Algebra 101This post is a part of [
GMAT MATH BOOK]
created by: shrouded1ScopeManipulation of various algebraic expressions
Equations in 1 & more variables
Dealing with non-linear equations
Algebraic identities
Notation & AssumptionsIn this document, lower case roman alphabets will be used to denote variables such as
a,b,c,x,y,z,wIn general it is assumed that the GMAT will only deal with real numbers (\(\mathbb{R}\)) or subsets of \(\mathbb{R}\) such as Integers (\(\mathbb{Z}\)), rational numbers (\(\mathbb{Q}\)) etc
Concept of variablesA variable is a place holder, which can be used in mathematical expressions. They are most often used for two purposes :
(a)
In Algebraic Equations : To represent unknown quantities in known relationships. For eg : "Mary's age is 10 more than twice that of Jim's", we can represent the unknown "Mary's age" by x and "Jim's age" by y and then the known relationship is \(x = 2y + 10\)
(b)
In Algebraic Identities : These are generalized relationships such as \(\sqrt{x^2} = |x|\), which says for any number, if you square it and take the root, you get the absolute value back. So the variable acts like a true placeholder, which may be replaced by any number.
Basic rules of manipulation- When switching terms from one side to the other in an algebraic expression + becomes - and vice versa.
Eg. \(x+y=2z \Rightarrow x=2z-y\) - When switching terms from one side to the other in an algebraic expression * becomes / and vice versa.
Eg. \(4*x=(y+1)^2 \Rightarrow x=\frac{(y+1)^2}{4}\) - you can add/subtract/multiply/divide both sides by the same amount. Eg. \(x+y = 10z \Rightarrow \frac{x+y}{43}=\frac{10z}{43}\)
- you can take to the exponent or bring from the exponent as long as the base is the same.
Eg 1.\(x^2+2=z \Rightarrow 4^{x^2+2}=4^z\)
Eg 2.\(2^{4x}=8^{y} \Rightarrow 2^{4x}=2^{3y} \Rightarrow 4x=3y\)
It is important to note that all the operations above are possible not just with constants but also with variables themselves. So you can "add x" or "multiply with y" on both sides while maintaining the expression. But what you need to be very careful about is when dividing both sides by a variable.
When you divide both sides by a variable (or do operations like "canceling x on both sides") you implicitly assume that the variable cannot be equal to 0, as division by 0 is undefined. This is a concept shows up very often on GMAT questions.
Degree of an expressionThe degree of an algebraic expression is defined as the highest power of the variables present in the expression.
Degree 1 : Linear
Degree 2 : Quadratic
Degree 3 : Cubic
Degree 4 : Bi-quadratic
Egs : \(x+y\) the degree is 1
\(x^3+x+2\) the degree is 3
\(x^3+z^5\) the degree of x is 3, degree of z is 5, degree of the expression is 5
Solving equations of degree 1 : LINEARDegree 1 equations or linear equations are equations in one or more variable such that degree of each variable is one. Let us consider some special cases of linear equations :
One variableSuch equations will always have a solution. General form is \(ax=b\) and solution is \(x=(b/a)\)
One equation in Two variablesThis is not enough to determine x and y uniquely. There can be infinitely many solutions.
Two equations in Two variablesIf you have a linear equation in 2 variables, you need at least 2 equations to solve for both variables. The general form is :
\(ax + by = c\)
\(dx + ey = f\)
If \((a/d) = (b/e) = (c/f)\) then there are infinite solutions. Any point satisfying one equation will always satisfy the second
If \((a/d) = (b/e) \neq (c/f)\) then there is no such x and y which will satisfy both equations. No solution
In all other cases, solving the equations is straight forward, multiply eq (2) by a/d and subtract from (1).
More than two equations in Two variablesPick any 2 equations and try to solve them :
Case 1 : No solution --> Then there is no solution for bigger set
Case 2 : Unique solution --> Substitute in other equations to see if the solution works for all others
Case 3 : Infinite solutions --> Out of the 2 equations you picked, replace any one with an un-picked equation and repeat.
More than 2 variablesThis is not a case that will be encountered often on the GMAT. But in general for n variables you will need at least n equations to get a unique solution. Sometimes you can assign unique values to a subset of variables using less than n equations using a small trick. For example consider the equations :
\(x + 2y + 5z = 20\)
\(x + 4y + 10z = 40\)
In this case you can treat \(2y+z\) as a single variable to get :
\(x + (2y+5z) = 20\)
\(x + 2*(2y+5z) = 40\)
These can be solved to get x=0 and 2y+5z=20
There is a common misconception that you need n equations to solve n variables. This is not true.Solving equations of degree 2 : QUADRATICThe general form of a quadratic equation is \(ax^2+bx+c=0\)
The equation has no solution if \(b^2<4ac\)
The equation has exactly one solution if \(b^2=4ac\)
This equation has 2 solutions given by \(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\) if \(b^2>4ac\)
The sum of roots is \(\frac{-b}{a}\)
The product of roots is \(\frac{c}{a}\)
If the roots are \(r_1\) and \(r_2\), the equation can be written as \((x-r_1)(x-r_2)=0\)
A quick way to solve a quadratic, without the above formula is to factorize it :
Step 1> Divide throughout by coeff of x^2 to put it in the form \(x^2+dx+e=0\)
Step 2> Sum of roots = -d and Product = e. Search for 2 numbers which satisfy this criteria, let them be f,g
Step 3> The equation may be re-written as (x-f)(x-g)=0. And the solutions are f,g
Eg. \(x^2+11x+30=0\)
The sum is -11 and the product is 30. So numbers are -5,-6
\(x^2+11x+30=x^2+5x+6x+30=x(x+5)+6(x+5)=(x+5)(x+6)\)
Solving equations with DEGREE>2You will never be asked to solved higher degree equations, except in some cases where using simple tricks these equations can either be factorized or be reduced to a lower degree or both. What you need to note is that an equation of degree n has at most n unique solutions.
FactorizationThis is the easiest approach to solving higher degree equations. Though there is no general rule to do this, generally a knowledge of algebraic identities helps. The basic idea is that if you can write an equation in the form :
\(A*B*C=0\)
where each of A,B,C are algebraic expressions. Once this is done, the solution is obtained by equating each of A,B,C to 0 one by one.
Eg. \(x^3 + 11x^2 + 30x = 0\)
\(x * (x^2 + 11x + 30) = 0\)
\(x * (x+5) * (x+6) = 0\)
So the solution is x=0,-5,-6
Reducing to lower degreeThis is useful sometimes when it is easy to see that a simple variable substitution can reduce the degree.
Eg. \(x^6 -3x^3 + 2 = 0\)
Here let \(y=x^3\)
\(y^2-3y+2=0\)
\((y-2)(y-1)=0\)
So the solution is y=1 or 2 or x^3=1 or 2 or x=1 or \(\sqrt[3]{2}\)
Other tricksSometimes we are given conditions such as the variables being integers which make the solutions much easier to find. When we know that the solutions are integral, often times solutions are easy to find using just brute force.
Eg. \(a^2+b^2=116\) and we know a,b are integers such that a<b
We can solve this by testing values of a and checking if we can find b
a=1 b=root(115) Not integer
a=2 b=root(112) Not integer
a=3 b=root(107) Not integer
a=4 b=root(100)=10a=5 b=root(91) Not integer
a=6 b=root(80) Not integer
a=7 b=root(67) Not integer
a=8 b=root(52)<a
So the answer is (4,10)
Algebraic IdentitiesThese can be very useful in simplifying & solving a lot of questions :
- \((x+y)^2=x^2+y^2+2xy\)
- \((x-y)^2=x^2+y^2-2xy\)
- \(x^2-y^2=(x+y)(x-y)\)
- \((x+y)^2-(x-y)^2=4xy\)
- \((x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\)
- \(x^3+y^3=(x+y)(x^2+y^2-xy)\)
- \(x^3-y^3=(x-y)(x^2+y^2+xy)\)