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2z-28 = -26
(2+x+y+6+z) += 9x,y and z are variables, there are 3.
3xz
-2z + 3
2z+17=21 subtract 17 from both sides. 2z=4 divide both sides by 2 z=2
2z-28 = -26
There are infinitely many solutions. One linear equation in two variables cannot be solved to give a single answer.
The sum of the given expression would depend on the values of the variables which have not been given.
2z(5z - 1)(25z^2 + 5z + 1)
9x + 2y = -11 I'm sorry to say this, but it is absolutely "impossible" to figure variables out with only one equality. You need one more of that to get those two variables (if you have only one variable it is perfectly fine and you can do it)
(2+x+y+6+z) += 9x,y and z are variables, there are 3.
(-3) x (-2z - 7) = 6z + 21 = 3 (2z + 7)
3xz
-2z + 3
16(2z-3)
(1) y2-2zy-2y-2z-3 = y2 -2y (z+1) - (2z+3) Delta = [-(z+1)]2 - (- (2z+3)) = z2+2z+1+2z+3= z2+2z+4 = (z+2)2 (If used the reduced form as b=-2(z+1) is even, b'=-(z+1) and delta=b'2-ac) y=(z+1) ± (z+2) so y = 2z+3 or y =-1 So the (1) can be written (y-2z-3)(y+1)
To solve for two unknown variables (x and y) you require two independent equations,