Sat practice Test #8 Answer Explanations

Choice B is correct. Subtracting the same number from each side of 
an inequality gives an equivalent inequality. Hence, subtracting 1 from 
each side of the inequality 2x > 5 gives 2x − 1 > 4. So the given system 
of inequalities is equivalent to the system of inequalities y > 2x − 1 and 
2x − 1 > 4, which can be rewritten as y > 2x − 1 > 4. Using the transitive 
property of inequalities, it follows that y > 4.

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Choice A is incorrect because there are points with a y-coordinate 
less than 6 that satisfy the given system of inequalities. For example, 
(3, 5.5) satisfies both inequalities. Choice C is incorrect. This may 
result from solving the inequality 2x > 5 for x, then replacing x with y. 
Choice D is incorrect because this inequality allows y-values that are 
not the y-coordinate of any point that satisfies both inequalities. For 
example, y = 2 is contained in the set y > 3 
__ 
2 ; however, if 2 is substituted 
into the first inequality for y, the result is x < 3 
__ 
2 . This cannot be true 
because the second inequality gives x > 5 
__ 
2 .
QUESTION 7
Choice B is correct. Subtracting 4 from both sides of √ 
_
2x + 6 + 4 = x + 3 
isolates the radical expression on the left side of the equation as 
follows: √ 
_
2x + 6 = x – 1. Squaring both sides of √ 
_
2x + 6 = x – 1 yields 
2x + 6 = x 
2
− 2x + 1. This equation can be rewritten as a quadratic 
equation in standard form: x 
2
– 4x – 5 = 0. One way to solve this 
quadratic equation is to factor the expression x 
2
− 4x – 5 by identifying 
two numbers with a sum of –4 and a product of –5. These numbers are 
–5 and 1. So the quadratic equation can be factored as (x – 5)(x + 1) = 0. 
It follows that 5 and –1 are the solutions to the quadratic equation. 
However, the solutions must be verified by checking whether 5 and 
–1 satisfy the original equation, √ 
_
2x + 6 + 4 = x + 3. When x = –1, the 
original equation gives
√ 
_
2(−1) + 6 + 4 = (−1) + 3, or 6 = 2, which is 
false. Therefore, –1 does not satisfy the original equation. When x = 5, 
the original equation gives
√ 
_
2(5) + 6 + 4 = 5 + 3, or 8 = 8, which is true. 
Therefore, x = 5 is the only solution to the original equation, and so 
the solution set is {5}.
Choices A, C, and D are incorrect because each of these sets contains 
at least one value that results in a false statement when substituted 
into the given equation. For instance, in choice D, when 0 is substituted 
for x into the given equation, the result is
√ 
_
2(0) + 6 + 4 = (0) + 3, 
or √ 
_
6 + 4 = 3. This is not a true statement, so 0 is not a solution to the 
given equation.
QUESTION 8