A linear equation is an equation whose graph is a straight line. A linear equation in one variable is an equation that simlpy involves x. A linear equation is any equation that can be written in the form ax + b = 0. There are no terms involving x2, x3, x1/2 etc. Each term has a degree of at most 1. All operations, such as addition or multiplication, involve only x and numerical constants. 3x + 4 = 5 is an example of a linear equation. 2(x+1) = 6(x-4) is also a linear equation. These equations can be solved very easily by performing algebraic operations to the equation to isolate x.
A linear equation in two variables is, as the name suggests, an equation that involves 2 variables. The standard form of this type of equation is Ax + By = C, where A,B and C are real numbers. For example, 3x + y = 7 is a linear equation in two variables. y = 2x + 1/3 is also an example, since it can be rewritten as 2x - y = -1/3 ( or equivalently 6x - 3y = -1 ).
Linear equation in one variable properties.
1. If a = b then a+c = b+c.
2. If a = b then a -c = b-c.
3. If a = b then ac = bc.
4. If a=b then a/b = b/c.
Linear equations in two variables can also be expressed in the slope-intercept form y = mx + b.
The slope of a line, represented by the variable m, is defined as the ratio of change in values of y to change in value of x. The slope is also known as rise over run. For any two points (x1 ,y1), (x2 ,y2) on a line L, the formula for calculating the slope of L is:
m = (y2 - y1)/(x2 - x1)
Two lines are parallel if they have equal slopes. Parallel lines never cross each other. The distance between two parallel lines is always the same for every point along the lines.
Two lines are perpendicular, meaning their angle of intersection is 90°, if their slopes are negative reciprocals of each other. For lines L1 and L2 with slopes m1 and m2, respectively,
m1m2 = -1
Example 1: A calculator has been marked up 15% and is being sold for $78.50. How much did the store pay the manufacturer of the calculator?
Solution: First, let’s define p to be the cost that the store paid for the calculator. The stores markup on the calculator is 15%. This means that 0.15p has been added on to the original price (p) to get the amount the calculator is being sold for. In other words, we have the following equation p + 0.15p = 78.50 that we need to solve for p. Doing this gives, 1.15p = 78.50 therefore p = 78.5/1.15.
The store paid $68.26 for the calculator.
Example 2: A shirt is on sale for $15.00 and has been marked down 35%. How much was the shirt being sold for before the sale?
Solution: Let’s start with defining p to be the price of the shirt before the sale. It has been marked down by 35%. This means that 0.35p has been subtracted off from the original price. Therefore, the equation (and solution) is,
p - 0.35p = 15
0.65p = 15
p = 15/0.65 = 23.07
A quadratic equation in one unknown is an equation of the form ax2 + bx + c =0, where a is not equal to 0. When we solve a linear equation, we may transpose the terms and leave the unknown on one side of the equation. However, this is often not the case for a quadratic equation. There are other methods to solve a quadratic equation, e.g. by factorization, by completing the square and by the quadratic formula. Furthermore, for a linear equation in the form of mx + n =0, where m is not 0, there is always a solution x = -n/m , which is a real number. On the contrary, a quadratic equation may have two real roots, one double root or no real roots.
Solving quadratic equations
There are several methods to solve a quadratic equation. Some quadratic expressions can be factorized and then the equation is easy to solve.
1. If pq = 0, then p=0 or q=0.
If quadratic equation ax2 + bx + c =0 can be factorized into (px+q)(rx+s) =0,
then we have px + q = 0 or rx + s = 0, which gives x =-q/p<!--[endif]--> or x = -s/r<!--[endif]--> respectively.
2. Method of completing square.
The method of completing the square is to change the equation from the form
x2 + b/a x = -c/a
x2 + 2.1/2.b/a x + (b/2a)2= -c/a + (b/2a)2
(x + b/2a)2 = -c/a + (b/2a)2
and thus p = b/2a and q = -c/a + (b/2a)2 . If q > 0 then x + p = +(-)q or x = -p +(-)q.
If q = 0, then x = -p. If q < 0, equation has no real roots. The result x = -p +(-)q still holds but the roots will be complex.
3. If ax2 + bx + c =0 and a is not equal to 0, the roots of the equation is given by
-b+_b2-4ac
x=-----------------------------
2a
Examples
Let A and B be the roots of a quadratic equation ax2 + bx + c =0 with a not equal to 0 then
A + B = - b/a and
AB = c/a
For a quadratic equation ax2 + bx + c =0 with a not equal to 0, the discriminant D = b2 - ac.
1. If D > 0, the equation has two distinct real roots.
2. if D = 0, the equation has one double real root.
3. if D < 0, the equation has no real root, it has two distinct unreal roots.
Assignment
1. Given the quadratic equation x2 - (A - 3) x - (A - 2) = 0, for what value of A will the sum of the squares of the roots be zero?
2. A quadratic with integral coefficients has two distinct positive integers as roots, the sum of its coefficients is prime and it takes the value -55 for some integer. The sum of the roots is
3. How many real r are there such that the roots of x2 + rx + 6r = 0 are both integers?
