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Answer:
[tex](4,5)\ or\ x=4,y=5[/tex]
Step-by-step explanation:
[tex]We\ are\ given:\\y=2x-3\\y-9=-x\\Now,\ lets\ separate\ all\ the\ constant\ terms\ to\ the\ RHS\ and\ the\ variable\\ terms\ to\ the\ LHS.\\Hence,\\y-2x=-3\\y+x=9\\By\ subtracting\ the\ two\ equations:\\(y-2x)-(y+x)=(-3)-(9)\\Hence\ lets\ simplify\ this\ main\ equation: \\y-2x-(y+x)=-3-9\\y-2x-y-x=-12\\-3x=-12\\x=\frac{-12}{-3}=4\\\\Now,\ lets\ consider\ the\ second\ equation\ in\ the\ solution:\\y-9=-x\\Substituting\ x=4,\\y-9=-(4)\\y-9=-4\\y=-4+9\\y=5\\[/tex]
[tex]Now,\\We\ get\ the\ solution\ to\ this\ system\ of\ equations:\\x=4,y=5\\By\ putting\ this\ as\ a\ co-ordinate\ pair (x,y):\\(4,5)[/tex]
Answer:
The solution of the system of equations will be:
[tex]y=5,\:x=4[/tex]
Step-by-step explanation:
Given the system of equation
[tex]\begin{bmatrix}y=2x-3\\ y-9=-x\end{bmatrix}[/tex]
Arrange equation variables for elimination
[tex]\begin{bmatrix}y-2x=-3\\ y+x=9\end{bmatrix}[/tex]
so
[tex]y+x=9[/tex]
[tex]-[/tex]
[tex]\underline{y-2x=-3}[/tex]
[tex]3x=12[/tex]
so the system of equations becomes
[tex]\begin{bmatrix}y-2x=-3\\ 3x=12\end{bmatrix}[/tex]
solve 3x=12 for x
[tex]3x=12[/tex]
Divide both sides by 3
[tex]\frac{3x}{3}=\frac{12}{3}[/tex]
[tex]x=4[/tex]
[tex]\mathrm{For\:}y-2x=-3\mathrm{\:plug\:in\:}x=4[/tex]
[tex]y-2\cdot \:4=-3[/tex]
[tex]y-8=-3[/tex]
Add 8 to both sides
[tex]y-8+8=-3+8[/tex]
[tex]y=5[/tex]
Thus, the solution of the system of equations will be:
[tex]y=5,\:x=4[/tex]