Question

What choice is a solution to the system of equations? y=2x-3 y-9=-x

Answer 1

`(4,5)\ or\ x=4,y=5`

Explanation:

`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=(-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\\`

`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)`

Answer 2

The solution of the system of equations will be:

`y=5,\:x=4`

Explanation:

Given the system of equation

`\begin{bmatrix}y=2x-3\\ y-9=-x\end{bmatrix}`

Arrange equation variables for elimination

`\begin{bmatrix}y-2x=-3\\ y+x=9\end{bmatrix}`

so

`y+x=9`

`-`

`\underline{y-2x=-3}`

`3x=12`

so the system of equations becomes

`\begin{bmatrix}y-2x=-3\\ 3x=12\end{bmatrix}`

solve 3x=12 for x

`3x=12`

Divide both sides by 3

`(3x)/(3)=(12)/(3)`

`x=4`

`\mathrm{For\:}y-2x=-3\mathrm{\:plug\:in\:}x=4`

`y-2\cdot \:4=-3`

`y-8=-3`

Add 8 to both sides

`y-8+8=-3+8`

`y=5`

Thus, the solution of the system of equations will be:

`y=5,\:x=4`