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3 years ago

18x3=54 How do I get a distributive property

1 answer:

3 0

Its like
a*(b + c) = [a*b + a*c]
or (b + c)*a = [b*a + c*a]
let 18 = 8 + 10, here b = 8 & c = 10, & a = 3
writing it like (8 + 10)*3 = [8*3 + 10*3] = 24 + 30 = 54.

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Jerry was using matrices to solve the system of three equations. He has shown all his steps. Did he make a mistake if so in what

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There are several ways to solve a system of equation; one of these ways is the use of matrix.

<em>Jerry made a mistake at step 2</em>

From the attachment, the step 1 is represented as:

$\mathbf{\frac 12R_1 \left[\begin{array}{cccc}1&1&1&0\\5&3&-2&-4\\3&2&1&1\end{array}\right] }$

The equation in step 2 is given as:

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So, we have:

$\mathbf{R_2 = 5\left[\begin{array}{cccc}1&1&1&0\end{array}\right] } - \left[\begin{array}{cccc}5&3&-2&-4\end{array}\right] }$

Expand

$\mathbf{R_2 = \left[\begin{array}{cccc}5&5&5&0\end{array}\right] } - \left[\begin{array}{cccc}5&3&-2&-4\end{array}\right] }$

Subtract corresponding cells

$\mathbf{R_2 = \left[\begin{array}{cccc}0&2&7&4\end{array}\right] }$

So, the new row 2 elements should be:

$\mathbf{ \left[\begin{array}{cccc}0&2&7&4\end{array}\right] }$

However, the row 2 elements of Jerry's steps are:

$\mathbf{R_2 = \left[\begin{array}{cccc}0&-2&-7&-4\end{array}\right] }$

This means that:

Jerry made a mistake; and the mistake is at step 2

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Attached the solution with work shown.

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