Algebra Know-It-ALL

654 Worked-Out Solutions to Exercises: Chapters 11 to 19

7. For reference, here again is the second equation stated in Prob. 1:

2 x− 5 y−z=− 1

The solutions for x and y can be plugged into this equation, and it can then be solved for

z in steps, as follows:

2 × (−209/223)+ 5 × (−18/223)−z=− 1

−418/223+ 90/223 −z=− 1

−328/223−z=− 1

−z=− 1 + 328/223

−z= 105/223

z=−105/223

This is the same solution we obtained in the chapter text.

8. All of these equations are in SI form. We know that their graphs must all be straight
   lines, because they are linear equations. All the slopes are different, but all the
   y-intercepts are equal to 1. Therefore, although no two of the lines coincide, all four
   of them pass through the point (0, 1). We can conclude that the system therefore has
   a unique solution: x= 0 and y= 1.


9. Let’s go through each equation, plugging in x= 0 and y= 1, and then grinding out the
   arithmetic. Here’s the first equation:

y=−x+ 1

1 =− 0 + 1

1 = 0 + 1

1 = 1

Check! Now the second equation:

y=− 2 x+ 1

1 =− 2 × 0 + 1

1 = 0 + 1

1 = 1

Check! Now the third:

y= 3 x+ 1

1 = 3 × 0 + 1

1 = 0 + 1

1 = 1