Write the matrix equation for the system, and then solve.
Just write out the values. We see we have a `2x` so the value is 2. We have `4y` so the value is 4. And it continues. You would leave the value in their place.\[\left[\begin{matrix}2 & 4 &|~3 \\ 2 & 3 & |~1\end{matrix}\right]\] That is how one looks.
then what?
Use the equation: X=(A^-1)B Then find the determinant: 6-8=-2 (Sorry, this is suppose to have brackets, but I don't know how to do that...) [2 4 X [3 Then find the inverse: A-^1=1/det [2 4 2 3] Y = 1] 2 3] A^-1= 1/-2 x [2 4 2 3] Inverse = [-1 -2 -1 -3/2] X [-1 -2 [3 Y = -1 -3/2] 1] x, y = (-5, 4.5)
I'm pretty sure this is right, but I didn't check over my answer.
thats not one of the options
It's not? Could you write out what the options are?
I'm sorry, for the answer, it's actually (-5, -4.5) or (-5, -9/2)
well, i actually did the problem in another lesson and saw its actually -2.5, 2
Oh, okay, I probably messed up somewhere in the middle. Sorry! I'm glad you know the answer though!!
its okay im so glad you actually helped me tho!
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