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ipn [44]
3 years ago
5

Determine all possibilities for the solution set of a system of 2 equations in 2 unknowns. I. No solutions whatsoever. II. One a

nd only one solution. III. Many solutions.
Mathematics
1 answer:
Minchanka [31]3 years ago
4 0

Answer:

II. One and only one solution

Step-by-step explanation:

Determine all possibilities for the solution set of a system of 2 equations in 2 unknowns. I. No solutions whatsoever. II. One and only one solution. III. Many solutions.

Let assume the equation is given as;

x + 3y = 11 .... 1

x - y = -1 ....2

Using elimination method

Subtract equation 1 from 2

(x-x) + 3y-y = 11-(-1)

0+2y = 11+1

2y = 12

y = 12/2

y = 6

Substitute y = 6 into equation 2:

x-y = -1

x - 6 = -1

x = -1 + 6

x = 5

Hence the solution (x, y) is (5, 6)

<em>Hence we can say the equation has One and only one solution since we have just a value for x and y</em>

<em />

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Charge of Ralph:

Charge = $24  + $33 per hour

Let the time worked be = x.

For x hours, the cost = 33*x = 33x

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I hope this helped
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3 years ago
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Answer:

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3 years ago
Read 2 more answers
How do you solve this?​
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\text{L.H.S}\\\\=\begin{vmatrix} 1 & bc& b+c\\ 1 & ca& c+a\\ 1 &ab&a+b\end{vmatrix}\\\\\\=\begin{vmatrix} 1 & bc& b+c-(a+b+c)\\ 1 & ca& c+a-(a+b+c)\\ 1 &ab&a+b-(a+b+c)\end{vmatrix}~~~~~~~~~~~~~~~;c_3\rightarrow c_3 -(a+b+c)c_1\\\\\\=\begin{vmatrix} 1 & bc& -a\\ 1 & ca& -b\\ 1 &ab&-c\end{vmatrix}\\\\\\=\dfrac 1{abc}\begin{vmatrix} a & abc& -a^2\\ b & abc& -b^2\\ c &abc&-c^2\end{vmatrix}\\\\\\=-\dfrac {abc}{abc}\begin{vmatrix} a & 1& a^2\\ b & 1& b^2\\ c &1&c^2\end{vmatrix}\\

=-1\begin{vmatrix} a & 1& a^2\\ b & 1& b^2\\ c &1&c^2\end{vmatrix}\\\\\\=-1 \times -1 \begin{vmatrix} 1 & a& a^2\\ 1 & b& b^2\\ 1 &c&c^2\end{vmatrix}\\\\\\=\begin{vmatrix} 1 & a& a^2\\ 1 & b& b^2\\ 1 &c&c^2\end{vmatrix}\\\\\\=\text{R.H.S}\\\\\text{Showed.}

3 0
2 years ago
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Answer:

Step-by-step explanation:

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8 0
3 years ago
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Answer:mean

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