How can I solve this algebraic system of equations?

Rework one of the equations so you get x or y equal to something then substitute in the other equation.

In this case, rewrite your first eqt. 5x-y=-1 to read y = something so, moving 5x over and multiply by -1,

y=5x+1, substitute this value for y in the second equation,

15x=2(5x+1) , then expand to: 15x=10x+2 then solve for x.

5x=2 or x=2/5 , substitute that value back into either of the first equations. So in eq. 1,

5(2/5)-y=-1 which is the same as 2-y=-1 which is the same as y=3

Therefore x=(2/5) and y=(3). Double check buy substituting these values in the second equation.

get one term (either one) in one eq to = term in other eq. you can divide or multiply.

multiply 3* { 5x - y = -1 } >>> 15x -3y = -3

lucky us , the second eq also has a 15x, and 15x = 15x

so [2y] -3y = -3 .... y = 3 ... plug that back into both eq and see if you get the same value for x

OR .... [ 15x - 2y = 0 ] * (1/2) becomes 7.5x - y ... becomes 7.5x = y

then, in eq 1 5x - [7.5x] = -1 ... solve for x; using that value, solve for y

15x=2y, turns into X=2/15y.

Sub it in for the X in 5X-Y= -1