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Nataly [62]
3 years ago
12

What is the solution of the system. Use the substution method. 3x+2y=11, x-2=-4y

Mathematics
1 answer:
liq [111]3 years ago
3 0

Answer:

x=\frac{10}{3} and y=\frac{1}{2}

Step-by-step explanation:

  • To use the substitution method, we should take one of the equations, and obtain an expression for one of the variables from this equation. Let's take the first one: 3x-2y = 11. We obtain an expression for x (it could be for y as well):  3x = 11 + 2 y ⇒ x=\frac{11}{3}+\frac{2}{3} \times{y}.
  • Now, we should replace the expression obtain for x in the first equation, on the second equation, as follow
  • x-2=4y⇒ (\frac{11}{3}+\frac{2}{3}\times{y})-2=4\times{y}
  • Then rearranging terms \frac{11}{3}-2=4\times{y}-\frac{2}{3}\times{y}
  • This expression results in y=\frac{1}{2}
  • Finally, using the firs equation, in which we obtained an expression from x, x=\frac{11}{3}-\frac{2}{3} \times{y}, we replace y=1/2 and get x=\frac{11}{3}- \frac{2}{3}\times{\frac{1}{2} }, then x=\frac{10}{3}

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Eliminate the x and y
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Step-by-step explanation:

Eliminate the x:

\left\{\begin{array}{ccc}x-2y=6\\3x+y=4\end{array}\right\\\\\text{Multiply by (-3)}\\\\\left\{\begin{array}{ccc}-3(x-2y)=(-3)(6)\\3x+y=4\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}-3x+6y=-18\\3x+y=4\end{array}\right}\\.\qquad\qquad7y=-14\\.\qquad\qquad y=-2

Eliminate the y:

\left\{\begin{array}{ccc}x-2y=6\\3x+y=4\end{array}\right\\\\\text{Multiply by 2}\\\\\left\{\begin{array}{ccc}x-2y=6\\2(3x+y)=(2)(4)\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}x-2y=6\\6x+2y=8\end{array}\right}\\\\.\qquad7x=14\\.\qquad x=2

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Explain how you can find the missing number and 3 4/5 divided by something equals 2 5/7 then find the missing number
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let's firstly, convert the mixed fractions to improper, and then do equation.


\bf \stackrel{mixed}{3\frac{4}{5}}\implies \cfrac{3\cdot 5+4}{5}\implies \stackrel{improper}{\cfrac{19}{5}}
~\hfill
\stackrel{mixed}{2\frac{5}{7}}\implies \cfrac{2\cdot 7+5}{7}\implies \stackrel{improper}{\cfrac{19}{7}}
\\\\[-0.35em]
\rule{34em}{0.25pt}


\bf \cfrac{~~\frac{19}{5}~~}{x}=\cfrac{19}{7}\implies \cfrac{~~\frac{19}{5}~~}{\frac{x}{1}}=\cfrac{19}{7}\implies \cfrac{19}{5}\cdot \cfrac{1}{x}=\cfrac{19}{7}\implies \cfrac{19}{5x}=\cfrac{19}{7}
\\\\\\
\stackrel{}{\cfrac{7}{5x}=\cfrac{19}{19}}\implies \cfrac{7}{5x}=1\implies 7=5x\implies \cfrac{7}{5}=x\implies 1\frac{2}{5}=x

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