Systems by simple substitution. Game on.

Substitute $y = 7$ into $y = 2x - 5$: $7 = 2x - 5$, so $12 = 2x$ and $x = 6$. The solution is $(6, 7)$.

Substitute $y = 4$ into $y = -2x + 10$: $4 = -2x + 10$, so $-6 = -2x$ and $x = 3$. The solution is $(3, 4)$.

Substitute $y = 14$ into $y = 5x - 1$: $14 = 5x - 1$, so $15 = 5x$ and $x = 3$. The solution is $(3, 14)$.

Substitute $y = 3$ into $y = 2x + 9$: $3 = 2x + 9$, so $-6 = 2x$ and $x = -3$. The solution is $(-3, 3)$.

Set the expressions for $y$ equal: $3x - 4 = x$, so $2x = 4$ and $x = 2$. Then $y = 2$. The solution is $(2, 2)$.

Set the expressions for $y$ equal: $-x + 8 = 3x$, so $8 = 4x$ and $x = 2$. Then $y = 3(2) = 6$. The solution is $(2, 6)$.

Set the expressions for $y$ equal: $4x + 5 = -x$, so $5x = -5$ and $x = -1$. Then $y = -(-1) = 1$. The solution is $(-1, 1)$.

Set the expressions for $y$ equal: $6x - 2 = 4x$, so $2x = 2$ and $x = 1$. Then $y = 4(1) = 4$. The solution is $(1, 4)$.