Rational Equations

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Rational Equations

Definition:

A rational equation is an equation that involves at least one rational expression, which is a fraction
where the numerator and/or the denominator are polynomials. Solving rational equations typically
involves finding values for the variable that make the equation true, ensuring these values do not
make any denominators equal to zero.

Essential Formulas to Remember:

1. Basic Rational Expression:

P (x) R(x)
=
​ ​

Q(x) S(x)

where P (x), Q(x), R(x), and S(x) are polynomials.

2. Cross-Multiplication (when there are two rational expressions):

P (x) × S(x) = Q(x) × R(x)

3. Finding a Common Denominator:

To solve:

P (x) R(x)
+​ = T (x)
​

Q(x) S(x)

Multiply each term by the least common denominator (LCD) to eliminate fractions.
4. Checking for Extraneous Solutions:
Solutions must be checked to ensure they do not make any denominator zero, as these are not
valid solutions.

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 Key Properties and Rules:

1. Identify the LCD: To solve a rational equation, first find the least common denominator (LCD) of
all the rational expressions involved.
2. Eliminate the Denominator: Multiply both sides of the equation by the LCD to clear the
fractions.
3. Simplify and Solve: Once the equation is free of fractions, simplify and solve the resulting
polynomial equation.
4. Check Solutions: Always substitute the solutions back into the original equation to ensure they
do not cause any denominator to become zero (i.e., check for extraneous solutions).

Advanced Examples:

Example 1: Solve the Rational Equation

3 2x
=
x−2 x+1
​ ​

Step 1: Cross-Multiply

3(x + 1) = 2x(x − 2)

Step 2: Expand and Simplify

3x + 3 = 2x2 − 4x

2x2 − 7x − 3 = 0

Step 3: Solve the Quadratic Equation

−b± b2 −4ac
Use the quadratic formula x = 2a
with
​

​ a = 2, b = −7, and c = −3:

7± 49 + 24 7 ± 73
x= =
​ ​

4 4
​ ​

The solutions are:

7+ 73 7− 73
x= and x =
​ ​

4 4
​ ​

Step 4: Check for Extraneous Solutions

Ensure neither solution makes the original denominators zero.

Example 2: Solve the Rational Equation

2x 3x
+ =1
x+3 x−4
​ ​

Step 1: Find the LCD

LCD is (x + 3)(x − 4).
Step 2: Multiply Each Term by the LCD

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 2x(x − 4) + 3x(x + 3)
= (x + 3)(x − 4)
(x + 3)(x − 4)
​

Simplifies to:

2x(x − 4) + 3x(x + 3) = (x + 3)(x − 4)

Step 3: Expand and Simplify

2x2 − 8x + 3x2 + 9x = x2 − x − 12

5x2 + x + 12 = 0

Step 4: Solve the Quadratic Equation

Use the quadratic formula with a = 5, b = 1, and c = 12. The discriminant is negative, so there are
no real solutions.
Conclusion: No real solutions for this equation.

Example 3: Solve the Rational Equation

x 2 7
+ = 2
x−1 x+2 x +x−2
​ ​ ​

Step 1: Factor the Denominators

x2 + x − 2 = (x − 1)(x + 2)

Step 2: Multiply by the LCD, (x − 1)(x + 2)

x(x + 2) + 2(x − 1) = 7

Simplify:

x2 + 2x + 2x − 2 = 7

x2 + 4x − 9 = 0

Step 3: Solve the Quadratic Equation

Use the quadratic formula with a = 1, b = 4, and c = −9:

−4 ± 16 + 36 −4 ± 52
x= =
​ ​

2 2
​ ​

Simplify:

−4 ± 2 13
x= = −2 ± 13
​

2
​ ​

Step 4: Check for Extraneous Solutions

Ensure neither solution makes any original denominator zero.

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 This structured approach covers all essential aspects of rational equations for your GMAT preparation.
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Radical Equations

Definition:

A radical equation is an equation in which the variable appears under a radical sign, such as a square
root, cube root, or any other root. Solving these equations involves isolating the radical and then
eliminating it by raising both sides of the equation to an appropriate power.

Essential Formulas to Remember:

1. Basic Radical Equation:

n
P (x) = Q(x)
​

where P (x) is an expression involving the variable and n is the index of the root.
2. Isolating the Radical:

P (x) = Q(x) or n
P (x) = Q(x)
​

The goal is to isolate the radical expression before removing the radical by raising both sides to
the corresponding power.
3. Eliminating the Radical (Square Root Example):

( P (x))2 = (Q(x))2
​ ⇒ P (x) = Q(x)2

4. Checking for Extraneous Solutions:

After solving, always substitute solutions back into the original equation to check for extraneous
solutions, which may arise due to squaring both sides of the equation.

Key Properties and Rules:

1. Isolate the Radical: The first step is always to isolate the radical expression on one side of the
equation.
2. Raise to a Power: Once the radical is isolated, raise both sides of the equation to a power that
matches the index of the radical to eliminate it. For example, if dealing with a square root, square
both sides.
3. Simplify and Solve: After eliminating the radical, simplify the resulting equation and solve for
the variable.
4. Check for Extraneous Solutions: Because raising both sides to a power can introduce
extraneous solutions, always substitute back into the original equation to verify that the
solutions are valid.
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 Advanced Examples:

Example 1: Solve the Radical Equation

2x + 3 = x − 1
​

Step 1: Isolate the Radical

The radical is already isolated on the left side.
Step 2: Square Both Sides

( 2x + 3)2 = (x − 1)2
​

2x + 3 = x2 − 2x + 1

Step 3: Rearrange and Solve the Quadratic Equation

x2 − 4x − 2 = 0

Use the quadratic formula:

4± 16 + 8 4 ± 24
x= =
​ ​

2 2
​ ​

Simplifies to:

x=2± 6 ​

Step 4: Check for Extraneous Solutions

Substitute x = 2 + 6 and x = 2 − 6 into the original equation to verify they do not result in a
​ ​

false statement.

Example 2: Solve the Radical Equation

3x + 7 + 2 = x
​

Step 1: Isolate the Radical

3x + 7 = x − 2
​

Step 2: Square Both Sides

( 3x + 7)2 = (x − 2)2
​

3x + 7 = x2 − 4x + 4

Step 3: Rearrange and Solve the Quadratic Equation

x2 − 7x − 3 = 0

Use the quadratic formula:

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 7± 49 + 12 7 ± 61
x= =
​ ​

2 2
​ ​

Step 4: Check for Extraneous Solutions

Substitute the solutions back into the original equation to check for validity.

Example 3: Solve the Radical Equation

x+6+ ​
2x − 1 = 5 ​

Step 1: Isolate One of the Radicals

x+6=5− ​
2x − 1 ​

Step 2: Square Both Sides

( x + 6)2 = (5 −
​ 2x − 1)2

Expanding the right side:

x + 6 = 25 − 10 2x − 1 + (2x − 1)
​

Simplifies to:

x + 6 = 24 + 2x − 10 2x − 1 ​

Step 3: Isolate the Remaining Radical and Square Again

10 2x − 1 = x − 18
​

Square both sides again:

100(2x − 1) = (x − 18)2

Step 4: Solve the Resulting Equation

Expand and simplify:

200x − 100 = x2 − 36x + 324

Rearrange into a quadratic equation:

x2 − 236x + 424 = 0

Use the quadratic formula to solve for x.

Step 5: Check for Extraneous Solutions
Substitute the solutions back into the original equation to verify which, if any, are valid.

This approach should cover all the essential aspects of radical equations for your GMAT preparation.
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 Exponential Equations

Definition:

An exponential equation is an equation in which the variable appears in the exponent. Solving these
equations typically involves using logarithms to rewrite the equation in a form where the variable can
be isolated and solved.

Essential Formulas to Remember:

1. Basic Exponential Form:

af (x) = b

where a is the base and f (x) is an expression involving the variable.

2. Logarithmic Conversion:
If af (x) = b, then:

f (x) = loga (b) ​

or using natural logarithms:

ln(b)
f (x) =
ln(a)

3. Exponential Growth/Decay:

y = y0 ⋅ ekt
​

where y0 is the initial amount, k is the growth (or decay) rate, and t is time.
​

4. Key Logarithmic Identities:

loga (ax ) = x
​

log(a ⋅ b) = log(a) + log(b)

log ( ab ) = log(a) − log(b)
​

log(ax ) = x ⋅ log(a)

Key Properties and Rules:

1. Isolate the Exponential Term: The first step is to isolate the exponential expression on one side
of the equation.
2. Apply Logarithms: Once the exponential term is isolated, take the logarithm of both sides
(typically using natural logarithms or base-10 logarithms) to bring down the exponent.
3. Solve for the Variable: After applying the logarithm, solve for the variable by isolating it.
4. Check Solutions: Substitute the solutions back into the original equation to verify that they
satisfy the equation.

Advanced Examples:

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 Example 1: Solve the Exponential Equation

32x = 81

Step 1: Express the Right Side with the Same Base

Notice that 81 = 34 . So the equation becomes:

32x = 34

Step 2: Equate the Exponents

Since the bases are the same, equate the exponents:

2x = 4

Step 3: Solve for x

x=2

Step 4: Verify the Solution

Substitute x = 2 back into the original equation to check:

32(2) = 81 which is true.

Example 2: Solve the Exponential Equation

5x+2 = 32x

Step 1: Apply Logarithms to Both Sides

Take the natural logarithm (ln) of both sides:

ln(5x+2 ) = ln(32x )

Step 2: Use the Logarithmic Power Rule

(x + 2) ln(5) = 2x ln(3)

Step 3: Expand and Rearrange the Equation

x ln(5) + 2 ln(5) = 2x ln(3)

x ln(5) − 2x ln(3) = −2 ln(5)

x(ln(5) − 2 ln(3)) = −2 ln(5)

Step 4: Solve for x

−2 ln(5)
x=
ln(5) − 2 ln(3)
​

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 Step 5: Simplify and Verify the Solution
Calculate and verify the solution in the original equation.

Example 3: Solve the Exponential Equation

23x−1 = 7x+2

Step 1: Apply Logarithms to Both Sides

Take the natural logarithm (ln) of both sides:

ln(23x−1 ) = ln(7x+2 )

Step 2: Use the Logarithmic Power Rule

(3x − 1) ln(2) = (x + 2) ln(7)

Step 3: Expand and Rearrange the Equation

3x ln(2) − ln(2) = x ln(7) + 2 ln(7)

3x ln(2) − x ln(7) = ln(2) + 2 ln(7)

Step 4: Factor and Solve for x

x(3 ln(2) − ln(7)) = ln(2) + 2 ln(7)

ln(2) + 2 ln(7)
x=
3 ln(2) − ln(7)
​

Step 5: Simplify and Verify the Solution

Calculate and verify the solution in the original equation.

This approach should help you master exponential equations for your GMAT preparation. Let me know
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Logarithmic Equations

Definition:

A logarithmic equation is an equation that involves a logarithm with a variable inside its argument.
Solving these equations typically involves using properties of logarithms to combine or simplify the
logarithmic expressions and then converting the logarithmic form into an exponential form to solve
for the variable.

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 Essential Formulas to Remember:

1. Basic Logarithmic Form:

loga (x) = b
​
⇒ ab = x

where a is the base, x is the argument, and b is the exponent.

2. Logarithm of a Product:

loga (xy) = loga (x) + loga (y)

​ ​ ​

3. Logarithm of a Quotient:

loga ( ) = loga (x) − loga (y)

x
​ ​ ​ ​

4. Logarithm of a Power:

loga (xb ) = b ⋅ loga (x)

​ ​

5. Change of Base Formula:

logb (x)
loga (x) =
​

logb (a)
​

where b is any positive number (commonly b = 10 or b = e).

Key Properties and Rules:

1. Isolate the Logarithmic Term: Start by isolating the logarithmic term on one side of the
equation.
2. Combine Logarithms: If there are multiple logarithms, use logarithmic properties to combine
them into a single logarithm.
3. Convert to Exponential Form: Once the logarithmic expression is isolated, convert the equation
to its equivalent exponential form.
4. Solve for the Variable: After converting to an exponential equation, solve for the variable.
5. Check for Extraneous Solutions: Logarithms are only defined for positive arguments, so check
that the solutions do not result in the logarithm of a non-positive number.

Advanced Examples:

Example 1: Solve the Logarithmic Equation

log2 (x − 3) = 4
​

Step 1: Convert to Exponential Form

x − 3 = 24

Step 2: Simplify

x − 3 = 16

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 Step 3: Solve for x

x = 19

Step 4: Check for Validity

Ensure that x = 19 does not make the argument of the logarithm negative or zero:

log2 (19 − 3) = log2 (16) = 4 which is correct.

​ ​

Example 2: Solve the Logarithmic Equation

log3 (x + 4) + log3 (x − 1) = 2
​ ​

Step 1: Combine Logarithms

log3 ((x + 4)(x − 1)) = 2

​

Step 2: Convert to Exponential Form

(x + 4)(x − 1) = 32

x2 + 3x − 4 = 9

Step 3: Solve the Quadratic Equation

x2 + 3x − 13 = 0

Use the quadratic formula:

−3 ± 9 + 52 −3 ± 61
x= =
​ ​

2 2
​ ​

Step 4: Check for Validity

Ensure that the solutions do not make the arguments of the logarithms negative or zero.

Example 3: Solve the Logarithmic Equation

log5 (x2 − 6x) = log5 (4)

​ ​

Step 1: Since the Bases are the Same, Set the Arguments Equal

x2 − 6x = 4

Step 2: Rearrange and Solve the Quadratic Equation

x2 − 6x − 4 = 0

Use the quadratic formula:

6± 36 + 16 6 ± 52
x= =
​ ​

2 2
​ ​

Simplifies to:
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 x=3± 13 ​

Step 3: Check for Validity

Substitute the solutions back into the original equation to ensure the arguments remain positive:

x2 − 6x > 0

Only positive results are valid.

This approach covers the essential steps to master logarithmic equations for your GMAT preparation.
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Absolute Value Equations

Definition:

An absolute value equation is an equation where the absolute value of a variable expression is set
equal to a number or another expression. The absolute value of a number is its distance from zero on
the number line, which means it is always non-negative. Solving absolute value equations typically
involves considering both the positive and negative cases of the expression inside the absolute value.

Essential Formulas to Remember:

1. Basic Absolute Value Equation:

∣x∣ = a

This equation has two solutions: x = a or x = −a.

2. Absolute Value with Variable Expressions:

∣f (x)∣ = g(x)

This equation has two cases: f (x) = g(x) or f (x) = −g(x).

3. Absolute Value Inequalities:
∣x∣ < a implies −a < x < a
∣x∣ > a implies x < −a or x >a

Key Properties and Rules:

1. Isolate the Absolute Value: The first step is to isolate the absolute value expression on one side
of the equation.
2. Set Up Two Cases: Once the absolute value is isolated, set up two separate equations: one
where the expression inside the absolute value equals the positive value and one where it equals
the negative value.

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 3. Solve Both Equations: Solve each equation separately to find the possible solutions.
4. Check for Extraneous Solutions: Not all solutions derived from the process will necessarily
satisfy the original equation, so substitute them back to ensure validity.

Advanced Examples:

Example 1: Solve the Absolute Value Equation

∣2x − 5∣ = 9

Step 1: Set Up Two Cases

Case 1: 2x − 5 = 9

2x = 14 ⇒ x=7

Case 2: 2x − 5 = −9

2x = −4 ⇒ x = −2

Step 2: Check Both Solutions

Substitute x = 7 and x = −2 back into the original equation:

∣2(7) − 5∣ = ∣14 − 5∣ = ∣9∣ = 9 (valid)

∣2(−2) − 5∣ = ∣ − 4 − 5∣ = ∣ − 9∣ = 9 (valid)

Step 3: Final Answer

The solutions are x = 7 and x = −2.

Example 2: Solve the Absolute Value Equation

∣3x + 2∣ = ∣x − 4∣

Step 1: Set Up Two Cases

Case 1: 3x + 2 =x−4

2x = −6 ⇒ x = −3

Case 2: 3x + 2 = −(x − 4)

3x + 2 = −x + 4

1
4x = 2 ⇒ x=
2
​

Step 2: Check Both Solutions

1
Substitute x = −3 and x = 2
​back into the original equation:

∣3(−3) + 2∣ = ∣ − 9 + 2∣ = ∣ − 7∣ = 7

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 ∣ − 3 − 4∣ = ∣ − 7∣ = 7 (valid)

1 3 7 7
∣3 ( ) + 2∣ = +2 = =
2 2 2 2
​ ​ ​ ​ ​ ​ ​ ​

1 1 8 −7 7
−4 = − = = (valid)
2 2 2 2 2
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​

Step 3: Final Answer

The solutions are x = −3 and x = 12 . ​

Example 3: Solve the Absolute Value Equation

∣4x − 1∣ + 3 = 8

Step 1: Isolate the Absolute Value

∣4x − 1∣ = 5

Step 2: Set Up Two Cases

Case 1: 4x − 1 = 5

3
4x = 6 ⇒ x=
2
​

Case 2: 4x − 1 = −5

4x = −4 ⇒ x = −1

Step 3: Check Both Solutions

Substitute x = 32 and x = −1 back into the original equation:
​

3
∣4 ( ) − 1∣ + 3 = ∣6 − 1∣ + 3 = 5 + 3 = 8 (valid)
2
​

∣4(−1) − 1∣ + 3 = ∣ − 4 − 1∣ + 3 = ∣ − 5∣ + 3 = 5 + 3 = 8 (valid)

Step 4: Final Answer

3
The solutions are x = 2
​and x = −1.

This organized approach should help you understand and solve absolute value equations for your
GMAT preparation. Let me know if you need further clarification or if there's another topic you'd like to
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Inequalities and Equations

Definition:

Inequalities are mathematical expressions that describe the relative size or order of two values. Unlike
equations, which assert that two expressions are equal, inequalities show that one expression is either
greater than, less than, greater than or equal to, or less than or equal to another expression. Solving
inequalities involves finding all possible values of the variable that make the inequality true.

Types of Inequalities:

1. Linear Inequalities:
Involves expressions of the form ax + b > c, ax + b < c, ax + b ≥ c, or ax + b ≤ c.
2. Quadratic Inequalities:
Involves expressions of the form ax2 + bx + c > 0, ax2 + bx + c < 0, ax2 + bx + c ≥ 0,
or ax2 + bx + c ≤ 0.
3. Rational Inequalities:
p(x)
Involves inequalities where the expressions are rational functions, such as q(x) ​
> 0 or
p(x)
q(x) < 0.
​

4. Absolute Value Inequalities:

Involves inequalities that include absolute value expressions, such as ∣ax + b∣ > c or
∣ax + b∣ ≤ c.

Essential Properties and Rules:

1. Addition/Subtraction Rule:
You can add or subtract the same number on both sides of an inequality without changing
the inequality's direction.
If a > b, then a + c > b + c and a − c > b − c.
2. Multiplication/Division Rule:
You can multiply or divide both sides of an inequality by a positive number without
changing the inequality's direction.
If you multiply or divide both sides by a negative number, you must reverse the inequality's
direction.
If a > b and c > 0, then ac > bc. If c < 0, then ac < bc.
3. Transitive Property:
If a > b and b > c, then a > c.
4. Reciprocal Rule (For Rational Inequalities):
If a > b > 0, then a1 < 1b .
​ ​

Advanced Examples:

Example 1: Solve the Linear Inequality

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 3x − 5 ≤ 7

Step 1: Isolate the Variable

Add 5 to both sides:

3x ≤ 12

Step 2: Solve for x

Divide both sides by 3:

x≤4

Step 3: Interval Notation

The solution is x ≤ 4, which can be written in interval notation as (−∞, 4].

Example 2: Solve the Quadratic Inequality

x2 − 4x − 5 > 0

Step 1: Factor the Quadratic Expression

(x − 5)(x + 1) > 0

Step 2: Determine the Critical Points

The critical points are x = 5 and x = −1.
Step 3: Test Intervals
Test the sign of the expression in the intervals determined by the critical points:

x < −1 (choose x = −2): (x − 5)(x + 1) > 0 is positive.

−1 < x < 5 (choose x = 0): (x − 5)(x + 1) > 0 is negative.
x > 5 (choose x = 6): (x − 5)(x + 1) > 0 is positive.

Step 4: Write the Solution

The solution is x < −1 or x > 5, which can be written as (−∞, −1) ∪ (5, ∞).

Example 3: Solve the Rational Inequality

x−3
≤0
x+2
​

Step 1: Determine the Critical Points

Set the numerator and denominator to zero to find the critical points:

Numerator: x − 3 =0⇒x=3
Denominator: x + 2 = 0 ⇒ x = −2

Step 2: Test Intervals

Test the sign of the expression in the intervals determined by the critical points:

x < −2 (choose x = −3): x−3

x+2 is positive.
​

−2 < x < 3 (choose x = 0): x−3

x+2
is negative.
​

x > 3 (choose x = 4): x+2 is positive.

x−3
​

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 Step 3: Include the Critical Points
The inequality is less than or equal to zero, so include x = 3 but exclude x = −2 since the
denominator cannot be zero.
Step 4: Write the Solution
The solution is (−2, 3].

Example 4: Solve the Absolute Value Inequality

∣2x − 1∣ > 3

Step 1: Set Up Two Cases

Case 1: 2x − 1 > 3

2x > 4 ⇒ x>2

Case 2: 2x − 1 < −3

2x < −2 ⇒ x < −1

Step 2: Combine the Solutions

Since this is a "greater than" inequality, the solution is the union of the two cases:

x < −1 or x > 2

Step 3: Write the Solution in Interval Notation

The solution is (−∞, −1) ∪ (2, ∞).

This approach should give you a solid understanding of how to solve different types of inequalities for
your GMAT preparation. Let me know if you need more examples or help with another topic!

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Equations with Variables on Both Sides

Definition:

Equations with variables on both sides involve algebraic expressions where the variable appears on
both sides of the equation. The goal is to manipulate the equation to isolate the variable on one side,
simplifying the equation to a form where the solution can be easily found.

Steps to Solve:

1. Simplify Both Sides (if necessary):

Combine like terms on each side.
Distribute any multiplication over addition or subtraction.
2. Move Variables to One Side:

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 Add or subtract terms to get all the variable terms on one side of the equation and all the
constant terms on the other side.
3. Isolate the Variable:
Once the variable is on one side, use addition, subtraction, multiplication, or division to
solve for the variable.
4. Check the Solution:
Substitute the solution back into the original equation to ensure it satisfies the equation.

Key Properties and Rules:

Addition/Subtraction:
You can add or subtract the same quantity from both sides of the equation without changing the
equality.
Multiplication/Division:
You can multiply or divide both sides of the equation by the same non-zero quantity without
changing the equality.
Distributive Property:
a(b + c) = ab + ac, which can help simplify the equation.
Combining Like Terms:
Combine terms with the same variable and degree to simplify the equation.

Advanced Examples:

Example 1: Solve the Equation

3x + 5 = 2x + 11

Step 1: Move Variables to One Side

Subtract 2x from both sides:

3x − 2x + 5 = 11

Simplify:

x + 5 = 11

Step 2: Isolate the Variable

Subtract 5 from both sides:

x=6

Step 3: Check the Solution

Substitute x = 6 back into the original equation:

3(6) + 5 = 2(6) + 11

18 + 5 = 12 + 11 which simplifies to 23 = 23 (valid)

Final Answer: The solution is x = 6.

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 Example 2: Solve the Equation

4(2x − 3) = 2x + 6

Step 1: Distribute to Simplify

Distribute 4 on the left side:

8x − 12 = 2x + 6

Step 2: Move Variables to One Side

Subtract 2x from both sides:

6x − 12 = 6

Step 3: Isolate the Variable

Add 12 to both sides:

6x = 18

Divide by 6:

x=3

Step 4: Check the Solution

Substitute x = 3 back into the original equation:

4(2(3) − 3) = 2(3) + 6

4(6 − 3) = 6 + 6

4(3) = 12 which simplifies to 12 = 12 (valid)

Final Answer: The solution is x = 3.

Example 3: Solve the Equation

5x − 2(3x − 4) = 3(x + 2) + 4

Step 1: Distribute to Simplify

Distribute −2 and 3:

5x − 6x + 8 = 3x + 6 + 4

Simplify:

−x + 8 = 3x + 10

Step 2: Move Variables to One Side

Add x to both sides:

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 8 = 4x + 10

Step 3: Isolate the Variable

Subtract 10 from both sides:

−2 = 4x

Divide by 4:

1
x=−
2
​

Step 4: Check the Solution

Substitute x = − 12 back into the original equation:
​

1 1 1
5 (− ) − 2 (3 (− ) − 4 ) = 3 (− + 2 ) + 4
2 2 2
​ ​ ​

Simplify both sides to check:

5 3
− + 14 = + 4
2 2
​ ​

23 23
= (valid)
2 2
​ ​

Final Answer: The solution is x = − 12 .

​

This methodical approach to solving equations with variables on both sides will be helpful for your
GMAT preparation. Let me know if you have any other topics you'd like to explore!

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ChatGPT said:

Equations Involving Fractions

Definition:

Equations involving fractions have one or more terms that are fractions. These equations can be
solved by eliminating the fractions, often by multiplying both sides by the least common denominator
(LCD) of all the fractions involved.

Steps to Solve:

1. Find the Least Common Denominator (LCD):

Identify the denominators in the equation and determine the least common denominator
(LCD).
2. Multiply Both Sides by the LCD:

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 Multiply every term on both sides of the equation by the LCD to eliminate the fractions.
3. Simplify the Equation:
After eliminating the fractions, simplify the resulting equation by combining like terms and
isolating the variable.
4. Solve the Simplified Equation:
Solve for the variable using algebraic techniques.
5. Check for Extraneous Solutions:
Substitute your solution back into the original equation to ensure it does not make any
denominator equal to zero, which would be undefined.

Key Properties and Rules:

Multiplying Both Sides by the LCD:

This step is crucial to eliminate fractions and simplify the equation.
Cross-Multiplication (for Proportions):
If the equation is a proportion (i.e., a fraction equals another fraction), you can cross-multiply to
solve for the variable.
Simplification and Distribution:
Ensure you distribute correctly when multiplying by the LCD and combine like terms to simplify.

Advanced Examples:

Example 1: Solve the Fraction Equation

2x 1 5
− =
3 4 6
​ ​ ​

Step 1: Find the Least Common Denominator (LCD)

The denominators are 3, 4, and 6. The LCD is 12.
Step 2: Multiply Both Sides by the LCD

2x 1 5
12 × ( − ) = 12 ×
3 4 6
​ ​ ​

4(2x) − 3(1) = 2(5)

Step 3: Simplify the Equation

8x − 3 = 10

Step 4: Solve for x

Add 3 to both sides:

8x = 13

Divide by 8:

13
x= = 1.625
8
​

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 Step 5: Check the Solution
Substitute x = 13
8
back into the original equation to ensure it is correct.
​

13
Final Answer: The solution is x = 8
. ​

Example 2: Solve the Fraction Equation

x x 5
+ =
2 3 6
​ ​ ​

Step 1: Find the Least Common Denominator (LCD)

The denominators are 2, 3, and 6. The LCD is 6.
Step 2: Multiply Both Sides by the LCD

x x 5
6×( + )=6×
2 3 6
​ ​ ​

3x + 2x = 5

Step 3: Simplify the Equation

5x = 5

Step 4: Solve for x

Divide both sides by 5:

x=1

Step 5: Check the Solution

Substitute x = 1 back into the original equation:
1 1 3 2 5
+ = + = (valid)
2 3 6 6 6
​ ​ ​ ​ ​

Final Answer: The solution is x = 1.

Example 3: Solve the Fraction Equation

3 2
= +1
x+2 x−1
​ ​

Step 1: Move All Terms to One Side

Rewrite the equation to have a single fraction on each side:

3 2
− =1
x+2 x−1
​ ​

Step 2: Find the Least Common Denominator (LCD)

The denominators are x + 2 and x − 1. The LCD is (x + 2)(x − 1).
Step 3: Multiply Both Sides by the LCD

3 2
(x + 2)(x − 1) × ( − ) = (x + 2)(x − 1) × 1
x+2 x−1
​ ​

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 Simplify:

3(x − 1) − 2(x + 2) = (x + 2)(x − 1)

Step 4: Simplify and Solve

Distribute and combine like terms:

3x − 3 − 2x − 4 = x2 + x − 2

Simplify further:

x − 7 = x2 + x − 2

Move all terms to one side:

x2 + x − x + 2 + 7 = 0

x2 + 9 = 0

Factor and solve the quadratic equation.

Final Answer: The solution involves complex roots x = ±i.

This structured approach should help you master equations involving fractions for the GMAT. Let me
know if you need more examples or assistance with another topic!

You said:

ChatGPT said:

Word Problems Involving Equations

Definition:

Word problems involving equations require translating a real-world situation into an algebraic
equation, which can then be solved to find the answer. These problems often involve identifying
relationships between quantities and expressing them mathematically.

Steps to Solve Word Problems:

1. Read and Understand the Problem:

Identify what is being asked.
Determine the unknown(s) and assign a variable(s) to represent them.
2. Translate the Problem into an Equation:
Write down the relationships described in the problem using algebraic expressions.
Set up an equation based on these relationships.
3. Solve the Equation:
Use algebraic methods to solve the equation for the unknown variable(s).
4. Check the Solution:
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 Substitute your solution back into the context of the problem to ensure it makes sense.
5. Answer the Question:
Write a complete sentence that answers the question posed by the problem.

Types of Word Problems:

1. Number Problems:
Involves relationships between numbers, often requiring you to find a specific number or
set of numbers.
2. Age Problems:
Involves relationships between ages at different times.
3. Motion Problems:
Involves relationships between distance, rate, and time.
4. Work Problems:
Involves relationships between work done, rate of work, and time.
5. Mixture Problems:
Involves mixing different substances or quantities to achieve a certain concentration or
quantity.

Advanced Examples:

Example 1: Number Problem

A number is three times another number. If the sum of these two numbers is 48, find the numbers.
Step 1: Assign Variables
Let x be the first number. Then the second number is 3x.
Step 2: Set Up the Equation

x + 3x = 48

Step 3: Solve the Equation

Combine like terms:

4x = 48

Divide by 4:

x = 12

The second number is 3x = 3(12) = 36.

Step 4: Check the Solution
The sum of 12 and 36 is indeed 48.
Final Answer: The numbers are 12 and 36.

Example 2: Age Problem

John is 4 years older than twice the age of his sister Mary. If the sum of their ages is 28, how old are
they?
Step 1: Assign Variables
Let m be Mary's age. Then John's age is 2m + 4.
Step 2: Set Up the Equation

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 m + (2m + 4) = 28

Step 3: Solve the Equation

Combine like terms:

3m + 4 = 28

Subtract 4 from both sides:

3m = 24

Divide by 3:

m=8

John's age is 2(8) + 4 = 16 + 4 = 20.

Step 4: Check the Solution
The sum of 8 and 20 is 28.
Final Answer: Mary is 8 years old, and John is 20 years old.

Example 3: Motion Problem

A car travels 50 miles faster than a bus. If the car covers 300 miles in the same time it takes the bus to
cover 150 miles, what is the speed of each vehicle?
Step 1: Assign Variables
Let the speed of the bus be b miles per hour. Then the speed of the car is b + 50 miles per hour.
Step 2: Use the Relationship Distance = Rate × Time
For the bus:

150
Time = ​

b
For the car:

300
Time =
b + 50
​

Since the times are the same:

150 300
=
b + 50
​ ​

b
Step 3: Solve the Equation
Cross-multiply:

150(b + 50) = 300b

Distribute:

150b + 7500 = 300b

Subtract 150b from both sides:

7500 = 150b

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 Divide by 150:

b = 50

The car's speed is b + 50 = 50 + 50 = 100 miles per hour.

Step 4: Check the Solution
The bus covers 150 miles in 150
50
​ = 3 hours, and the car covers 300 miles in 300
100
​ = 3 hours.
Final Answer: The bus travels at 50 miles per hour, and the car travels at 100 miles per hour.

Example 4: Mixture Problem

How many liters of a 40% alcohol solution must be mixed with 60 liters of a 70% alcohol solution to
obtain a 50% alcohol solution?
Step 1: Assign Variables
Let x be the number of liters of the 40% alcohol solution.
Step 2: Set Up the Equation Based on Total Alcohol Content
The total alcohol from the 40% solution is 0.4x.
The total alcohol from the 70% solution is 0.7(60).
The total mixture is x + 60 liters at 50% alcohol, so:

0.4x + 0.7(60) = 0.5(x + 60)

Step 3: Solve the Equation

Distribute:

0.4x + 42 = 0.5x + 30

Subtract 0.4x from both sides:

42 = 0.1x + 30

Subtract 30 from both sides:

12 = 0.1x

Divide by 0.1:

x = 120

Step 4: Check the Solution

Substitute back to check if the resulting mixture gives 50% alcohol:

0.4(120) + 42 = 48 + 42 = 90

The total solution is 120 + 60 = 180 liters, and 0.5(180) = 90.

Final Answer: 120 liters of the 40% alcohol solution must be mixed.

This structured approach will help you tackle word problems involving equations for the GMAT. If you
need further explanations or more examples, feel free to ask!

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feel free to ask! ​

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It seems like I can’t create the Word document right now. Please try again later, or if you need help
with something else, feel free to ask! ​
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