# Linear systems

#### helpmeddddd

Orginal 2 equations are:

Equation 1: 3x + y - 2z = -7

Equation 2: -az = x + y + 6 rearranged to -x-y+3z=6 ? or x + y - 3z = -6 ? Either way, I get the same results.

1) A new set of three equations is formed using the original Equation 1 and Equation 2 and the third equation is formed by multiplying all the coefficients and the constant in Equation 1 by 3,
so the third equation is: 9x + 3y - 6z = -21

3x + y - 2z = -7
-x - y + 3z = 6
9x + 3y - 6z = -21

3x + y - 2z = -7, -x - y + 3z = 6 , 9x + 3y - 6z = -21 - Wolfram|Alpha ?

2) A new set of three equations is formed using the original Equation 1 and Equation 2 and the third equation is formed by using Equation 1 with the constant changed from -7 to 12.

3x + y - 2z = -7
-x - y + 3z = 6
3x + y - 2z = 12

I get no solution?

#### skeeter

Math Team
Equation 1: 3x + y - 2z = -7

Equation 2: -az = x + y + 6 reranged to -x - y + 3z = 6 ? or x + y - 3z = -6 ? Either way, I get the same results.
If equation 2 is actually $-3z = x +y + 6$ (you typed "-a" as the coefficient for z), then equation 2 should be $x+y+3z = -6$

#### helpmeddddd

if equation 2 is actually $-3z = x+y+6$ (you typed "-a" as the coefficient for z), then equation 2 should be $x+y+3z = -6$

Ok yes a =-3 shouldn't it be $x+y-3z = -6$

#### skeeter

Math Team
Ok yes a =-3 shouldn't it be $x+y−3z=−6$
you originally typed equation 2 as $-az = x+y+6$ ... now you say $a = -3 \implies -a = -(-3) = 3$

now make sure ... was equation (2) $3z = x+y+6$ or $-3z=x+y+6$ ?

#### mrtwhs

Orginal 2 equations are:

Equation 1: 3x + y - 2z = -7

Equation 2: -az = x + y + 6 reranged to -x-y+3z=6 ? or x + y - 3z = -6 ? either way I get the same results.

1)A new set of three equations is formed using the original Equation 1 and Equation 2 and the third equation is formed by multiplying all the coefficients and the constant in Equation 1 by 3.
so the third equation is: 9x+3y-6z=-21

3x + y - 2z = -7
-x-y+3z=6
9x+3y-6z=-21

3x + y - 2z = -7, -x-y+3z=6 , 9x+3y-6z=-21 - Wolfram|Alpha ?
Are you trying to solve two equations with three unknowns (for unique values of x,y,z) by making a third equation from the first two? Won't happen! One equation in three unknowns represents a plane in three dimensions. Two equations in three unknowns will either be the same plane, parallel planes, or they will intersect in a line (like where two walls come together). Any linear combination of these two planes will be a third plane that is either coincident, parallel, or will intersect in the same common line. Try it with a simpler example. Would you expect x = 3 to have a different solution from 2x = 6?

2)A new set of three equations is formed using the original Equation 1 and Equation 2 and the third equation is formed by using Equation 1 with the constant change from -7 to 12.

3x + y - 2z = -7
-x-y+3z=6
3x + y - 2z = 12

I get no solution?
Are you surprised by this? Look at the first and third equations. If you can't see it from the first and third then try this example. Solve these two equations for a single value of x ... x=1 and x=2.

#### helpmeddddd

you originally typed equation 2 as $-az = x+y+6$ ... now you say $a = -3 \implies -a = -(-3) = 3$

now make sure ... was equation (2) $3z = x+y+6$ or $-3z=x+y+6$ ?
It was $-3z=x+y+6$