Mathematics: When would someone need a calculate area

Step-by-step explanation:Question-1:by order pair we obtain:cancel 2p from the i equation to get a certain value of p:now substitute the value of p to the second equation:simplify parentheses:cancel 1 from both sides:divide both sides by 2:question-2:by order pair we obtain:cancel out y from the second equation:substitute the value of x to the first equation:simplify:divide both sides by 3:substitute the value of y to the second equation which yields:Question-3:by order pair we obtain;rearrange:subtract...

When would someone need a calculate area - id13188735, imhacking2048

Ответ:

netflixacc0107

19.01.2021

Mathematics

Step-by-step explanation:

Question-1:

by order pair we obtain:

$\displaystyle \begin{cases} \displaystyle 3p = 2p - 1 \dots \dots i\\2q - p = 1 \dots \dots ii\end{cases}$

cancel 2p from the i equation to get a certain value of p:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q - p = 1 \end{cases}$

now substitute the value of p to the second equation:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q - ( - 1) = 1 \end{cases}$

simplify parentheses:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q + 1= 1 \end{cases}$

cancel 1 from both sides:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q = 0\end{cases}$

divide both sides by 2:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\q = 0\end{cases}$

question-2:

by order pair we obtain:

$\displaystyle \begin{cases} \displaystyle 2x - y= 3 \dots \dots i\\3y= x + y \dots \dots ii\end{cases}$

cancel out y from the second equation:

$\displaystyle \begin{cases} \displaystyle 2x - y= 3 \dots \dots i\\ x = 2y \dots \dots ii\end{cases}$

substitute the value of x to the first equation:

$\displaystyle \begin{cases} \displaystyle 2.2y-y= 3 \\ x = 2y \end{cases}$

simplify:

$\displaystyle \begin{cases} \displaystyle 3y= 3 \\ x = 2y \end{cases}$

divide both sides by 3:

$\displaystyle \begin{cases} \displaystyle y= 1 \\ x = 2y \end{cases}$

substitute the value of y to the second equation which yields:

$\displaystyle \begin{cases} \displaystyle y= 1 \\ x = 2 \end{cases}$

Question-3:

by order pair we obtain;

$\displaystyle \begin{cases} \displaystyle 2p + q = 2 \dots \dots i\\3q + 2p = 3 \dots \dots ii\end{cases}$

rearrange:

$\displaystyle \begin{cases} \displaystyle 2p + q = 2 \\2p + 3q= 3 \end{cases}$

subtract and simplify

$\displaystyle \begin{array}{ccc} \displaystyle 2p + q = 2 \\2p + 3q= 3 \\ \hline - 2q = - 1 \\ q = \dfrac{1}{2} \end{array}$

substitute the value of q to the first equation:

$\displaystyle 2.p+ \frac{1}{2} = 2$

make q the subject of the equation:

$\displaystyle p = \frac{3}{4}$

hence,

$\displaystyle q = \frac{1}{2} \\ p = \frac{3}{4}$

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When would someone need a calculate area

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