And second eq is 2a+ 25d.
Multiply the first eq by cofficient of a from second eq . i.e.,
2a+14d=22 and
 2a+25d=0 
By subtracting both equations we get 
d=-2.
