We presume you are interested in the solution to a pair of linear equations. However, the second one seems to be missing a character, so we'll make one up.

Suppose the equations are

2x - 10y = 30

2x + a*5y = 15, where a is either +1 or -1 depending on the original intent.

We can subtract the second equation from the first (to cancel x terms).

(2x - 10y) - (2x + 5ay) = (30) - (15)

y(-10-5a) = 15 (collect terms)

y = -15/(10+5a) = -3/(2+a) (divide by -10-5a, then reduce the fraction by removing a factor of 5)

We know from the first equation that

2x - 10y = 30

2x = 30 + 10y (add 10y to both sides)

x = 15 + 5y (divide both sides by 2)

x = 15 + 5(-3/(2+a)) (substitute for y)

x = 15(1 - 1/(2+a)) (factor out 15)

x = 15(2+a - 1)/(2+a) (convert mixed fraction to "improper" fraction)

x = 15(1+a)/(2+a) (simplify)

Our solution is (x, y) = (15(1+a)/(2+a), -3/(2+a))

This has two values.

If the second equation was originally

2x + 5y = 15

we have a = 1, and the solution (x, y) = (15*2/3, -3/3) = (10, -1)

If the second equation was originally

2x - 5y = 15

we have a = -1, and the solution (x, y) = (15*0/1, -3/1) = (0, -3)

Suppose the equations are

2x - 10y = 30

2x + a*5y = 15, where a is either +1 or -1 depending on the original intent.

We can subtract the second equation from the first (to cancel x terms).

(2x - 10y) - (2x + 5ay) = (30) - (15)

y(-10-5a) = 15 (collect terms)

y = -15/(10+5a) = -3/(2+a) (divide by -10-5a, then reduce the fraction by removing a factor of 5)

We know from the first equation that

2x - 10y = 30

2x = 30 + 10y (add 10y to both sides)

x = 15 + 5y (divide both sides by 2)

x = 15 + 5(-3/(2+a)) (substitute for y)

x = 15(1 - 1/(2+a)) (factor out 15)

x = 15(2+a - 1)/(2+a) (convert mixed fraction to "improper" fraction)

x = 15(1+a)/(2+a) (simplify)

Our solution is (x, y) = (15(1+a)/(2+a), -3/(2+a))

This has two values.

If the second equation was originally

2x + 5y = 15

we have a = 1, and the solution (x, y) = (15*2/3, -3/3) = (10, -1)

If the second equation was originally

2x - 5y = 15

we have a = -1, and the solution (x, y) = (15*0/1, -3/1) = (0, -3)