- Thread starter frctl
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First subtract \(\displaystyle \begin{pmatrix}3\\4\end{pmatrix}\) from both sides then the first equation is

\(\displaystyle 1 \cdot x_1 + 2 \cdot x_2 = 6\)

I leave you to figure out the second equation

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View attachment 17443

I require help with this system the way it is presented

Where does the x_{1}and x_{2}go or with which coefficient?

\(\left[ {\begin{array}{*{20}{c}} 1&2 \\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\View attachment 17443

I require help with this system the way it is presented

Where does the x_{1}and x_{2}go or with which coefficient?

y \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3 \\ 4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

9 \\ 9 \end{array}} \right]\) Subtract from both sides.

\(\left[ {\begin{array}{*{20}{c}} 1&2 \\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\

y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

6 \\ 5 \end{array}} \right]\) Find the inverse.

\(\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 1&2 \\ 1&1 \end{array}} \right]^{-1} \left[ {\begin{array}{*{20}{c}} 6 \\ 5 \end{array}} \right]\)

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Now check your answer!Ah wait if I multiply them out I obtain

x_{1}= 4

x_{2}= 1

and be quick about it!Now check your answer!

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Go on, frctl, don't be obtuse. Substitute the values into the equation in the OP. See if they work.How do I check my answer in this case

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That's not correct, frctl. You already posted your solution: xthe matrix

x_{1}and x_{2}

The second equation in post #6 shows the matrix containing elements 6 and 5.

We need to be more careful with language.