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Exemplo 5.28 de Elementary Algebra, 5.3 Solve Systems of Equations by Elimination
Solve Systems of Equations by Elimination: exemplo 5.28
Problema e solução em inglês, como no livro original.
Solve the system by elimination.
Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.
Solução
In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by a constant to get the opposites.
Both equations are in standard form. To get opposite
coefficients of y, we will multiply the first equation by 2
and the second equation by 3.Simplify. Add the two equations to eliminate y. Solve for x.
Substitute x = 0 into one of the original equations.Solve for y. Write the solution as an ordered pair. The ordered pair is (0, −3). Check that the ordered pair is a solution to
both original equations.The solution is (0, −3). What other constants could we have chosen to eliminate one of the variables? Would the solution be the same?
Sua vez
O Try It do livro logo depois deste exemplo: mesma ideia, outros números. Só a resposta é dada.
Solve the system by elimination.
Como foi?