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Exercise 2.3 · Q29

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

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Step 1: x+3y=2, 3x+5y=4x+3y=2,\ 3x+5y=4 becomes [1335][xy]=[24]\begin{bmatrix}1&3\\3&5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}2\\4\end{bmatrix}.

Step 2: Use R2→R2−3R1R_2\to R_2-3R_1: new row 2 of AA is (3−3(1), 5−3(3))=(0,−4)(3-3(1),\ 5-3(3))=(0,-4); new entry of BB is 4−3(2)=−24-3(2)=-2. …

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