Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
5. Working With Sets
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Exercise 13 Page 197

Start by solving the inequality.

{ y | y≥ 4}

Practice makes perfect
Before we can write the inequality in set-builder notation, we have to solve the inequality by isolating y.
4y+7≥23
4y≥ 16
y≥ 4
To use set-builder notation we need a variable, which in this case is y, and a statement describing the elements of the set. Knowing that y≥4 is the solution to the inequality, we can write the solution in set-builder notation. { y | y≥ 4} This can be read as the set includes any value of y such that y is greater than or equal to 4.

Extra

Why It Works?

Let's focus on why we could subtract 7 to both sides of the inequality, then divide both sides by 4, and the final answer is correct. First, we will recall the Subtraction Property of Inequality.

Subtraction Property of Inequality

Subtracting the same number from both sides of an inequality creates an equivalent inequality. This equivalent inequality will have the same solution as the original inequality.

Because of this, subtracting 7 from both sides of the inequality does not change the answer. The next property we used is the Division Property of Inequality.

Division Property of Inequality

Dividing both sides of an inequality by the same number creates an equivalent inequality. However, if the number is positive, the inequality sign remains the same. If the number is negative, the inequality sign needs to be reversed to create an equivalent inequality.

Notice that in our solution we divided both sides of the equation by a positive number, 4. This means, that the inequality y ≥ 4 has the same answer set as the original inequality 4y ≥ 16.