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When simplifying algebraic expressions, we perform operations within parentheses first and then exponents and then multiplication and then division and then addition and lastly subtraction. This can be remembered by the mnemonic:

Please Excuse My Dear Aunt Sally

When solving equations, however, we apply the mnemonic in reverse order: SADMEP. This is often expressed as follows: inverse operations in inverse order. The goal in solving an equation is to isolate the variable on one side of the equal sign (usually the left side). This is done by identifying the main operation—addition, multiplication, etc.—and then performing the opposite operation.



Solve the following equation for x:  2x + y = 5

Solution: The main operation is addition (remember addition now comes before multiplication, SADMEP), so subtracting y from both sides yields
2x + y – y = 5 – y


Simplifying yields
2x = 5 – y


The only operation remaining on the left side is multiplication.  Undoing the multiplication by dividing both sides by 2 yields


Canceling the 2 on the left side yields




Solve the following equation for x: 3x – 4 = 2(x – 5)

Solution: Here x appears on both sides of the equal sign, so let’s move the x on the right side to the left side. But the x is trapped inside the parentheses. To release it, distribute the 2:
3x – 4 = 2x – 10


Now, subtracting 2x from both sides yields
x – 4 = –10


Finally, adding 4 to both sides yields
x = –6


We often manipulate equations without thinking about what the equations actually say. The GRE likes to test this oversight. Equations are packed with information. Take for example the simple equation 3x + 2 = 5. Since 5 is positive, the expression 3x + 2 must be positive as well. An equation means that the terms on either side of the equal sign are equal in every way. Hence, any property one side of an equation has the other side will have as well. Following are some immediate deductions that can be made from simple equations.




yx = 1

y > x

y2 = x2

y = ± x, or . That is, x and y can differ only in sign.

y3 = x3

y = x

y = x2

y ≥ 0

y > 0
Both x and y are positive or both x and y are negative.

x2 + y2 = 0

y = x = 0

3y = 4x and x > 0

y > x and y is positive.

3y = 4x and x < 0

y < x and y is negative.

 y ≥ 0 and x ≥ –2

y = 2x

y is even

y = 2x + 1

y is odd

yx = 0

y = 0, or x = 0, or both


Note: In Algebra, you solve an equation for, say, y by isolating y on one side of the equality symbol. On the test, however, you are often asked to solve for an entire term, say, 3 – ψ by isolating it on one side.



If a + 3a is 4 less than b + 3b, then a – b =

A.  –4
B.  –1
C.  1/5
D.  1/3

E.  2

Translating the sentence into an equation gives a + 3a = b + 3b – 4


Combining like terms gives 4a = 4b – 4


Subtracting 4b from both sides gives 4a – 4b = –4


Finally, dividing by 4 gives a – b = –1


Hence, the answer is (B).

Note: Sometimes on the test, a system of 3 equations will be written as one long “triple” equation. For example, the three equations x = y, y = z, x = z, can be written more compactly as x = y = z.



If w ≠ 0 and , what is the value of w – x in terms of y ?

A. 2y

E. y

The equation  
stands for three equations: w = 2x, , and . From the last equation, we get ; and from the second equation, we get . Hence, 


Hence, the answer is (B).


Note: Often on the test, you can solve a system of two equations in two unknowns by merely adding or subtracting the equations—instead of solving for one of the variables and then substituting it into the other equation.



If p and q are positive, p2 + q2 = 16, and p2 – q2 = 8, then q =

A  2
B  4
C  8

Subtract the second equation from the first: p2 + q2 = 16


(–) p2 – q2 = 8
2q2 = 8


Dividing both sides of the equation by 2 gives q2 = 4


Finally, taking the square root of both sides gives q = ±2


Hence, the answer is (A).

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