Equations Strategies

Question 1

Solving Three Equations With Three Variables:

Answer 1

As with two equations with two variables use combination and substitution to solve.

Question 2

Complex Absolute Value Equations:

Answer 2

When absolute value questions have two or more variables and/or contain two or more absolute value expressions it becomes more complicated.

There are two primary types of questions:

1. Equation contains two or more variables in more than one absolute value expression, usually lacks constant.

These generally test concepts of positives and negatives and the way to solve is to test cases.

2. Equation with ONE variable, at least one constant in more than one absolute value expression.

In these use an algebraic approach.

E.g. if |x-2| = |2x – 3|, what are the possible values for x?

For these question usually you would have to test these four cases:

1. Positive/positive
2. Positive/negative
3. Negative/positive
4. Negative/negative

But you would see that case 1 and 4 yield the same equation and so do 2 and 3.

So: You only have to test two cases: one in which neither expression changes sign, one in which one expression changes sign.

So in the case above test use these two cases to find the values for x:

1. (x – 2) = (2x - 3) x = 1
2. (x – 2) = -(2x – 3) x = 5/3

With complex absolute value questions you have to also check the validity of each value for the variables you found. Pluck in the values and see if the expression is still true. Absolute value equations can sometimes yield results that are not valid.

Question 3

Integer Constraints:

Answer 3

Some GMAT questions only allow solutions that are integers or mention integer constraints in the stem.

E.g. 2y – x = 2xy and x ≠ 0. If x and y are integers, which of the following could be equal to y?

For these use backsolving. Plug in values for y given in answer choices (all integers since stem says it has to be integer) and see for which x is also an integer.

Question 4

Using FOIL with Square Roots:

Answer 4

With square roots problems like these:

(√8 - √3) (√8 + √3)

use foil to simplify:

8 - √24 + √24 – 3 = 5

Question 5

Quadratic Formula:

Answer 5

Usually you can solve quadratic equations by factoring or square rooting to find the values for x or whatever the variable is.

But for some questions you can’t use factoring because there are no two values that multiplied and added or subtracted yield the values needed. In that case you can use the quadratic formula to find x. It’s for a top score. The quadratic formula works for all quadratic equations.

Quadratic Formula:

X = -b +- √b^2 – 4ac/2a

E.g. for x^2 + 8x + 13

Just plug the values for the constants from the equation into the formula to find the values for x.

X = - 8 +- √8^2 – 4x1x13/2x1
X = - 8 +-√64 – 52/2
X = - 8 +- √12/2
X = -8 +- √4x3/2
X = -8 +- 2√3/2
X = -8/2 +- √3
X = - 4 +- √3

So the two values for x are: - 4 - √3 and - 4 + √3

REMEMBER: if the value under the root (b^2 – 4ac is negative it means that there are NO values for x.

E.g. if x^2 – 6x + 11 = 0

Using the formula we get for under the root: 36 – 44 = - 8

That means there are no solutions for x.

If the value under the root is positive it means there will be two solutions for x as usual in quadratic equations.

Question 6

Fractions with Square Roots in Denominator:

Answer 6

When there are square roots in the denominator of fractions do this:

1. if only a square root in denominator multiply by the square root:

E.g. 4/√2 = 4 √2/ √2x√2 = 4 √2/2 = 2√2

2. If square root and another term in denominator you have to change the sign of the second term and multiply by that.

E.g. 4/3-√2 = 4 (3+√2)/(3-√2) (3+√2) = 12 + 4√2/9 + 3√2 - 3√2 - 2 = 12 + 4√2/9 - 2 = 12 + 4√2/7