ANSWERS: 2

I'm not going to solve the entire problem for you, but I'll give you an explanation that should allow you to do it yourself. First create a matrix with the coefficients on the variables of each of the three equations. Place the solutions in the 4th column, creating a 3x4 matrix. Second, use row operations on the matrix until the first three columns are in reduced echelon form. The first row will be 1 0 0 followed by a number, the second row will be 0 1 0 followed by a number, and the third row will be 0 0 1 followed by a number. At this point you have solved the system. x, y, and z are equal to the final numbers in rows one, two, and three, respectively. Ask in comments if you need more help.

This is just like the old fashioned elimination you used to do, but now you have one more variable and one more equation. You start by choosing two equations and one variable to eliminate. You figure out by what you must multiply each equation in order to add them together and have one variable dissappear. Let's choose to use the last two equations and eliminate y because that looks easiest: eqn 2: 2xy+8z=13 (2)(2xy+8z)=(13)(2) 4x+2y16z=26 Now we can add equations 2 and 3 to eliminate y. eqn 2: 4x+2y16z=26 eqn 3: 5x2y+7z=20 + ___________________ x+0y9z=6 x9z=6 (we know this is true because the amount we added to each side of equation 2 was equal) Now, we will use our new equation and our first equation to eliminate another variable. new eqn: x9z=6 (3)(x9z)=(6)(3) 3x+27z=18 We are ready to add equation 1 and the new equation. eqn one: 3x+4y+5z=18 new eqn: 3x+27z=18 + _____________________ 0x+32z=36 32z=36 Now, you can use this newest equation to solve for z. Once you find z, you can substitute it into the "new equation" because it only has two variables and you will know one of them. With the "new equation", you can solve for x. Once you find x, you can use any of your original three equations to solve for y. Once you find all three variables, check them in all three equations to make sure you did it right.
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