Given:
[tex]5x+y=-10[/tex][tex]y=x^2-2x-8[/tex]We can solve for the value of x through substitution.
We will substitute the value of y from the second equation, to the first equation.
[tex]5x+y=-10[/tex][tex]5x+(x^2-2x-8)=-10[/tex][tex]5x+x^2-2x=-10+8[/tex][tex]x^2+3x=-2[/tex][tex]x^2+3x+2=0[/tex]We then get the factors of this equation.
(x + 1)(x + 2) = 0
Next, we will solve for the value of x.
[tex]x+1=0[/tex][tex]x=-1[/tex][tex]x+2=0[/tex][tex]x=-2[/tex]With these two values of x, we can now find the values of y through substituting the values of x to either equations.
Using the first equation, given that:
x = -1
x = -2
[tex]x=-1[/tex][tex]5x+y=-10[/tex][tex]5(-1)+y=-10[/tex][tex]-5+y=-10[/tex][tex]y=-10+5[/tex][tex]y=-5[/tex]Therefore, this will give us a solution of (-1, -5).
Next,
[tex]x=-2[/tex][tex]5x+y=-10[/tex][tex]5(-2)+y=-10[/tex][tex]-10+y=-10[/tex][tex]y=-10+10[/tex][tex]y=0[/tex]And this will give us a solution of (-2, 0).
Therefore, all the possible solutions for this system of equations are (-1, -5) and (-2, 0).
