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1 year ago

Please help me !! (no this isnt a test/exam)

Mathematics

1 answer:

stiv31 [10]1 year ago

8 0

Answer:

infinitely many

Step-by-step explanation:

Add 12-y to both sides of the second equation to put it into standard form like the first equation:

y +(12 -y) = (x -12) +(12 -y)

12 = x - y . . . simplify

x - y = 12 . . . . second equation in standard form

We see this is identical to the first equation. That means every solution of the first equation is also a solution of the second equation. There are infinitely many solutions.

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Solve the system of equations by finding the reduced row echelon form of the augmented matrix. Write the solutions for x and y i

Gwar [14]

Answer:

The solutions for the given system of equations are:

$\left \{ {{x=-5} \atop {y=12-4z}} \right.$

Step-by-step explanation:

Given the equation system:

$\left \{ {{3x+y+4z=-3} \atop {-x+y+4z=17}} \right.$

We obtain the following matrix:

$\left[\begin{array}{cccc}3&1&4&-3\\-1&1&4&17\end{array}\right]$

Step 1: Multiply the fisrt row by 1/3.

$\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\-1&1&4&17\end{array}\right]$

Step 2: Sum the first row and the second row.

$\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&\frac{4}{3} &\frac{16}{3}&16\end{array}\right]$

Step 3: Multiply the second row by 3/4.

$\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&1 &4&12\end{array}\right]$

Step 4: Multiply the second row by -1/3 and sum the the first row.

$\left[\begin{array}{cccc}1&0 &0&-5\\0&1 &4&12\end{array}\right]$

The result of the reduced matrix is:

$\left \{ {{x=-5} \atop {y+4z=12}} \right.$

This is equal to:

$\left \{ {{x=-5} \atop {y=12-4z}} \right.$

These are the solutions for the system of equations in terms of z, where z can be any number.

6 0

2 years ago

What is 2(x+14)+(2x-14

lawyer [7]

2x+28+2x-14
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2 years ago

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Alecsey [184]

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WHAT IS 24 DIVIDED BY 48 WITH WORK PLEASE

kotegsom [21]

So,

$\frac{24}{48}$

Write in prime factored form.
$\frac{2*2*2*3}{2*2*2*2*3}$

Cross-cancel
$\frac{1}{2}$

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3 years ago

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