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tino4ka555 [31]
3 years ago
10

(07.05) Which of these is a simplified form of the equation 7x + 4 = 5 + 3x + 1x? (4 points) 3x = 9 3x = 1 11x = 9 11x = 1

Mathematics
1 answer:
slava [35]3 years ago
4 0
Okay first you need to combine like terms together (3x and 1x)

7x + 4 = 5 + 4x

now you need x only on one side, you could subtract either 7x or 4x 

3x + 4 = 5

subtract 4 over

3x = 1 is the answer 



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Solve the 3 × 3 system shown below. Enter the values of x, y, and z. X + 2y – z = –3 (1) 2x – y + z = 5 (2) x – y + z = 4 (3)
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<h2>Answer:</h2>

\boxed{x=1, \y=-1, \ z=2}

<h2>Step-by-step explanation:</h2>

We will use the Gaussian elimination method to solve this problem. To do so, let's follow the following steps:

Step 1: Let's multiply first equation by −2. Next, add the result to the second equation. So:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\~~ x&-~~~~~ y&+~~~~ z&~=~4\end{array}

Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&-~~~3~ y&+~~2~ z&~=~7\end{array}

Step 3: Let's multiply second equation by −35, Next, add the result to the third equation. Therefore:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&&+~~\frac{ 1 }{ 5 }~ z&~=~\frac{ 2 }{ 5 }\end{array}

Step 4: solve for z, then for y, then for x:

\frac{ 1 }{ 5 } ~ z & = \frac{ 2 }{ 5 } \\ \\ \boxed{z & = 2}

-5y+3z &= 11\\-5y+3\cdot 2 &= 11\\ \\ \boxed{y &= -1}

By substituting y=-1 \ and \ z=2 into the first equation, we get the x. So:

x+2(-1)-2=-3 \\ \\ x-2-2=-3 \\ \\ \boxed{x=1}

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2. Parallel lines are the same distance apart everywhere. Each line will have two parallel lines at some given distance from it, one on each side. Here, the separation distance from AB is 1 cm, so your locus of points is the two lines parallel AB that are 1 cm from it on either side.

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