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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
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Tutti i libri di testoBlitzer 8th EditionCh. 5 - Systems of Equations and InequalitiesProblema 9
Capitolo 6, Problema 9

In Exercises 5–18, solve each system by the substitution method. {x=4y−2x=6y+8\(\begin{cases}\) x = 4y - 2 \\ x = 6y + 8 \(\end{cases}\)

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Since both equations are equal to \(x\), set the right-hand sides of the equations equal to each other: \(4y - 2 = 6y + 8\).
Next, solve the equation \(4y - 2 = 6y + 8\) for \(y\). Start by subtracting \$4y$ from both sides to get $-2 = 2y + 8$.
Then, subtract 8 from both sides to isolate the term with \(y\): \(-2 - 8 = 2y\), which simplifies to \(-10 = 2y\).
Divide both sides by 2 to solve for \(y\): \(y = \frac{-10}{2}\).
Finally, substitute the value of \(y\) back into either original equation (for example, \(x = 4y - 2\)) to find the corresponding value of \(x\).

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System of Linear Equations

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Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one variable, making it easier to solve. It is especially useful when one variable is already isolated.
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Solving linear equations means finding the value(s) of the variable(s) that make the equation true. This often involves simplifying expressions, isolating variables, and performing arithmetic operations. Mastery of these skills is essential for solving systems effectively.
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