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I can solve from inspection (1,1,0) that is not what is needed here...must use "elimination"
I'm just confused on how to show my work because I know that x=1 y=1 z=0. But I figured that out by plugging in numbers for the equations. But it says use elimination... And I'm not really sure how to do that.
eliminate x from the first and last equation by adding them: -2x+2y+3z=0 2x +3y+3z=5 -------------------- 5y + 6z=5 First equation without x Now eliminate x again by subtracting the 2nd equation from the first one: -2x+2y+3z=0 -2x-y+z=-3 ------------------ 3y+2z=3 Second equation with out x Now focus on the first and second equation with out x 5y + 6z =5 and 3y+ 2z =3 We will now eliminate z, we do this by multiplying the second equation by 3 5y + 6z = 5 9y + 6z = 9 subtract getting -4y = -4 y = -4/-4 = 1 Now that you have a value for y, simply substitute that value for y in one of the equations that have no x and solve for z. finally subtitute the values obtained for y and z in one of the original equations and solve for x.
The process of "elimination" is to select one of the variables (in this case x, y, or z) and use algebraic manipulation to get the coefficients of the selected variable (in this case I selected x) equal and to either add or subtract (depends on the sign) and get a result of 0 thus "eliminating" that variable in the resulting sum or differencel
Thank you so much!!! @radar
U r welcome, and the last thing you should do is verify your solution by checking that the values obtained will work in all of the provided original equations. Good luck with your studies.
Okay will do! Thanks again :)
In this case x=1, y=1 and z=0 will work.