ALGEBRA 2016-17 – UNIT 3 - SOLVING LINEAR EQUATIONS AND INEQUALITIES SOLVING LINEAR EQUATIONS (DAY 1) How to Solve Basic Equation: 7x + 5 = 40 -5 -5 Box VARIABLE TERM and Solve for variable 7x = 35 Get TERM by itself on one side of the = sign 7 7 Solve for VARIABLE BOXED TERM: x = 5 How to Solve Equation with ( ) 3y (8y 6) = 46 – – Circle # in front of ( ) **if a fraction, follow directions pg. 3 for fractions in equation.** 3y 8y + 6 = 46 – Use distributive property to get rid of -5y + 6 = 46 Solve remaining equation using steps from above -6 -6 -5y = 40 -5 -5 y = -8 the following: 4x + 3 = 11 2. 8y (5y + 2) = 16 – What if there is a variable on both sides?! Rover Red Rover bring the smaller x over.” “Red 9x 6 = 5x 15 + x 4. – –
ALGEBRA – 2016-17 UNIT 3 - SOLVING LINEAR EQUATIONS AND INEQUALITIES SOLVING LINEAR EQUATIONS (DAY 1) How to Solve Basic Equation: Box VARIABLE TERM and Solve for variable Get TERM by itself on one side of the = sign Solve for VARIABLE BOXED TERM: 7x + 5 = 40 -5 -5 7x = 35 7 7 x = 5 How to Solve Equation with ( ) Circle # in front of ( ) **if a fraction, follow directions pg. 3 for fractions in equation.** Use distributive property to get rid of Solve remaining equation using steps from above 3y – (8y – 6) = 46 3y – 8y + 6 = 46 -5y + 6 = 46 -6 -6 -5y = 40 -5 -5 y = -8 the following: 4x + 3 = 11 2. 8y – (5y + 2) = 16 What if there is a variable on both sides?! “Red Rover Red Rover bring the smaller x over.” 9x – 6 = 5x – 15 + x 4.