Solve this maths If (1;5)is a solution of the equations n²x+my=14 ny−mx=8 Find values of n and m

The given equations are:

1. $n^2x + my = 14$
2. $ny - mx = 8$

Substitute $x=1$ and $y=5$ into both equations:
For equation 1:

$$n^2(1) + m(5) = 14$$ $$n^2 + 5m = 14 \quad (*)$$

For equation 2:

$$n(5) - m(1) = 8$$ $$5n - m = 8 \quad (**)$$

From equation $(**)$, express $m$ in terms of $n$:

$$m = 5n - 8$$

Substitute $m = 5n - 8$ into $n^2 + 5m = 14$:

$$n^2 + 5(5n - 8) = 14$$ $$n^2 + 25n - 40 = 14$$ $$n^2 + 25n - 40 - 14 = 0$$ $$n^2 + 25n - 54 = 0$$

Factor the quadratic equation $n^2 + 25n - 54 = 0$. We need two numbers that multiply to -54 and add to 25. These numbers are 27 and -2.

$$(n + 27)(n - 2) = 0$$

This gives two possible values for $n$:

$$n + 27 = 0 \implies n = -27$$ $$n - 2 = 0 \implies n = 2$$

Using the relation $m = 5n - 8$:

Case 1: For $n = -27$

$$m = 5(-27) - 8$$ $$m = -135 - 8$$ $$m = -143$$

Case 2: For $n = 2$

$$m = 5(2) - 8$$ $$m = 10 - 8$$ $$m = 2$$

The values for $n$ and $m$ are:
$\boxed{n = -27, m = -143}$ or $\boxed{n = 2, m = 2}$