we write two equations to solve the numbers
[tex]\begin{gathered} a\times b=-15 \\ a+b=-14 \end{gathered}[/tex]
solve a from the second equation
[tex]a=-14-b[/tex]
and replace on the first
[tex]\begin{gathered} (-14-b)\times b=-15 \\ -14b-b^2=-15 \\ b^2+14b-15=0^{} \end{gathered}[/tex]
factor by any method to find b
[tex]b=1,-15[/tex]
now replace b on any equation to find a
[tex]\begin{gathered} a\times b=-15 \\ a\times1=-15 \\ a=-15 \end{gathered}[/tex][tex]\begin{gathered} a\times-15=-15 \\ a=1 \end{gathered}[/tex]
the values of and b must be -15 and 1
Second
write the equations
[tex]\begin{gathered} a\times b=-75 \\ a+b=-10 \end{gathered}[/tex]
solve a from the second equation
[tex]a=-10-b[/tex]
replace on first
[tex]\begin{gathered} (-10-b)\times b=-75 \\ -10b-b^2=-75 \\ b^2+10b-75=0 \end{gathered}[/tex]
factor to solve
[tex]b=5,-15[/tex]
we have two solutions, replace each solutions on any equations to know the true
[tex]\begin{gathered} a\times b=-75 \\ a\times5=-75 \\ a=-\frac{75}{5} \\ \\ a=-15 \end{gathered}[/tex][tex]\begin{gathered} a\times b=-75 \\ a\times-15=-75 \\ a=\frac{-75}{-15} \\ \\ a=5 \end{gathered}[/tex]
the values of a and b are 5 and -15
Third
[tex]\begin{gathered} a\times b=12 \\ a+b=7 \end{gathered}[/tex]
solve a from the second equation
[tex]a=7-b[/tex]
and replace on first
[tex]\begin{gathered} (7-b)\times b=12 \\ 7b-b^2=12 \\ b^2-7b+12=0 \end{gathered}[/tex]
and factor
as we can see the two values that come out of factoring will be the values a and b
[tex]b=3,4[/tex]
so the values of a and b must be 3 and 4
