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Stitz-Zeager_College_Algebra_e-book

# 1 1 3 0 1 2 1 1 3 3 3 1 replace r3 with 24 r3 19 1 6

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Unformatted text preview: check our answer graphically, we graph the lines 2x − y = 1 and y = 3 and verify that they intersect at (2, 3). 2. To solve the second system, we use the addition method to eliminate the variable x. We take the two equations as given and ‘add equals to equals’ to obtain 3x + 4y = −2 + (−3x − y = 5) 3y = 3 This gives us y = 1. We now substitute y = 1 into either of the two equations, say −3x − y = 5, to get −3x − 1 = 5 so that x = −2. Our solution is (−2, 1). Substituting x = −2 and y = 1 into the ﬁrst equation gives 3(−2) + 4(1) = −2, which is true, and, likewise, when we check (−2, 1) in the second equation, we get −3(−2) − 1 = 5, which is also true. Geometrically, the lines 3x + 4y = −2 and −3x − y = 5 intersect at (−2, 1). 8.1 Systems of Linear Equations: Gaussian Elimination 451 y y 4 2 (2, 3) 1 (−2, 1) 2 x −4 −3 −2 −1 1 −1 −1 1 2 3 4 x −2 3x + 4y = −2 −3 x − y = 5 2x − y = 1 y=3 3. The equations in the third system are more approachable if we clear denominators....
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