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Andrew [12]
1 year ago
8

Jada is solving the equation shown below. Negative one-half (x + 4) = 6 Which is a possible first step to begin to simplify the

equation? Select two options. Divide both sides of the equation by –2. Subtract 4 from both sides of the equation. Multiply both sides of the equation by –2. Distribute –2 over (x + 4). Distribute Negative one-half over (x + 4).
Mathematics
1 answer:
elena-14-01-66 [18.8K]1 year ago
3 0

The two options that are the possible first step to begin to simplify the equation are (c) Multiply both sides of the equation by –2. and (d). Distribute Negative one-half over (x + 4).

<h3>How to determine which two options are the possible first step to begin to simplify the equation?</h3>

The equation is given as:

Negative one-half (x + 4) = 6

Rewrite the above equation properly

So, we have

-1/2(x + 4) = 6

A possible first step is to multiply both sides of the equation -1/2(x + 4) = 6 by -2

So, we have

x + 4 = -12

Another possible first step is to distribute the expression -1/2(x + 4) in the equation

So, we have

-1/2x - 1/2 * 4 = 6

Hence, the two options that are the possible first step to begin to simplify the equation are (c) Multiply both sides of the equation by –2. and (d). Distribute Negative one-half over (x + 4).

Read more about equations at:

brainly.com/question/2972832

#SPJ1

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In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
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Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

k^2-k-2=0

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Case k = −1:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]

This gives the infinite many solution.

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