Algebra2 Part4 Chapter1

Service Description

(Note)- It does not matter which side the variables on a special side nor the constant on any special side. It just matters that you solve it for the ending value of the variable and then it can easily be moved from one side to the other. "Some equations that may not appear to be linear at first become so while solving them." Example 1.3 Solve x(x+3) +5 = x to the exponent of 2 +7x = 10 for x Solution x( x +3) +5 = x to the exponent of 2 +7x -10Because of the x to the exponent term, this equation does not appear to be linear. x to the exponent of 2 +3x +5 = x to the exponent of 2 +7x -10 -x to the exponent of 2 -3x-5 =-x to the exponent of 2 - 7x -10 15/4 =4x/4 The x to the exponent of 2 terms on both sides canceled out, x = 3.75 leaving a linear equation. Example 1.4 Sole x +7/2x =4/5 for x Solution x +7/2x =4/5 Linear equations do not normally have the variable in a denominator. 5(x +7) = 4(2x) Cross-multiplication (an application of the multiplication property) changes this to a linear equation. 5x + 35 =35 = 8x -5x -5x ------------------------ 35/3 = 3x/3 x = 35/3 = 11.67 example 1.5 Solve 4 - (-x) =x + 5 for x Solution 4 - (3 - x) = x +5 No value of x makes this equation true. It is an example 4 -3 + x = x + 5 of an equation with 'no solution'. x + 1 = x + 5 1=5 Example 1.6 Solve 2(x + 1) - (x - 2) = x + 4 Solution 2(x + 1) - (x - 2) = x + 4 This is true for any value of x. The solution 2x +2 - x + 2 = x + 4 set is all real numbers. This equation is an x +4 = x + 4 example of an 'identity'. 4 = 4