Answer:

Step-by-step explanation:
To find x₁ and x₂ :
![\left[\begin{array}{ccc}-4&1\\5&4\\\end{array}\right] \times \left[\begin{array}{ccc}x_1\\x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%261%5C%5C5%264%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%5C%5Cx_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%5C%5C-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step 1
Multiply first 2 x 2 matrix with 2 x 1 vector, we get
![\left[\begin{array}{ccc}-4x_1&+ x_2\\5x_1&+ 4x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4x_1%26%2B%20%20x_2%5C%5C5x_1%26%2B%20%204x_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%5C%5C-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5Cend%7Barray%7D%5Cright%5D)
Step 2
Add the 2 x 1 matrices on LHS, we get
![\left[\begin{array}{ccc}-4x_1&+x_2&+11\\5x_1&+4x_2&-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4x_1%26%2Bx_2%26%2B11%5C%5C5x_1%26%2B4x_2%26-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5Cend%7Barray%7D%5Cright%5D)
Step 3,
we get

and

Step 4,
Simplify, we get

Step 5,
multiply eqn(1) by 4
we get

Step 6,
eqn (2) - eqn(3)
we get

substituting in eqn (1), we get

so, we get

Therefore

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