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Kamila [148]
3 years ago
14

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

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
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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Solve the quadratic equation 4x2 − 121 = 0. Verify your answer using a difference-of-squares factoring method.
Alenkasestr [34]

Answer:

Option d) is correct

That is x equals plus or minus start fraction 11 over two end fraction

Step-by-step explanation:

Given quadratic equation is 4x^2-121=0

To write the given quadratic equation by using a difference-of-squares factoring method:

4x^2-121=0

The above equation can be written as

4x^2-11^2=0

(2x)^2-11^2=0

The above equation is in the form of difference-of-squares

Therefore the given quadratic equation can be written in the form of difference-of-squares

by factoring method is (2x)^2-11^2=0

(2x+11)(2x-11)=0 (which is in the form a^2-b^2=(a+b)(a-b) )

2x+11=0   or   2x-11=0

x=\frac{-11}{2}    or   2x=11

x=\frac{-11}{2}    or   x=\frac{11}{2}

x=\pm \frac{11}{2}

Therefore option d) is correct

That is x equals plus or minus start fraction 11 over two end fraction

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Step-by-step explanation:

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