Verified answer

If you solve for it, and get a valid statement. Like x=x, or 1=1, or the like, then it is all real numbers (because any number you plug in for x will result in a true statement.

If it comes out to an invalid statement like 1=2, 3=7, or the like, then it is no real solution, because no number could make that statement true.

No solution.

There are two ways to differentiate between the two.

"No solution" is a statement, that appears when an irrational equation is formed, when the two equations are solved algebraically, like -9=8 or when the two equations have the same slope (they are parallel and therefore do not intersect. Because of this, they do not have a solution (s) of linear equations and thus there is no solution).

"All solution" is a statement that appears when a rational equation is formed, when the two equations are solved algebraically, but instead like a normal equation like x=3, and unlike irrational equations, the equation has the same numbers, like 3=3. Also, if the two equations have the same equation, they are the same line therefore they will have the same solutions. Becuause of this, the "all solution statement" is formed.

If you distribute it ends up as -3x-9 = -3x-8. Since you know this can't be true (-9 does not = -8), then the answer is no solution.

If it had been -3x-9 = -3x-9, that would be an "all real numbers" answer because any number you plug in for x would work.

- 3(x + 3) = - 3(x + 1) - 5

- 3x - 9 = - 3x - 3 - 5

0x = 1.

no solution.

−3(x + 3) = −3(x + 1) − 5

-3x-9=-3x-3-5

-9=-8

That's is not possible in Real Numbers!!

I hope that help!!

a million) 4-3x<=8+x 4+8<=x+3x 12<=4x 3<=x or l [3,plus infinity) 2) 3x-3>-x+a million 3x+x>a million+3 4x>4 x>a million or (a million, plus infinity) 3) 3(x-a million)=<4x 3x-3<=4x -3<=x or [-3, plus infinity) 4)a million/2x<x+a million -a million<x/2 -2<x or (-2, plus infinity) 5)-a million/2x>a million-x x/2>a million x>2 or x indoors the era (2, plus infinity) s

no solution