It depends what 6(p - 4) - 30p is supposed to equal. Substituting p = -1, gives: 6((-1) - 4) - 30(-1) = 6×-5 - 30×-1 = -30 + 30 = 0 Only if 6(p - 4) - 30p = 0 is p = -1 a solution. Otherwise it is not a solution.
No, an equation with integer coefficients does not always have an integer solution. For example, the equation (x + 1 = 2) has an integer solution, (x = 1), but the equation (2x + 3 = 1) has no integer solution since (x = -1) is not an integer. Solutions depend on the specific equation and its constraints, and rational or real solutions may exist instead.
No. That is because 2 - 8x is an algebraic expression, not an equation. And since it is not an equation, it does not have a solution.
Yes and sometimes it can have more than one solution.
Without an equality sign it cannot be classed as an equation therefore it follows that there is no solution.
It depends what 6(p - 4) - 30p is supposed to equal. Substituting p = -1, gives: 6((-1) - 4) - 30(-1) = 6×-5 - 30×-1 = -30 + 30 = 0 Only if 6(p - 4) - 30p = 0 is p = -1 a solution. Otherwise it is not a solution.
An equation that has no solution is called an equation that has no solution.
It is 1/30.
There are 3 30p's in £1. But you still have 10p left over needed to make the pound.
1+1=2
No. That is because 2 - 8x is an algebraic expression, not an equation. And since it is not an equation, it does not have a solution.
Yes and sometimes it can have more than one solution.
Without an equality sign it cannot be classed as an equation therefore it follows that there is no solution.
0
Without any equality and not knowing the plus or minus values of 1 and 45 it is not an equation so therefore it has no solution.
It is a linear equation in one variable r. The solution to the equation is r = 1.
Without an equality sign it can't be considered to be an equation and so therefore it has no solution.