3 years ago

How to solve y+z=4 and y-z=8 using elimination

Mathematics

1 answer:

Levart [38]3 years ago

4 0

Hope this helps with ur question

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To find the equation we will put the values of x and y in the equations and then check if LHS=RHS.

$\large\red{\sf{y=3x-1}}$

$\begin{gathered}\\ \sf\longmapsto 9=3(1)-1\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9=3-1\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9≠2 \end{gathered}$

$\large\green{\sf{y=5x+4}}$

$\begin{gathered}\\ \sf\longmapsto 9=5(1)+4\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9=5+4\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9=9\end{gathered}$

$\large\blue{\sf{y=-2x-8}}$

$\begin{gathered}\\ \sf\longmapsto 9=-2(1)-8\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9=-2-8\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9≠10\end{gathered}$

$\large\pink{\sf{y=6x-8}}$

$\begin{gathered}\\ \sf\longmapsto 9=6(1)-8\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9=6-8\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 9≠-2\end{gathered}$

Hence, Option B. y=5x+4 satisfies the system (1,9)

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