Let's check each of the cases to determine the solution of the problem
case A) we have
[tex]x+10=-5[/tex]
Substitute the value of [tex]x=-5[/tex] in the equation
[tex](-5)+10=-5[/tex]
[tex]5=-5[/tex] ------> the equation is not true
therefore
The value of [tex]x=-5[/tex] is not a solution of the equation
case B) we have
[tex]-x^{2} +50=25[/tex]
Substitute the value of [tex]x=-5[/tex] in the equation
[tex]-(-5)^{2} +50=25[/tex]
[tex]-(25) +50=25[/tex]
[tex]25=25[/tex] ------> the equation is true
therefore
The value of [tex]x=-5[/tex] is a solution of the equation [tex]-x^{2} +50=25[/tex]
case C) we have
[tex]\left|2x\right|=10[/tex]
Substitute the value of [tex]x=-5[/tex] in the equation
[tex]\left|2(-5)\right|=10[/tex]
[tex]\left|-10\right|=10[/tex]
[tex]10=10[/tex] -----> the equation is true
therefore
The value of [tex]x=-5[/tex] is a solution of the equation [tex]\left|2x\right|=10[/tex]
case D) we have
[tex]x < 0[/tex]
Substitute the value of [tex]x=-5[/tex] in the inequality
[tex]-5 < 0[/tex] --------> the inequality is true
therefore
The value of [tex]x=-5[/tex] is a solution of the inequality [tex]x < 0[/tex]
case E) we have
[tex]2x <10[/tex]
Substitute the value of [tex]x=-5[/tex] in the inequality
[tex]2*(-5) < 10[/tex]
[tex]-10 < 10[/tex] --------> the inequality is true
therefore
The value of [tex]x=-5[/tex] is a solution of the inequality [tex]2x <10[/tex]
the answer is
[tex]-x^{2} +50=25[/tex]
[tex]\left|2x\right|=10[/tex]
[tex]x < 0[/tex]
[tex]2x <10[/tex]
