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WARRIOR [948]
3 years ago
12

What are the steps to this equation to -2x-4+5x=8

Mathematics
2 answers:
balandron [24]3 years ago
7 0

Answer:

x=4

Step-by-step explanation:

3x=12

x=4

laila [671]3 years ago
6 0

Answer:

x = 4

Explanation:

−2x − 4 + 5x = 8

[ Simplify both sides of the equation ]

−2x − 4 + 5x = 8

−2x + −4 + 5x = 8

(−2x + 5x) + (−4) = 8 (Combine Like Terms)

3x + −4 = 8

3x − 4 = 8

[ Add 4 to both sides ]

3x − 4 + 4 = 8 + 4

3x = 12

[ Divide both sides by 3 ]

3x / 3 = 12 / 3

x = 4

- PNW

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olga55 [171]

Answer:

x = - 4

Step-by-step explanation:

First do distributive property

5x-15=10x+5

then add 15

5x=10x+20

-10x

-5x=20

divide 5

x=-4

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Answer:

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(a+b)(c+d)=ac+ad+bc+bd\\(3m-5)(5m-2)=3m*5m+3m(-2)-5*5m(-2)\\=3m*5m+3m(-2)-5*5m(-2)\\=15m^2-31m+10

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2 years ago
Please answer and explain. 25pts
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3 years ago
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
svet-max [94.6K]

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

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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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