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Equations

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:
PEMDAS
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.
 
Example

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
 

Example

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
 
 
 
Note, students often mistakenly add 2x to both sides of this equation because of the minus symbol between 2x and 10. But 2x is positive, so we subtract it. This can be seen more clearly by rewriting the right side of the equation as –10 + 2x.

We often manipulate equations without thinking about what the equations actually say. The SAT 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.

 

Equation

Deduction

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 SAT, however, you are often asked to solve for an entire term, say, 3 – y by isolating it on one side.

Example

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

  1. –4
  2. –1
  3. 1/5
  4. 1/3
  5. 2
Solution
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 SAT, 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.
 
Example

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

  1. 2y
  2. y
Solution
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 SAT, 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.
 
Example

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

  1. 2
  2. 4
  3. 8
Solution
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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