Solving Equations

Is a statement that two algebraic expressions are equal. EXAMPLES 3x – 5 = 7, x 2 – x – 6 = 0, and 4x = 4 To solve a equation in x means to find all real values of x for which the equation is true.

The set of real numbers for which an algebraic expression is defined. Remember the real numbers are made up of the Natural, Whole, Integer, Rational and Irrational numbers. The real numbers can be represented on a number line with 0 as the origin.

An equation that is true for every real number in the domain of the variable EXAMPLE x 2 – 9 =(x-3)(x+3) is an Identity, because it is a true statement for any real value of x

An equation that is true for some (or even none) of the real numbers in the domain of the variable. EXAMPLE x 2 – 9 =0 is conditional because x =3 and x = -3 are the only values in the domain that satisfy the equation.

An equation in one variable that can be written in the standard form: ax + b = 0 where and a and b are real numbers and a ≠ 0 EXAMPLE 2x – 5 = 0

A linear equation has exactly one solution and the solution is found by isolating the variable on one side of the equation by a sequence of equivalent equations. EXAMPLE 2x – 4 = 10 2x – 4+ 4 = x = 14 ½(2x) = ½(14) X = 7

A solution that does not satisfy the original equation. EXAMPLE 1/(x-2)= 3/(x+2) – 6x/(x 2 -4) x = -2 However, x = -2 yields a denominator of zero, so the original equation has no solution.

An equation in one variable written in the general form: ax 2 + bx + c = 0 Where a, b, and c are real numbers and a ≠ 0 EXAMPLE x 2 –x – 6 = 0

A quadratic equation has either one real solution, two real solutions or two conjugate imaginary solutions. METHODS  Factoring  Square Root Principle  Completing the Square  Quadratic Formula

EXAMPLE X 2 –x – 6 = 0 (x -3)(x+2) = 0 X – 3 = 0 or x + 2 = 0 X = 3 x = -2

EXAMPLE (x +3) 2 = 16 x + 3 = ± 4 X+3 = 4 or x + 3 = - 4 X = 1 x = -7

EXAMPLE x 2 + 6x = 5 x + 6x + (6/2) 2 = 5 + (6/2) 2 (x+3) 2 = x + 3 = ±  14 X = -3 ±  14

EXAMPLE 2x 2 + 3x -1 = 0 x = {(-3 ±[( (2)(-1))] ½ }/(2(2)) x = ¾ ±  17/4

 Polynomials of Higher Degree  3x 4 = 48x 2  x 3 – 3x 2 – 3x + 9 = 0  Equations Involving Radicals  (2x +7) ½ - x = 2  (x – 4) ⅔ = 25  Equations with Absolute Values  |x – 2| = 3  |x 2 – 3x| = -4x + 6

Solving Inequalities

Solve as if it were an equality, but remember to reverse the sign of the inequality whenever you multiply or divide by a negative number. Write the answer using interval notation. EXAMPLE 5x – 7 > 3x + 9 x > 8 so the solution set is (8, ∞) meaning 8 is not part of the solution, but all real numbers larger than 8 to infinity is the solution

Solve as if it were an equality, but isolate the variable as the middle term or solve as two separate inequalities EXAMPLE -3 ≤ 6x – 1 < 3 -⅓ ≤ x <⅔ Solution interval is [-⅓, ⅔) [ ] means closed interval; ( ) means open interval

Write equivalent inequalities and solve. EXAMPLE |x + 3| ≥ 7 (a) x + 3 ≥ 7 or (b) – (x + 3) ≥7 x ≥ 4 or x ≤ - 10 The solution interval is (-∞, -10]  [4, ∞)

Write equivalent inequalities and solve. EXAMPLE x 2 – x – 6 < 0 Solve as if equality and find the critical values (x -3) (x + 2) = 0 then x =3, or x = -2 are the critical values Check intervals (- ∞, -2) if x = -3 then (-3) 2 –(-3) – 6 is positive (-2, 3) if x = 0 then (0) 2 – (0) – 6 is negative (3, ∞) if x = 4 then (4) 2 - (4) – 6 is positive The solution interval is (-2, 3)

Find the critical values which occur for values making the numerator 0 or the denominator undefined. (right hand side must be 0 first) EXAMPLE (2x – 7) / (x – 5) ≤ 3 Simplify to (-x +8)/(x-5) ≤ 0, thus critical values are 8 and 5 Check the intervals (-∞, 5) if x = 4 then inequality is negative (5,8) if x = 6 then inequality is positive (8, ∞) if x = 9 then inequality is negative The solution interval is (-∞, 5)  [8, ∞)

Remember the domain is the set of all x values for which the expression is defined EXAMPLE (64 -4x 2 ) ½ Since this is a square root expression, the expression must be larger than 0. So find the critical values for (64 -4x 2 ) ½ ≥ 0 and test each of the intervals to find the solution set. The critical numbers are 4 and - 4

Graphical Representation of Data

consists of two perpendicular number lines, dividing the plane into four regions called quadrants

X-AXIS - the horizontal number line Y-AXIS - the vertical number line ORIGIN - the point where the x-axis and y-axis cross

ORDERED PAIR - a unique assignment of real numbers to a point in the coordinate plane consisting of one x-coordinate and one y-coordinate (-3, 5), (2,4), (6,0), (0,-3)

Coordinate Plane

D = [(x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ] ½ EXAMPLE (-2,1) and (3,4) D =  34

Use the distance formula and Pythagorean Theorem to determine a right triangle. EXAMPLE (2,1), (4,0) and (5,7)

Find the midpoint of a line segment that joins two points in a coordinate plane. M = [(x 1 + x 2 )/ 2,(y 1 + y 2 )/2] EXAMPLE (-5,-3) and (9,3) M = (2,0)

Graphs of Equations

Solutions points that have zeros as either the x-coordinate or the y-coordinate are called intercepts because they are the points where the graph intersects the x-axis or y-axis

 To find the x-intercepts, let y be zero and solve the equation for x.  To find the y-intercepts, let x be zero and solve the equation for y

 y = x  y = x 3 – 4x  y 2 = x + 4

Types of symmetry  x – axis if (x,y) then (x,-y)  y – axis If (x,y) then (-x, y)  Origin If (x,y) then (-x, -y)

If replacing y with –y yields an equivalent equation then symmetric about x-axis If replacing x with –x yields an equivalent equation then symmetric about y-axis If replacing x with –x and y with –y yields an equivalent equation then symmetric about the origin

The standard form of the equation of a circle is: (x – h) 2 +(y-k) 2 = r 2 where (x,y) is a point on the circle, (h,k) is the center and r is the radius

Find the equation of a circle whose center is (-1,2) with point (3,4) on the circle. Find the radius by using the distance formula (x +1) 2 + (y-2) 2 = 20

Linear Equations in Two Variables

is the ratio of vertical change to the horizontal change. The variable m is used to represent slope.

m = change in y-coordinate change in x-coordinate Or m = rise run Formula for Slope

m = y 2 – y 1 x 2 – x 1 or  y  x ***this is important Slope of a Line

Slope Intercept form y= mx + b, where m is the slope, b is y-intercept Point-Slope form y – y 1 = m(x- x 1 ) Writing Linear Equations in Two Variables

Parallel Lines - are two distinct nonvertical lines having identical slopes; m 1 = m 2 Perpendicular Lines – are two nonvertical lines whose slopes are the negative reciprocal of each other Parallel and Perpendicular Lines