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Ch. R - Algebra Review
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161, 9780135440780, 9780136881117, 9780135440827, 9780135924891Non è quello che usi tu?Cambia libro di testo
Capitolo 1, Problema R.6.33

Determine whether each equation is an identity, a conditional equation, or a contradiction. Give the solution set. See Example 4. 4(2x + 7) = 2x + 22 + 3(2x + 2)

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Start by expanding both sides of the equation to simplify the expressions. Use the distributive property: multiply 4 by each term inside the parentheses on the left side, and multiply 3 by each term inside the parentheses on the right side. This gives you: \(4(2x + 7) = 2x + 22 + 3(2x + 2)\) becomes \(8x + 28 = 2x + 22 + 6x + 6\).
Next, combine like terms on the right side of the equation. Add the terms involving \(x\) and the constant terms separately: \(2x + 6x = 8x\) and \(22 + 6 = 28\). So the equation now looks like \(8x + 28 = 8x + 28\).
Now, analyze the simplified equation. Since both sides are identical expressions, this suggests the equation might be an identity. To confirm, subtract \(8x + 28\) from both sides to see if the equation reduces to a true statement.
After subtracting, you get \(0 = 0\), which is always true regardless of the value of \(x\). This means the original equation holds for all real numbers \(x\).
Therefore, conclude that the equation is an identity, and the solution set is all real numbers, often written as \(\{ x \mid x \in \mathbb{R} \}\).

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