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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 41

In Exercises 31–42, solve by the method of your choice. Identify systems with no solution and systems with infinitely many solutions, using set notation to express their solution sets. 2x = 3y + 4 4x = 3 - 5y

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Rewrite both equations in a standard form to make them easier to work with. For the first equation, 2x = 3y + 4, subtract 3y and 4 from both sides to get 2x - 3y = 4. For the second equation, 4x = 3 - 5y, add 5y to both sides and subtract 3 to get 4x + 5y = 3.
Choose a method to solve the system: substitution or elimination. Here, elimination might be efficient. Multiply the first equation by 2 to align the coefficients of x: 2(2x - 3y) = 2(4) which simplifies to 4x - 6y = 8.
Now subtract the second equation 4x + 5y = 3 from the new equation 4x - 6y = 8 to eliminate x. This gives (4x - 6y) - (4x + 5y) = 8 - 3, simplifying to -11y = 5.
Solve for y by dividing both sides by -11: y = \(\frac{5}{-11}\) = -\(\frac{5}{11}\).
Substitute the value of y back into one of the original equations, for example 2x - 3y = 4, to solve for x. Replace y with -\(\frac{5}{11}\) and solve the resulting equation for x.

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