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x D 0 =D 78=26 3; answer; the value of x: To nd the value of y we must now rework step 3 to nd the new value of D 0 ; thus   5 D   3
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 7   125 21 104 25 
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thus, repeating step 4, y D 0 =D 104=26 4, answer, the value of y. To check the correctness of the answers, go back to the original two equations and set x 3 and y 4. Doing this we nd that 5 3 2 4 7 which is true ; 3 3 4 4 25 which is true ; showing that the answers are correct. It should be noted that x 3 and y 4 are the only values of x and y that simultaneously satisfy both of the given equations. 41. Step 1 is already satis ed. Next, for step 2 we have (note that x 1 times x)  1   D 1  1 1 1 2  1   1   1 
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Now, using property 7, to each element of column 3 add the corresponding element of column 1 multiplied by 1, thus getting  1   D 1  1 thus, D 6. Step 3: Let us solve for the values of x, y, and z, in that order, as follows.  6   0  3 1 1 2 6  1   1   1    0 5 3       5 1 1  0 3   3   1 1   3 2 1  4; answer; value of x: 6 6  1 2   1  1 0  2 1  2 0   1  6 2 
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(Note: to get the second determinant above from the rst, to each element of row 1 add the corresponding element of row 3 multiplied by 2:
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Next,  1   1  1  1   1  1   1   1     0 1   1   3 1   1 6   1 6   1     1 0  1   2 3   1 6 6 0 0 3 6 5 1 2 6  3   1   1   0   0  3    1 3    3   1 1  1; answer; value of y: 6    1 5    3   1 1 3; answer; value of z: 6
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Hence the answers are x 4, y 1, and z 3. 42. Step 1 is satis ed. Next, for step 2 we have     1 1 1  1 1 1   1    1 5   D 3 4 2   0   4     0 4 5   0 4 5  thus, D 25. Step 3: Let us solve for the values of x, y, and z, in that order, as follows. First, for the value of x, the value of D 0 is    4 1 1   D 0  2 4 2     1 4 5  Now take the following steps: to each element of row 3 add the corresponding element of row 2 multiplied by 1; then, to each element of row 2 add the corresponding element of row 1 multiplied by 4. Doing this, the above expression for D 0 becomes        4 1 1 1  4 1       0  2 4 2   18 0 6   18 6  120 D   1 7      1 0 7   1 0 7  Hence, x D 0 =D 120= 25 4:8, answer, value of x. Next, for the value of y,  1  0 D 3  0 thus, y D 0 =D 4 2 1  1  2  75  5 
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