Step-by-step explanation:
5=x+29
collect like terms
x=29-5
x=24
the answer is 24
hope this helps
Solve your equation step-by-step.
2(x−3)−17=13−3(x+2)
Simplify both sides of the equation.
2(x−3)−17=13−3(x+2)
Simplify
2x−23=−3x+7
Add 3x to both sides.
2x−23+3x=−3x+7+3x
5x−23=7
Add 23 to both sides.
5x−23+23=7+23
5x=30
Divide both sides by 5.
5x/5 = 30/5
x=6
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
![\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)
The third row tells us that
. Then in the second row,

and in the first row,

Answer:
9
Step-by-step explanation:
equation: QRI+IRS=QRS
calculate it and you get x=9