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 1:
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
$\frac{2x}{2}$ = $\frac{5-y}{2}$
Canceling the 2 on the left side yields
x = $\frac{5-y}{2}$
Example 2:
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.
*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.
Equation | Deduction | |
y – x = 1 | y > x | |
y^{2} = x^{2} | y = ± x, or |y| = |x| . That is, x and y can differ only in sign. | |
y^{3} = x^{3} | y = x | |
y = x^{2} | y ≥ 0 | |
$\frac{y}{{x}^{2}}=1$ | y > 0 | |
$\frac{y}{{x}^{3}}=2$ | Both x and y are positive or both x and y are negative. | |
x^{2} + y^{2} = 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. | |
$y=\sqrt{x+2}$ | y ≥ 0 and x ≥ –2 | |
y = 2x | y is even | |
y = 2x + 1 | y = 2x + 1 y is odd | |
yx = 0 | y = 0 or x = 0, or both |
In Algebra, you solve an equation for, say, y by isolating y on one side of the equality symbol. On the GRE, 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 = ?
(A) –4
(B) –1
(C) 1/5
(D) 1/3
(E) 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).
Sometimes on the GRE, 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 w = 2x = √ 2 y, what is the value of w – x in terms of y?
(A) 2y
(B) $\frac{\sqrt{2}}{2}$y
(C) √ 2y
(D) $\frac{4}{\sqrt{2}}$y
(E) y
The equation w = 2x = $\sqrt{2y}$ stands for three equations: w= 2x, 2x = $\sqrt{2y}$, and w = $\sqrt{2y}$. From the last equation, we get w = $\sqrt{2y}$; and from the second equation, we get x= $\frac{\sqrt{2}}{2}$ y. Hence, w - x = $\sqrt{2y}$ - $\frac{\sqrt{2}}{2}$y = $\frac{2}{2}$$\sqrt{2y}$ - $\frac{\sqrt{2}}{2}$y = $\frac{2\sqrt{2y}-\sqrt{2y}}{2}$= $\frac{\sqrt{2y}}{2}$. Hence, the answer is (B).
Often on the GRE, 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, p^{2} + q^{2} = 16, and p^{2} - q^{2} = 8, then q =?
(A) 2
(B) 4
(C) 8
(D) 2 √2
(E) 2 √6
Subtract the second equation from the first:
p^{2} + q^{2} = 16
(–) p^{2} - q^{2} = 8
2q^{2} = 8
Dividing both sides of the equation by 2 gives
q^{2} = 4
Finally, taking the square root of both sides gives
q = ±2
Hence, the answer is (A).
Method of Substitution (Four-Step Method)
Although on the GRE you can usually solve a system of two equations in two unknowns by merely adding or subtracting the equations, you still need to know a standard method for solving these types of systems.
The four-step method will be illustrated with the following system:
2x + y = 10
5x – 2y = 7
1) Solve one of the equations for one of the variables:
Solving the top equation for y yields y = 10 – 2x.
2) Substitute the result from Step 1 into the other equation:
Substituting y = 10 – 2x into the bottom equation yields 5x – 2(10 – 2x) = 7.
3) Solve the resulting equation:
5x – 2(10 – 2x) = 7
5x – 20 + 4x = 7
9x – 20 = 7
9x = 27
x = 3
4) Substitute the result from Step 3 into the equation derived in Step 1:
Substituting x = 3 into y = 10 – 2x yields y = 10 – 2(3) = 10 – 6 = 4.
Hence, the solution of the system of equations is the ordered pair (3, 4).