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nikklg [1K]
3 years ago
14

determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution. x+y=5 2x-y=-

3
Mathematics
2 answers:
max2010maxim [7]3 years ago
7 0

Answer:

(2/3, 13/3)    (Exactly one solution)

Step-by-step explanation:

Write these equations in a column:

 x+y=5

2x-y=-3

Note that we can eliminate y immediately by adding these two equations together. We get:

3x = 2, so that x = 2/3.

Substituting 2/3 for x in the first equation, we get:

2/3 + y = 5.  Clear out fractions by multiplying all three terms by 3:

2 + 3y = 15, or 3y = 13.  Then y = 13/3, and the solution is

(2/3, 13/3)      There is exactly one solution.

Murljashka [212]3 years ago
4 0

Answer:

Hi! The answer to your question is One Solution.

Step-by-step explanation:

☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆☆*: .。..。.:*☆

☆Brainliest is greatly appreciated!☆

Hope this helps!!

- Brooklynn Deka

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Rina8888 [55]

Answer:

4.255 ft²

Step-by-step explanation:

the area of the figure =

(2.25 × 1.5)+(1/8 ×3.14× 1.5²)

= 3.375 + 0.88

= 4.255 ft²

6 0
2 years ago
-3.8x - 5.9x = 223.1
katrin2010 [14]

Answer:

x=-23

Step-by-step explanation:

-3.8x-5.9x=223.1

-9.7x=223.1

x=223.1/-9.7

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3 years ago
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valentina_108 [34]

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3 0
3 years ago
Justin has $500 in a savings account at the beginning of the summer. He wants to have at least $300 in the account by the end of
SSSSS [86.1K]

Answer:

10 weeks

Step-by-step explanation:

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3 0
3 years ago
Explain how to multiply the following whole numbers 21 x 14
Lesechka [4]

Answer:

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

________

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

Step-by-step explanation:

Given

21\:\times \:14

Line up the numbers

\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

Multiply the bold numbers:    1×4=4

\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}

Multiply the bold numbers:    2×4=8

\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}

Multiply the top number by the bolded digit of the bottom number

\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}

Multiply the bold numbers:    1×1=1

\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}

Multiply the bold numbers:    2×1=2

\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}

Add the rows to get the answer. For simplicity, fill in trailing zeros.

\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}

adding portion

\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}

Add the digits of the right-most column: 4+0=4

\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}

Add the digits of the right-most column: 8+1=9

\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}

Add the digits of the right-most column: 0+2=2

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

Therefore,

\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}

________

\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}

6 0
3 years ago
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