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Liono4ka [1.6K]
3 years ago
7

Solve the pair of simultaneous equation 2-4(3x-y)=8(4-y)-39x and 8(4-y)-39x=6(y-x)

Mathematics
1 answer:
andre [41]3 years ago
8 0

Answer:

X = -(146/123) , Y = 21/41

Step-by-step explanation:

  • 2-4(3x-y) = 8(4-y)-39x

2-12x-4y = 32-8y-39x

39x-12x+8y-4y = 32-2

27x+4y = 30

27x+4y-30 = 0

  • 8(4-y) -39x = 6(y-x)

32-8y-39x = 6y-6x

32-8y-6y-39x+6x = 0

-33x-14y+32 = 0

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Solve the following 3 × 3 system. Enter the coordinates of the solution below.
love history [14]
The system is:

i)    <span>2x – 3y – 2z = 4
ii)    </span><span>x + 3y + 2z = –7
</span>iii)   <span>–4x – 4y – 2z = 10 

the last equation can be simplified, by dividing by -2, 

thus we have:

</span>i)    2x – 3y – 2z = 4
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The procedure to solve the system is as follows:

first use any pairs of 2 equations (for example i and ii, i and iii) and equalize them by using one of the variables:

i)    2x – 3y – 2z = 4   
iii)   2x +2y +z = -5 

2x can be written as 3y+2z+4 from the first equation, and -2y-z-5 from the third equation.

Equalize:  

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similarly, using i and ii, eliminate x:

i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7

multiply the second equation by 2:


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3y+2z+4=-6y-4z-14
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a)   5y+3z=-9 
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Remark: it is always a good attitude to check the answer, because often calculations mistakes can be made:

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