X-y+z=-4 3x+2y-z=5 -2x+3y-z=15 How do I solve this?

1 answer:

5 0

$1)\ \ \ x-y+z=-4\ \ \ \Rightarrow\ \ \ z=-4-x+y\\\\2)\ \ \ 3x+2y-z=5 \\.\ \ \ \ \Rightarrow\ \ 3x+2y-(-4-x+y)=5 \\.\ \ \ \ \Rightarrow\ \ 3x+2y+4+x-y=5 \\.\ \ \ \ \Rightarrow\ \ 4x+y=1\\\\3)\ \ \ -2x+3y-z=15\\.\ \ \ \ \Rightarrow\ \ -2x+3y-(-4-x+y)=15\\.\ \ \ \ \Rightarrow\ \ -2x+3y+4+x-y=15\\.\ \ \ \ \Rightarrow\ \ -x+2y=11\\--------------------\\$

$z=-4-x+y\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and\\ \left \{ {{4x+y=1\ \ \ \ } \atop {-x+2y=11\ /\cdot4}} \right. \\\\\left \{ {{4x+y=1\ \ \ \ } \atop {-4x+8y=44}} \right. \\-------\\y+8y=1+44\\9y=45\ /:9\\y=5\\\\-x+2y=11\ \ \ \Rightarrow\ \ \ x=2y-11\ \ \ \Rightarrow\ \ \ x=2\cdot5-11=-1\\\\z=-4-x+y=-4-(-1)+5=-4+1+5=2\\\\Ans.\ x=-1\ \ \ and\ \ \ y=5\ \ \ and\ \ \ z=2$

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Help me out asap!please and thank you

irga5000 [103]

Answer:

c. (5, -2)

Step-by-step explanation:

as we can clearly see, the 4th vertex has to be on the positive side of x.

therefore, all answer options with negative x are out.

(2, -5) is almost at D (-1, 5). that cannot be right for a rectangle.

that leaves only c as right answer.

FYI - how to get this without predefined answer portions ?

the x- difference of B to A is 4 (from -5 to -1). the x-difference from D to C must be the same (4). 1 + 4 = 5.