WHEELS
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Solve:
- Let x be the number of motorcycles
- Let y be the number of vans
[tex]\large \tt \red{given} \begin{cases} \sf \: x + y = 6 \tt \red{ \: \: \: \: \: \: \: \: \: \: (eq.1)} \\ \sf \: 2x + 4y = 16 \: \: \: \tt \red{(eq.2)} \end{cases} \\ \\ [/tex]
Find the the value of x as the substitution. Use the first equation.
[tex] \large \begin{cases} \begin{align} \sf \: x + y &= \sf 6 \\ \sf \: x& = \sf6-y\end{align} \end{cases}[/tex]
Now substitute x as 6 - y. Use the second equation.
[tex] \begin{align}&\large \begin{cases} \begin{align} \sf \: 2x + 4y&= \sf 16 \\ \sf \: 2(6 - y) + 4y& = \sf16 \\ \sf \: 12 - 2y + 4y& = \sf16 \\ \sf \: 12 + 2y& = \sf16 \end{align} \end{cases} \\ & \large \begin{cases} \begin{align} \sf \: 2y& = \sf16 - 12 \\ \sf \: 2y& = \sf4 \\ \sf \: \frac{ \cancel2y}{ \cancel2} & = \sf \frac{4}{2} \\ \sf \orange{ y}& \orange{= \sf2} \end{align} \end{cases} \end{align}[/tex]
Use the first transposed equation, substitute y as 2.
[tex] \large \begin{cases} \begin{align} \sf \: x& = \sf6-y \\ \sf \: x& = \sf6-2 \\ \sf \: \orange{ x}& \orange{ = \sf4}\end{align} \end{cases}[/tex]
[tex] \: [/tex]
Final Answer:
[tex] \large \implies \sf \LARGE \orange{4 \: motorcycles}[/tex]
[tex] \large \implies \sf \LARGE \orange{2 \: vans}[/tex]
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