1) All of them equate to 0. The first and second equations are in first degree, while the last one is in second degree. Also, Eq3 is like a product of of Eq1 and Eq2.
2) For the Eq1, x=-7 only. For Eq2, x=4. For Eq3, both x=-7 and x=4 satisfy the equation.
3) To know if the values of x satisfy the equation, substitute them to the equations. If the equation is true, then it is a solution of that equation.
Eq1: x+7=0 Eq2: x-4=0 Eq3: x=-7 | (x+7)(x-4)=0 | x=4
-7+7=0 4-4=0 (x+7)(x-4)=0 (x+7)(x-4)=0
0=0 ✓ 0=0 ✓ (-7+7)(-7-4)=0 (4+7)(4-4)=0
0=0 ✓ 0=0 ✓
4) The solutions for Eq1 and Eq2 satisfy Eq3.
5) Yes, because to find the solutions for Eq3, we need to equate both of the factors to 0. Doing this gives us Eq1 and Eq2, thus, Eq1 And Eq2 are the necessary equations to solve for Eq3.
6) (x+7) (x-4) = 0 will be true if at least one of its factors are equal to 0.