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Exercise 2.3 · Q31

Q.Solve the following equations by reduction method. 5x+2y=4, 7x+3y=55x + 2y = 4,\ 7x + 3y = 5

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Step 1: 5x+2y=4, 7x+3y=55x+2y=4,\ 7x+3y=5 becomes [5273][xy]=[45]\begin{bmatrix}5&2\\7&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}4\\5\end{bmatrix}.

Step 2: Use R2→5R2−7R1R_2\to 5R_2-7R_1 (clears (2,1): 5(7)−7(5)=05(7)-7(5)=0): new row 2 of AA is (0, 5(3)−7(2))=(0,1)(0,\ 5(3)-7(2))=(0,1); new entry of BB is 5(5)−7(4)=25−28=−35(5)-7(4)=25-28=-3. …

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