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3 years ago

I have a quation about functions.

Mathematics

1 answer:

sweet-ann [11.9K]3 years ago

6 0

Step by step...shows how to get -22/7

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Solve the system. find x,y, and z please show all steps<br><br> x-y+z=1<br> x+y+3z=-3<br> 2x-y+2z=0

Mariulka [41]

Solving the system we get, x= -1, y= -2 and z= 0

Step-by-step explanation:

Solving the equations to find x, y and z

$x-y+z=1\,\,\,eq(1)\\x+y+3z=-3\,\,\,eq(2)\\2x-y+2z=0\,\,\,eq(3)$

Adding eq(1) and eq(2)

$x-y+z=1\\x+y+3z=-3\\--------\\2x+4z=-2\,\,\,\,eq(4)\\$

Adding eq(2) and eq(3)

$x+y+3z=-3\\2x-y+2z=0\\----------\\3x+5z=-3\,\,\,eq(5)$

Multiplying eq(4) by 3 and eq(5) by 2

$6x+12z=-6\\6x+10z=-6\\-\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\\----------\\2z=0\\z=0$

So, value of z =0

Putting value of z in eq(4)

$2x+4z=-2\\2x+4(0)=-2\\2x=-2\\x=-1$

So, value of x = -1

Now putting value of z=0 and x =-1 in equation 1

$x-y+z=1\\-1-y+0=1\\-y=1+1\\-y=2\\y=-2$

So, value of y = -2

So, solving the system we get, x= -1, y= -2 and z= 0

Keywords: Solve the system

Learn more about Solve the system at:

#learnwithBrainly

7 0

3 years ago

Find the LCM (Least Common Factor) of 12 and 27 using prime factorizations

Aneli [31]

The LCM (least common factor) of 12 and 27 is the number: 3

5 0

3 years ago

Adrian is a songwriter who collects royalties on his songs whenever they are played in a commercial or a movie. Adrian will earn

seraphim [82]

Answer:

Step-by-step explanation:

Let number of songs are played in a commercial be a

number of songs are played in a movie be b

Adrian earned a total of $1010 -> a*40 + b*130 = 1010 (1)

14 commercials and movies -> a+b = 14 (2)

(1) - (2)*40 = a*40 + b*130 - a*40 -b*40 = 1010 -14*40

-> b*90 = 450

-> b = 5

-> a = 14-5 = 9

6 0

4 years ago

Read 2 more answers

Solve 10e^2x - 5 =23^x for x.

yawa3891 [41]

$10e^{2x}-5=23e^x\qquad\text{subtract}\ 23e^x\ \text{from both sides}\\\\10e^{2x}-23e^x-5=0\\\\10(e^x)^2-23(e^x)-5=0\\\\\text{substitution:}\ e^x=t > 0\\\\10t^2-23t-5=0\\\\10t^2-25t+2t-5=0\\\\5t(2t-5)+1(2t-5)=0\\\\(2t-5)(5t+1)=0\iff2t-5=0\ \vee\ 5t+1=0\\\\2t=5\ \vee\ 5t=-1\\\\t=\dfrac{5}{2} > 0\ \vee\ t=-\dfrac{1}{5} < 0\\\\\text{therefore}\ e^x=\dfrac{5}{2}\to\ln e^x=\ln\left(\dfrac{5}{2}\right)\\\\\boxed{x=\ln\left(\dfrac{5}{2}\right)}$