first use any pairs of 2 equations (for example i and ii, i and iii) and equalize them by using one of the variables:
i) 2x – 3y – 2z = 4 iii) 2x +2y +z = -5
2x can be written as 3y+2z+4 from the first equation, and -2y-z-5 from the third equation.
Equalize:
3y+2z+4=-2y-z-5, group common terms: 5y+3z=-9
similarly, using i and ii, eliminate x:
i) 2x – 3y – 2z = 4 ii) x + 3y + 2z = –7
multiply the second equation by 2:
i) 2x – 3y – 2z = 4 ii) 2x + 6y + 4z = –14
thus 2x=3y+2z+4 from i and 2x=-6y-4z-14 from ii:
3y+2z+4=-6y-4z-14 9y+6z=-18
So we get 2 equations with variables y and z:
a) 5y+3z=-9 b) 9y+6z=-18
now the aim of the method is clear: We eliminate one of the variables, creating a system of 2 linear equations with 2 variables, which we can solve by any of the standard methods.
Let's use elimination method, multiply the equation a by -2:
a) -10y-6z=18 b) 9y+6z=-18 ------------------------ add the equations:
-10y+9y-6z+6z=18-18 -y=0 y=0,
thus : 9y+6z=-18 0+6z=-18 z=-3
Finally to find x, use any of the equations i, ii or iii:
Sample Response: No, the triangle is not possible. The triangle inequality rule states the sum of any two sides must be greater than the third side. 6 + 4 is 10, which is less than 11.
The area of the square is obtained by taking the square of the length of one side. Since it is said that the area of the square is less than 100 m^2, taking the square root of that where: sqrt (100) = 10, the side of that square should be less than 10 as well. Among the choices, D. 0 <y <10 (where y = length of side of square) is the correct answer.