Equations / Inequalities - GRE Quantitative Reasoning
Card 1 of 1560
and 
are both integers.
If 
, 
, and 
, which of the following is a possible value of 
?
If
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Take the values of y that are possible, i.e. 2 and 3, and plug them into the first inequality. First, plug in 2. 2 – 3x > 21. Subtract 2 from both sides, and then divide by –3. Don't forget that when you divide or multiply by a negative number in an inequality you must flip the inequality sign. Thus, x < –19/3. Now plug in 3. We find, following the same steps, that when y=3, x < –6. Thus –7 is the correct answer.
Take the values of y that are possible, i.e. 2 and 3, and plug them into the first inequality. First, plug in 2. 2 – 3x > 21. Subtract 2 from both sides, and then divide by –3. Don't forget that when you divide or multiply by a negative number in an inequality you must flip the inequality sign. Thus, x < –19/3. Now plug in 3. We find, following the same steps, that when y=3, x < –6. Thus –7 is the correct answer.
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Factor 
.
Factor
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We know the equation a2 – b2 = (a + b)(a – b) for the difference of squares. Since y_2_ is the square of y, and 4 is the square of 2, the correct answer is (y + 2)(y – 2).
We know the equation a2 – b2 = (a + b)(a – b) for the difference of squares. Since y_2_ is the square of y, and 4 is the square of 2, the correct answer is (y + 2)(y – 2).
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x2 – 9X + 18 = 0
Find x
x2 – 9X + 18 = 0
Find x
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factor the equation:
(x – 3)(x – 6) = 0
set each equal to 0
x = 3, 6
factor the equation:
(x – 3)(x – 6) = 0
set each equal to 0
x = 3, 6
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Solve for 
.
Solve for
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Factor the equation by finding two numbers that add to -3 and multiply to -28.
Factors of 28: 1,2,4,7,14,28
-7 and 4 work.
(x-7)(x+4) = 0
Set each equal to zero:
x=7,-4
Factor the equation by finding two numbers that add to -3 and multiply to -28.
Factors of 28: 1,2,4,7,14,28
-7 and 4 work.
(x-7)(x+4) = 0
Set each equal to zero:
x=7,-4
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25_x_2 – 36_y_2 can be factored into:
25_x_2 – 36_y_2 can be factored into:
Tap to reveal answer
This is the difference of squares. You must know this formula for the GRE!
_a_2 – _b_2 = (a – b)(a + b)
Here a = 5_x_ and b = 6_y_, so the difference of squares formula gives us (5_x_ – 6_y_)(5_x_ + 6_y_).
This is the difference of squares. You must know this formula for the GRE!
_a_2 – _b_2 = (a – b)(a + b)
Here a = 5_x_ and b = 6_y_, so the difference of squares formula gives us (5_x_ – 6_y_)(5_x_ + 6_y_).
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Simplify 
.
Simplify
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To begin, let's factor the first two terms and the second two terms separately.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
(z – 1) can be pulled out because it appears in both terms.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1) = (z – 1)(_z_2 – 9)
(_z_2 – 9) is a difference of squares, so we can use the formula _a_2 – _b_2 = (a – b)(a + b).
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
= (z – 1)(_z_2 – 9)
= (z – 1)(z – 3)(z + 3)
To begin, let's factor the first two terms and the second two terms separately.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
(z – 1) can be pulled out because it appears in both terms.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1) = (z – 1)(_z_2 – 9)
(_z_2 – 9) is a difference of squares, so we can use the formula _a_2 – _b_2 = (a – b)(a + b).
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
= (z – 1)(_z_2 – 9)
= (z – 1)(z – 3)(z + 3)
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Factor 3_u_4 – 24_uv_3.
Factor 3_u_4 – 24_uv_3.
Tap to reveal answer
First pull out 3u from both terms.
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember but pat yourself on the back for getting to such hard questions! The difference of cubes formula to remember is _a_3 – _b_3 = (a – b)(_a_2 + ab + b_2). In our problem, a = u and b = 2_v, so
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
= 3_u_(u – 2_v_)(u_2 + 2_uv + 4_v_2)
First pull out 3u from both terms.
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember but pat yourself on the back for getting to such hard questions! The difference of cubes formula to remember is _a_3 – _b_3 = (a – b)(_a_2 + ab + b_2). In our problem, a = u and b = 2_v, so
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
= 3_u_(u – 2_v_)(u_2 + 2_uv + 4_v_2)
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Factor .
Factor
Tap to reveal answer
We know the equation a2 – b2 = (a + b)(a – b) for the difference of squares. Since y_2_ is the square of y, and 4 is the square of 2, the correct answer is (y + 2)(y – 2).
We know the equation a2 – b2 = (a + b)(a – b) for the difference of squares. Since y_2_ is the square of y, and 4 is the square of 2, the correct answer is (y + 2)(y – 2).
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x2 – 9X + 18 = 0
Find x
x2 – 9X + 18 = 0
Find x
Tap to reveal answer
factor the equation:
(x – 3)(x – 6) = 0
set each equal to 0
x = 3, 6
factor the equation:
(x – 3)(x – 6) = 0
set each equal to 0
x = 3, 6
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Solve for .
Solve for
Tap to reveal answer
Factor the equation by finding two numbers that add to -3 and multiply to -28.
Factors of 28: 1,2,4,7,14,28
-7 and 4 work.
(x-7)(x+4) = 0
Set each equal to zero:
x=7,-4
Factor the equation by finding two numbers that add to -3 and multiply to -28.
Factors of 28: 1,2,4,7,14,28
-7 and 4 work.
(x-7)(x+4) = 0
Set each equal to zero:
x=7,-4
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25_x_2 – 36_y_2 can be factored into:
25_x_2 – 36_y_2 can be factored into:
Tap to reveal answer
This is the difference of squares. You must know this formula for the GRE!
_a_2 – _b_2 = (a – b)(a + b)
Here a = 5_x_ and b = 6_y_, so the difference of squares formula gives us (5_x_ – 6_y_)(5_x_ + 6_y_).
This is the difference of squares. You must know this formula for the GRE!
_a_2 – _b_2 = (a – b)(a + b)
Here a = 5_x_ and b = 6_y_, so the difference of squares formula gives us (5_x_ – 6_y_)(5_x_ + 6_y_).
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Simplify .
Simplify
Tap to reveal answer
To begin, let's factor the first two terms and the second two terms separately.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
(z – 1) can be pulled out because it appears in both terms.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1) = (z – 1)(_z_2 – 9)
(_z_2 – 9) is a difference of squares, so we can use the formula _a_2 – _b_2 = (a – b)(a + b).
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
= (z – 1)(_z_2 – 9)
= (z – 1)(z – 3)(z + 3)
To begin, let's factor the first two terms and the second two terms separately.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
(z – 1) can be pulled out because it appears in both terms.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1) = (z – 1)(_z_2 – 9)
(_z_2 – 9) is a difference of squares, so we can use the formula _a_2 – _b_2 = (a – b)(a + b).
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
= (z – 1)(_z_2 – 9)
= (z – 1)(z – 3)(z + 3)
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Factor 3_u_4 – 24_uv_3.
Factor 3_u_4 – 24_uv_3.
Tap to reveal answer
First pull out 3u from both terms.
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember but pat yourself on the back for getting to such hard questions! The difference of cubes formula to remember is _a_3 – _b_3 = (a – b)(_a_2 + ab + b_2). In our problem, a = u and b = 2_v, so
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
= 3_u_(u – 2_v_)(u_2 + 2_uv + 4_v_2)
First pull out 3u from both terms.
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember but pat yourself on the back for getting to such hard questions! The difference of cubes formula to remember is _a_3 – _b_3 = (a – b)(_a_2 + ab + b_2). In our problem, a = u and b = 2_v, so
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
= 3_u_(u – 2_v_)(u_2 + 2_uv + 4_v_2)
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Factor .
Factor
Tap to reveal answer
We know the equation a2 – b2 = (a + b)(a – b) for the difference of squares. Since y_2_ is the square of y, and 4 is the square of 2, the correct answer is (y + 2)(y – 2).
We know the equation a2 – b2 = (a + b)(a – b) for the difference of squares. Since y_2_ is the square of y, and 4 is the square of 2, the correct answer is (y + 2)(y – 2).
← Didn't Know|Knew It →
x2 – 9X + 18 = 0
Find x
x2 – 9X + 18 = 0
Find x
Tap to reveal answer
factor the equation:
(x – 3)(x – 6) = 0
set each equal to 0
x = 3, 6
factor the equation:
(x – 3)(x – 6) = 0
set each equal to 0
x = 3, 6
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Solve for .
Solve for
Tap to reveal answer
Factor the equation by finding two numbers that add to -3 and multiply to -28.
Factors of 28: 1,2,4,7,14,28
-7 and 4 work.
(x-7)(x+4) = 0
Set each equal to zero:
x=7,-4
Factor the equation by finding two numbers that add to -3 and multiply to -28.
Factors of 28: 1,2,4,7,14,28
-7 and 4 work.
(x-7)(x+4) = 0
Set each equal to zero:
x=7,-4
← Didn't Know|Knew It →
25_x_2 – 36_y_2 can be factored into:
25_x_2 – 36_y_2 can be factored into:
Tap to reveal answer
This is the difference of squares. You must know this formula for the GRE!
_a_2 – _b_2 = (a – b)(a + b)
Here a = 5_x_ and b = 6_y_, so the difference of squares formula gives us (5_x_ – 6_y_)(5_x_ + 6_y_).
This is the difference of squares. You must know this formula for the GRE!
_a_2 – _b_2 = (a – b)(a + b)
Here a = 5_x_ and b = 6_y_, so the difference of squares formula gives us (5_x_ – 6_y_)(5_x_ + 6_y_).
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Simplify .
Simplify
Tap to reveal answer
To begin, let's factor the first two terms and the second two terms separately.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
(z – 1) can be pulled out because it appears in both terms.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1) = (z – 1)(_z_2 – 9)
(_z_2 – 9) is a difference of squares, so we can use the formula _a_2 – _b_2 = (a – b)(a + b).
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
= (z – 1)(_z_2 – 9)
= (z – 1)(z – 3)(z + 3)
To begin, let's factor the first two terms and the second two terms separately.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
(z – 1) can be pulled out because it appears in both terms.
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1) = (z – 1)(_z_2 – 9)
(_z_2 – 9) is a difference of squares, so we can use the formula _a_2 – _b_2 = (a – b)(a + b).
_z_3 – z_2 – 9_z + 9 = (_z_3 – z_2) + (–9_z + 9) = _z_2(z – 1) – 9(z – 1)
= (z – 1)(_z_2 – 9)
= (z – 1)(z – 3)(z + 3)
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Factor 3_u_4 – 24_uv_3.
Factor 3_u_4 – 24_uv_3.
Tap to reveal answer
First pull out 3u from both terms.
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember but pat yourself on the back for getting to such hard questions! The difference of cubes formula to remember is _a_3 – _b_3 = (a – b)(_a_2 + ab + b_2). In our problem, a = u and b = 2_v, so
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
= 3_u_(u – 2_v_)(u_2 + 2_uv + 4_v_2)
First pull out 3u from both terms.
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember but pat yourself on the back for getting to such hard questions! The difference of cubes formula to remember is _a_3 – _b_3 = (a – b)(_a_2 + ab + b_2). In our problem, a = u and b = 2_v, so
3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]
= 3_u_(u – 2_v_)(u_2 + 2_uv + 4_v_2)
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Solve for x: x2 = 45 - 12x
Solve for x: x2 = 45 - 12x
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For quadratic equations, you need get the equation into the form:
Ax2 + Bx + C = 0. Therefore subtract 45 and add 12 x to both sides of x2 = 45 - 12x. This will give you:
x2 +12x - 45 = 0
Following this, you must factor the quadratic element. Since C is negative, your groups will have to be positive and negative. We know that 45 factors into: 1 * 45, 5 * 9, and 15 * 3. The correct pair is 15 and 3.
(x + 15) (x - 3) = 0
Now, we set each of these factors equal to 0 since we know that if either is 0, the whole equation will be zero (hence solving for the correct value of x):
x + 15 = 0; x = -15
x - 3 = 0; x = 3
Therefore x is -15 or 3.
For quadratic equations, you need get the equation into the form:
Ax2 + Bx + C = 0. Therefore subtract 45 and add 12 x to both sides of x2 = 45 - 12x. This will give you:
x2 +12x - 45 = 0
Following this, you must factor the quadratic element. Since C is negative, your groups will have to be positive and negative. We know that 45 factors into: 1 * 45, 5 * 9, and 15 * 3. The correct pair is 15 and 3.
(x + 15) (x - 3) = 0
Now, we set each of these factors equal to 0 since we know that if either is 0, the whole equation will be zero (hence solving for the correct value of x):
x + 15 = 0; x = -15
x - 3 = 0; x = 3
Therefore x is -15 or 3.
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