Linear Equations

Today’s topic deals with how to solve linear and quadratic equations. Before we begin this topic, I will first teach you how to identify a linear and a quadratic equation. A linear equation comes in the form, ax + b = c ……where a, b & c are any value. The main idea of identifying a linear equation is that, the highest power of x existing in the equation must be 1. A quadratic equation comes in the form ax^2 +bx + c = d ……where a, b, c & d are any value. The main idea of identifying a quadratic equation is that, the highest power of x existing in the equation must be 2.


Now let’s learn how to solve linear equations. This is easy when compared to solving quadratic or even cubic equations. Luckily for you, the S.A.T doesn’t give you any cubic equations to solve, but the quadratic equations are popular.

1) 2x +3 = 9……..what is x?

That question is simple. All you have to do is carry all the numbers on one side and keep x on the other side. So it look like
2x = 9 – 3 …….2x = 6………x = 3
but the questions in the S.A.T have a little addition.
1) 2x +3 = 9……..what is 4x – 3 ?.......you can solve for x like above and then substitute it into 4x – 3 to get 9, but why solve for x? They asked for 4x – 3. Let me show you my technique. As the above, when you got 2x = 6, multiply both sides by 2 so that you get, 4x =12….. and then minus 3 and you get 9. As I told you before a large percentage of the Math in the S.A.T doesn’t require you to solve for x. You have to manipulate the question for you to gain time in the exam. Linear equations appear in the first 5 questions of the exam.

2) 6 + 5(2x + 3) = 31…..another linear equation.

First you must expand the brackets and then solve it just like above……6 + 10x + 15 = 31
10x = 31 – 15 – 6
10x = 10

x = 1

That’s about it for linear equations. Wasn’t that easy? Now we are going to take a look at handling quadratic equations.