Question

Solve the following system of equations.

2x – 3y + z = –1
3x + 2y + 2z = –1
x – y – 3z = –4


(–1, 0, 1)

(1, 1, 4)

(–1, –1, –2)

(1, 2, 1)

Answer 1

Would i use elimination or substitution ?

Answer 2

well even i have never solved such a qn before so let me have a go -

2x – 3y + z = –1
3x + 2y + 2z = –1
x – y – 3z = –4

Now in eqn 3, x-y-3z = -4

y = x-3z+4

using this value of y in eqn 1,

2x - 3(x-3z+4) + z = -1

2x - (3x-9z+12) + z = -1

2x-3x+9z-12+z+1 = 0

i.e. 10z-x-11 = 0

Now using same value of y in eqn 2,

3x+2y+2z = -1,

3x + 2(x-3z+4) + 2z = -1,

3x+2x-6z+8+2z+1 = 0

5x-4z+9 = 0

so x-10z =-11 and 5x-4z = 9

i.e. 5x-50z = -55 and 5x-4z =- 9

i.e. -46z =- 46 or z = 1.

now 5x-4z = 9

5x-4 = -9

5x = -5 hence x = -1

Now using x = -1, z = 1 in y =x-3z+4,

y = -1 -3.1 + 4

=> -4+4 or y = 0

So solution set is (-1,0,1) i.e. A i have done this using substitution of y in terms of other 2 variables.

Answer 3

oh wow you made it look easy lol thanks(:!