solve this simultaneous equation: 2x + 3y = 19 (Eq1) and 3x + y = 11 (Eq2)
Begin by multiplying the second equation by three in order to get a 3y value: 9x + 3y = 33. We can now subtract equation one from equation two, 9x + 3y = 33 and 2x + 3y = 19 -->7x + 0y = 14. Resulting in, 7x = 14. Solving for x gives us x = 2. By substituting x = 2 into one of the original equations we can the solve for y, (3x2) + y = 11 -->6 + y = 11-->y = 5. Therefore x=2 and y=5.
TB
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