Sulalgtrig7e Isg 1 1
Download as PPT, PDF1 like209 views
This document provides examples of solving linear equations by manipulating the equations to isolate the variable. It shows adding or subtracting the same quantity from both sides, distributing terms, combining like terms, and dividing both sides by a coefficient to isolate the variable. The examples are worked through step-by-step and checked by substitution back into the original equation.
Sulalgtrig7e Isg 1 1
- 1. 2x − 3 = 1 N S 3x = 4 T IO U A E Q −1 = 2 − 4x A R IN E L 3x − 1 = 5 − 2 x
- 2. A linear equation in one variable is equivalent to an equation of the form: ax + b = 0 If we have a linear equation we can “manipulate” it to get it in this form. We just need to make sure that whatever we do preserves the equality (keeps both sides =) 3x + 1 = 4 We can add or subtract the same thing from both sides of the equation. -4 -4 -3 0 Notice this is the equation 3x − 3 = 0 above where a = 3 and b = -3. While this is in the general form for a linear equation, we often want to find all values of x so that the equation is true. You could probably do this one in your head and see that when x = 1 we’d have a true statement 0 = 0
- 3. We will want to find those values of x that make the equation true by isolating the x (this means get the x all by itself on one side of the equal sign) Since the x is in more than 2( x − 4 ) = −5( 2 x + 1) one place and inside of parenthesis the first thing we’ll do is get rid of parenthesis by distributing. 2 x − 8 = −10 x − 5 Now let’s get all constants +8 (terms without x’s) on the right +8 side. We’ll do this by adding 8 +3 to both sides. 2 x = −10 x + 3 We are ready to get all x terms on the left side by adding 10x to both sides. + 10x + 10x 1 Now get the x by itself by 12 x = 3 x= getting rid of the 12. 12x means 12 12 4 12 times x so we get rid of it by dividing both sides by 12.
- 4. Let’s check this answer by substituting it into the original equation to see if we get a true statement. 2( − 4 ) = −5( ) 1 1 x 4 ? 2 x + 1 Distribute and multiply 4 1 ? 1 − 8 = −5 + 1 Distribute 2 2 1 ? 5 −8 = − −5 Get a common denominator 2 2 1 16 ? 5 10 15 15 − =− − − =− 2 2 2 2 2 2 It checks!
- 5. y +1 3 y Solve the following equation: − = 3 4 2 When we see an equation with fractions, it is generally easiest to find the common denominator and multiply all terms by this common denominator. This will give you an equivalent “fraction free” equation to solve. 4 3 6 The common denominator is 12 so we’ll 12 y +1 3(12) y (12) multiply each term by 12. − = 3 4 2 Cancel all denominators 4( y + 1) − 9 = 6 y Here is our “fraction free” equation to solve 4y + 4 − 9 = 6y Distribute −5 = 2y Subtract 4y from both sides and combine like terms 5 − =y Divide both sides by 2 2
- 6. 2 3 1 Solve the following equation: − = 2 x −1 x +1 x −1 Let's get a "fraction free" equation by multiplying by the common denominator. Factor any denominators that will factor so you can determine the lowest common denominator. 2(x+1)(x-1)3(x+1)(x-1) 1(x+1)(x-1) − = The common denominator is (x+1)(x-1) x − 1 x + 1 ( x + 1)( x − 1) so multiply all terms by this Cancel all denominators 2( x + 1) − 3( x − 1) = 1 Here is our “fraction free” equation to solve Distribute 2 x + 2 − 3x + 3 = 1 Combine like terms − x +5 =1 subtract 5 from both sides − x = −4 divide both sides by -1 Make sure the answer would not cause you to divide by 0 in x=4 the original equation. (Only 1 and -1 would cause this so we are okay).