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nasty-shy [4]
3 years ago
12

Solve the system of equations:

Mathematics
1 answer:
Ahat [919]3 years ago
8 0

Answer:

Step-by-step explanation:

z = 3

2y + z = 1

2y + 3 = 1        {substituting z =3}

2y = 1 -3

2y = -2

y = - 2/2

y = -1

2x +3y +2z = 13

2x + 3 * (-1) + 2*3 = 13      { {substituting z =3} and y = -1}

2x -3 + 6 =13

2x + 3 = 13

2x  = 13 - 3

2x= 10

x = 10/2

x = 5

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Answer:

\frac{x-3}{2x-3}. hole or removable discontinuity at x=2

Step-by-step explanation:

Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

So let's start by factoring the first equation:

x^2-5x+6

Now let's find what ac is (it's just c since a=1...)

AC= 6

List factors of -6

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Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...

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Ok now let's factor the second equation:

2x^2-7x+6

Multiply a and c

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List factors of 12:

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Rewrite equation:

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Factor out GCF:

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Here we can factor out the x-2 and the simplified form is:

\frac{x-3}{2x-3}

So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.

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