ALGEBRA: Solve (Single) Equations
KEY CONCEPTS TO LEARN:
 Solve Equations
 Intuition 2x = 10, x = 5
 Algebra Rules
 Eventually, too complex to use just intuition
 Must resort to rules of algebra
 Do same operation on both entire sides
 generally inverse operations to “undo” complexity
 Isolate variable, x
 x = number e.g. x = 5
 x “in terms of” y, x = 2y – 3
 Do same operation on both entire sides
 What to do given different equation presentations?
 Base case: 2step equation 2x + 10 = 20
 Subtract 10 to “undo” +10 on same side as x, 2x = 10
 Divide by 2 to “undo” multiply by 2, 2x/2 = 10/2, x = 5
 Base case: 2step equation 2x + 10 = 20
 Solve (more complex) Equations

 Same Variable on both sides. e.g. 5x + 10 = 4x – 5
 Add/subtract variables to isolate x on one side
 e.g. 5x + 10 – 4x = 4x – 5 – 4x, x + 10 = 5
 Parentheses e.g. 2(x + 5) = (x – 5)
 Distribute to get rid of parentheses
 a(b + c) = ab + ac distributive property
 2x + 2•5 = x – (5) becomes 2x + 10 = x + 5
 * Be careful of signs * (huge source of mistakes)
 Distribute to get rid of parentheses
 Denominators/Fractions in equation e.g. 3/x = 10
 Multiply by the denominator(s) on both sides to get rid of the denominators/fractions e.g. x(3/x) = x(10), 3 = 10x
 1st step after simplification if not already simplified
 maybe find common denominator if fraction add/subtract
 maybe cross multiply if proportion equation a/b = c/d
 goal is to get an equation w/o fractions/denominators
 multiply by the entire denominator. e.g. 3/(x+1) = 10
 (x+1)(3/(x+1) = (x+1)10, 3 = 10(x+1)
 Multiply by the denominator(s) on both sides to get rid of the denominators/fractions e.g. x(3/x) = x(10), 3 = 10x
 Exponents, squares, or square roots
 Isolate the radical or squared term with the variable
 e.g. (x + 1)^{2} – 6 = 10, (x + 1)^{2} = 16
 Apply the inverse function (undo a square with a square root etc.)
 e.g. (x + 1) = ±4
 Same Variable on both sides. e.g. 5x + 10 = 4x – 5

 Check your result
 Plug your answer back into the equation
 make it a habit to avoid mistakes
 almost half of all mistakes result from bad habits, not bad math knowledge
 Plug your answer back into the equation