In matrix form, the system is given by
I'll use G-J elimination. Consider the augmented matrix
![\left[ \begin{array}{ccc|c} -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%20-1%20%26%201%20%26%20-1%20%26%20-20%20%5C%5C%202%20%26%20-1%20%26%201%20%26%2029%20%5C%5C%203%20%26%202%20%26%201%20%26%2029%20%5Cend%7Barray%7D%20%5Cright%5D)
• Multiply through row 1 by -1.
![\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%20-1%20%26%201%20%26%2020%20%5C%5C%202%20%26%20-1%20%26%201%20%26%2029%20%5C%5C%203%20%26%202%20%26%201%20%26%2029%20%5Cend%7Barray%7D%20%5Cright%5D)
• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.
![\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%20-1%20%26%201%20%26%2020%20%5C%5C%200%20%26%201%20%26%20-1%20%26%20-11%20%5C%5C%200%20%26%205%20%26%20-2%20%26%20-31%20%5Cend%7Barray%7D%20%5Cright%5D)
• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.
![\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%20-1%20%26%201%20%26%2020%20%5C%5C%200%20%26%201%20%26%20-1%20%26%20-11%20%5C%5C%200%20%26%200%20%26%203%20%26%2024%20%5Cend%7Barray%7D%20%5Cright%5D)
• Multiply row 3 by 1/3.
![\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%20-1%20%26%201%20%26%2020%20%5C%5C%200%20%26%201%20%26%20-1%20%26%20-11%20%5C%5C%200%20%26%200%20%26%201%20%26%208%20%5Cend%7Barray%7D%20%5Cright%5D)
• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.
![\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%20-1%20%26%201%20%26%2020%20%5C%5C%200%20%26%201%20%26%200%20%26%20-3%20%5C%5C%200%20%26%200%20%26%201%20%26%208%20%5Cend%7Barray%7D%20%5Cright%5D)
• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).
![\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cc%7D%201%20%26%200%20%26%200%20%26%209%20%5C%5C%200%20%26%201%20%26%200%20%26%20-3%20%5C%5C%200%20%26%200%20%26%201%20%26%208%20%5Cend%7Barray%7D%20%5Cright%5D)
Then the solution to the system is
If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix
The third row tells us that 
. Then in the second row,
and in the first row,
If you want to estimate to the tens, the answer is 680
If you want to estimate to the hundreds, the answer is 700
If you want to estimate to the thousands then the answer would be not an answer
The real answer without estimating is 681
If two figures are similar, the resulting corresponding angles are always congruent.
(For full answer you might have to go to the comments)
Answer: 28x+30
Explanation: we divide (3x^3-2x^2+4x-3) by (x^2+3x+3) Using long division
3x-11
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(x^2+3x+3) 3x^3-2x^2+4x-3
-(3x^3+9x^2+9x)
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-11x^2-5x-3
-(-11x^2-33x-33)
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28x+30
So our remainder will be 28x+30
Answer:
They are (6, 1) or (3, 2).
Step-by-step explanation:
The factor pairs of 6 are 1×6 and 2×3
So the parallelogram (height by length) could be:
1 by 6
2 by 3
3 by 2
6 by 1.
