Gateway Quiz Reminders: The next Gateway will be given in class next week or the week after )check your course calendar.) Before then, each student who did not score 8/8 on the first Gateway must go over that quiz one-on- one with a teacher or TA and get a signature from them on the bottom of the second page of the quiz. This can be done either in the open lab or in any class session after lecture. You must show your classroom teacher the signed quiz before you will be allowed to take the next version.

NOW PLEASE CLOSE YOUR LAPTOPS and prepare to take notes.

Sections 2.2 An algebraic equation is a statement that two expressions have equal value. Solving algebraic equations involves finding values for a variable that make the equation true. Equivalent equations are equations with the same solutions.

A linear equation in one variable can be written in the form ax + b = c, a  0 Examples: 3x + 6 = -10 ½x – 2 = ¼ 4x = 0 7y + 10 = 0 Note: “Linear” means that the variable is not raised to a power. So x 3 – 2x + 1 = 0 is NOT a linear equation.

A solution to an equation is a number that can be substituted in place of the variable that makes the equation a true statement. Examples: Is x = 5 a solution of the equation 2x – 10 = 0? Check: Sub 5 in for x: 2·5 – 10 = 10 – 10 = 0  So x = 5 IS a solution of 2x – 10 = 0. Is y = -3 a solution of the equation 3y + 10 = -2? Check: Sub -3 in for y: 3·(-3) + 10 = = 1, which is not equal to -2 So x = -3 IS NOT a solution of 3y + 10 = -2.

Example: x – 1 = 5 and 3x + 1 = 19 are equivalent equations because plugging in x = 6 makes both of them true. Equivalent equations are equations with the same solutions.

In sections 2.2 and 2.3 we will study two properties that can be applied to linear equations in order to find a solution (a number that makes the statement true when it is plugged in for the variable). These properties allow us to make complicated linear equations into simpler equivalent linear equations. NOTE: Our final goal in solving linear equations is to obtain an equivalent equation of the very simple form “ x = a number”. Addition Property of Equality: (Section 2.2) If a = b then a + c = b + c and a – c = b – c. Examples: If x = y, then x + 3 = y + 3 If x = y, then x - 7= y - 7 Multiplication Property of Equality: (Section 2.3) If a = b then ac = bc and a/c = b/c. Examples: If x = y, then 5x= 5y If x = y, then x/6= y/6

Solving linear equations in one variable 1) Simplify each side of equation by distributing where necessary and then combining like terms. 2) Get all variable terms on one side and number terms on the other side of equation (addition property of equality). 3) Get variable alone with no coefficient (i.e. an understood coefficient of 1) in front of it (multiplication property of equality). 4) ALWAYS check solution by substituting into original problem.

3a + 2a = 7 + 4a 5a = 7 + 4a (Simplify the left side) a = 7 (Simplify both sides) Examples of Solving Equations 5a + -4a = 7 + 4a + -4a (Add –4a to each side) Now CHECK your answer in 3a + 2a = 7 + 4a: Sub 7 in for a in the original equation: First side: 3·7 + 2·7 = = 35 Second side: 7 + 4·7 = = = 35 

Problem from today’s homework: Don’t forget: Always check your answer by plugging it back in for the variable in the original equation given in the problem to see if both sides of the equation come out to the same number.

Problem from today’s homework:

Don’t forget: Always check your answer by plugging it back in for the variable in the original equation given in the problem to see if both sides of the equation come out to the same number. (You can use “check answer” on homework, but you won’t have that available on quizzes, so you need to learn how to do this and practice it on your homework problems.)

15 – (6 – 7k) = 2 + 6k 15 – 6 + 7k = 2 + 6k (Use distributive property) 9 + 7k = 2 + 6k (Simplify the left side) k = -7 (Simplify both sides) 9 + 7k + -6k = 2 + 6k + -6k (Add –6k to each side) 9 + k = 2 (Simplify both sides) k = (Add –9 to both sides) Example of Solving Equation Remember to check your answer by plugging -7 in for k in the equation.

Problem from today’s homework:

Example of Solving Equation 3z – 1 = 26 3z = 27 (simplify both sides) z = 9 (simplify both sides) 3z – = (add 1 to both sides) (divide both sides by 3) (From section 2.3: Multiplication Property of Equality) Now check back into 3z – 1 = 26: 3∙9 – 1 = – 1 =26 

-y = 8 y = -8 (simplify both sides) Example of solving a simple equation using the multiplication property: (-1)(-y) = 8(-1) (multiply both sides by –1)

(multiply both sides by 7)(simplify both sides) Another example of solving an equation using the multiplication property:

(multiply both sides by fraction) (simplify both sides) Another example of solving an equation using the multiplication property:

Plan for solving more complicated linear equations: Steps for solving linear equations in one variable: 1) Multiply to clear fractions (if necessary). 2) Use the distributive property as needed. 3) Simplify each side of the equation by combining like terms. 4) Get all the variable (letter) terms on one side and the constants (pure number terms) on the other side of equation by using the addition property of equality. 5) Simplify each side again by combining like terms. 6) Get the variable completely by itself (get rid of the coefficient (the number in front of the letter) by using the multiplication property of equality. 7) Simplify the answer if needed. 8) Check solution by substituting into original problem.

Using these steps to solve a linear equation: Example: Solve the equation 2(x – 5) + 7x = 5 + 3(4 -x) Step 1: There are no fractions to clear. Step 2: Use the distributive property: 2∙x + 2∙(-5) + 7x = 5 + 3∙4 +3∙(-x) 2x – x = – 3x Step 3: Simplify both sides: 2x – x = – 3x 9x – 10 = 17 – 3x 1.Multiply to clear fractions (if necessary). 2.Use the distributive property as needed. 3.Simplify each side of the equation by combining like terms. 4.Get all the variable (letter) terms on one side and the constants (pure number terms) on the other side of equation by using the addition property of equality. 5.Simplify each side again by combining like terms. 6.Get the variable completely by itself (get rid of the coefficient (the number in front of the letter) by using the multiplication property of equality. 7.Simplify the answer if needed. 8.Check solution by substituting into original problem.

Example (continued): Solve 2(x – 5) + 7x = 5 + 3(4 -x) Step 3 left us with this equation: 9x – 10 = 17 – 3x Step 4: Get all variable terms on one side and all pure numbers on other. 9x – 10 = 17 – 3x add 3x to both sides add 10 to both sides +3x x 12x = 27 Step 5: Simplify each side again. Step 6: Divide both sides by 12 to get x by itself: 12x = Multiply to clear fractions (if necessary). 2.Use the distributive property as needed. 3.Simplify each side of the equation by combining like terms. 4.Get all the variable (letter) terms on one side and the constants (pure number terms) on the other side of equation by using the addition property of equality. 5.Simplify each side again by combining like terms. 6.Get the variable completely by itself (get rid of the coefficient (the number in front of the letter) by using the multiplication property of equality. 7.Simplify the answer if needed. 8.Check solution by substituting into original problem.

Example (continued): Solve 2(x – 5) + 7x = 5 + 3(4 -x) Step 6 left us with Step 7: Simplify the answer: ALWAYS DO THIS FINAL STEP : Check the solution by substituting the number back in for x in original equation. 1.Multiply to clear fractions (if necessary). 2.Use the distributive property as needed. 3.Simplify each side of the equation by combining like terms. 4.Get all the variable (letter) terms on one side and the constants (pure number terms) on the other side of equation by using the addition property of equality. 5.Simplify each side again by combining like terms. 6.Get the variable completely by itself (get rid of the coefficient (the number in front of the letter) by using the multiplication property of equality. 7.Simplify the answer if needed. 8.Check solution by substituting into original problem.

Example (continued): Solve 2(x – 5) + 7x = 5 + 3(4 -x) ALWAYS DO THIS FINAL STEP : Check the solution (9/4) by substituting the number back in for x in original equation. 2(x – 5) + 7x = 5 + 3(4 - x) Left side: 2(9/4 – 5) + 7∙9/4 = Right side: 5 + 3(4 – 9/4) = Use the order of operations and rules for working with fractions to calculate out both sides. Since both sides of the equation come out to the same number, we can be certain that 9/4 is the correct solution 1.Multiply to clear fractions (if necessary). 2.Use the distributive property as needed. 3.Simplify each side of the equation by combining like terms. 4.Get all the variable (letter) terms on one side and the constants (pure number terms) on the other side of equation by using the addition property of equality. 5.Simplify each side again by combining like terms. 6.Get the variable completely by itself (get rid of the coefficient (the number in front of the letter) by using the multiplication property of equality. 7.Simplify the answer if needed. 8.Check solution by substituting into original problem.

Problem from today’s homework:

Reminder: This homework assignment on section 2.2 is due at the start of next class period. Remember to go over your Gateway Quiz with a teacher or TA if you haven’t done so yet.

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