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2 years ago

Solve this its not 48

Mathematics

1 answer:

algol132 years ago

7 0

$\begin{cases} 3x+5y=50\\ x+5y=31\\[-0.5em] \hrulefill\\ x+5y=31\\ x = 31-5y \end{cases}\qquad \implies \stackrel{\textit{substituting in the 1st equation}}{3(31-5y)+5y=50} \\\\\\ 93-15y+5y=50\implies 93-10y=50\implies -10y=-43 \\\\\\ y=\cfrac{-43}{10}\implies \boxed{y=\cfrac{43}{10}} \\\\[-0.35em] ~\dotfill\\\\ x = 31-5y\implies x = 31-5\left( \frac{43}{10} \right)\implies x = 31-\cfrac{43}{2}\implies \boxed{x = \cfrac{19}{2}}$

$\stackrel{~\hfill \textit{\large total for 7x + 5y}}{7\left( \cfrac{19}{2} \right)+5\left( \cfrac{43}{10} \right)\implies \cfrac{133}{2}+\cfrac{43}{2}\implies \cfrac{133+43}{2}\implies \cfrac{176}{2}\implies 88}$

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3 / m + 4 - 4 / m = 6 (i) Show that this equation can be written as 6m^2 + 25m+16= 0

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Answer:

Proved Below

Step-by-step explanation:

$\sf \frac{3}{m+4} - \frac{4}{m} = 6\\LCM = m(m+4)\\Multiplying \ both \ sides \ by\ m(m+4)\\3m - 4(m+4) = 6m(m+4)\\3m-4m-16 = 6m^2+24m\\-m-16 = 6m^2+24m\\Adding\ m\ to\ both\ sides\\-16 = 6m^2+24m+m\\-16 = 6m^2+25m\\Adding \ 16 \ to \ both \ sides\\0 = 6m^2+25m+16\\OR\\6m^2+25m+16 = 0$

Hence, Proved that $\sf \frac{3}{m+4} - \frac{4}{m} = 6$ is equivalent to $\sf 6m^2+25m+16 = 0$ .

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3 years ago

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What is the justification for each step in the solution of the equation?

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For this case we have the following equation:

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Answer:

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Step-by-step explanation:

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$\boxed{gradient = \frac{y1 -y 2}{x1 - x2} }$

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$= \frac{6}{12}$

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= -1 ÷(½)

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