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Sedaia [141]
4 years ago
10

Solve by the method of your choice, select correct choice below if it is either a solution set, infinitely many solutions the so

lution set is (x,y)? or if there is no solution and the solution set is 0
7x-3y=6
2x+y=11
Mathematics
1 answer:
mafiozo [28]4 years ago
6 0

Answer: it is a solution set

Step-by-step explanation:

The given system of simultaneous linear equations is expressed as

7x-3y=6 - - - - - - - - - -1

2x+y=11 - - - - - - - - - - - 2

We would eliminate x by multiplying equation 1 by 2 and equation 2 by 7. It becomes

14x - 6y = 12

14x + 7y = 77

Subtracting, it becomes

- 13y = - 65

Dividing the left hand side and the right hand side of the equation by - 13, it becomes

- 13y/13 = - 65/- 13

y = 5

Substituting x = 5 into equation 2, it becomes

2x + 5 = 11

Subtracting 5 from the left hand side and the right hand side of the equation, it becomes

2x + 5 - 5 = 11 - 5

2x = 6

Dividing the left hand side and the right hand side of the equation by 2, it becomes

x = 6/2 = 3

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A

Step-by-step explanation:

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4 0
3 years ago
Both equations in a system of linear equations have a slope of 1/2. Does this system have infinitely many solutions? Explain. pl
tino4ka555 [31]

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Step-by-step explanation:

3 0
2 years ago
What is the solution to the given system of equations?
marin [14]
Hi!

See my image for the answer.

I put each of the choices into the equation, then solved it.

The answer is B. (4,-12)

Hope this helps! :)
-Peredhel

8 0
4 years ago
Read 2 more answers
Please solve the problem ​
jek_recluse [69]

Treat the matrices on the right side of each equation like you would a constant.

Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.

Then you can eliminate <em>Y</em> by taking the sum

4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>

==>   <em>X</em> = (4<em>A</em> + <em>B</em>)/11

Similarly, you can eliminate <em>X</em> by using

-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>

==>   <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11

It follows that

X=\dfrac4{11}\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\dfrac1{11}\begin{bmatrix}7&-10\\-7&11\end{bmatrix} \\\\ X=\dfrac1{11}\left(4\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\left(\begin{bmatrix}48&-12\\40&88\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\begin{bmatrix}55&-22\\33&99\end{bmatrix} \\\\ X=\begin{bmatrix}5&-2\\3&9\end{bmatrix}

Similarly, you would find

Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}

You can solve the second system in the same fashion. You would end up with

P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}

3 0
3 years ago
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