# 1-3 Solving Addition and Subtraction Equations Warm Up

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1-3 Solving Addition and Subtraction Equations Warm Up
Pre-Algebra Warm Up Write an algebraic expression for each word phrase. 1. a number x decreased by 9 2. 5 times the sum of p and 6 3. 2 plus the product of 8 and n 4. the quotient of 4 and a number c x  9 5(p + 6) 2 + 8n 4 c

Learn to solve equations using addition and subtraction.

Vocabulary equation solve solution inverse operation
isolate the variable Addition Property of Equality Subtraction Property of Equality

An equation uses an equal sign to show that two expressions are equal
An equation uses an equal sign to show that two expressions are equal. All of these are equations. 100 2 = 50 3 + 8 = 11 r + 6 = 14 24 = x – 7 To solve an equation, find the value of the variable that makes the equation true. This value of the variable is called the solution of the equation.

Determine which value of x is a solution of the equation.
Additional Example 1: Determining Whether a Number is a Solution of an Equation Determine which value of x is a solution of the equation. x + 8 = 15; x = 5, 7, or 23 Substitute each value for x in the equation. x + 8 = 15 ? 5 + 8 = 15 ? Substitute 5 for x. 13= 15 ? So 5 is not solution.

Determine which value of x is a solution of the equation. x + 8 = 15; x = 5, 7, or 23 Substitute each value for x in the equation. x + 8 = 15 ? 7 + 8 = 15 ? Substitute 7 for x. 15= 15 ? So 7 is a solution.

Determine which value of x is a solution of the equation. x + 8 = 15; x = 5, 7, or 23 Substitute each value for x in the equation. x + 8 = 15 ? = 15 ? Substitute 23 for x. 31= 15 ? So 23 is not a solution.

Addition and subtraction are inverse operations, which means they “undo” each other.
To solve an equation, use inverse operations to isolate the variable. This means getting the variable alone on one side of the equal sign.

To solve a subtraction equation, like y  15 = 7, you would use the Addition Property of Equality. ADDITION PROPERTY OF EQUALITY Words Numbers Algebra You can add the same number to both sides of an equation, and the statement will still be true. 2 + 3 = 5 x = y + 4 x = y + z 2 + 7 = 9

SUBTRACTION PROPERTY OF EQUALITY
There is a similar property for solving addition equations, like x + 9 = 11. It is called the Subtraction Property of Equality. SUBTRACTION PROPERTY OF EQUALITY Words Numbers Algebra You can subtract the same number from both sides of an equation, and the statement will still be true. 4 + 7 = 11 x = y  3 x = y  z 4 + 4 = 8

Subtract 10 from both sides. 0 + n = 8 n = 8
Additional Example 2A: Solving Equations Using Addition and Subtraction Properties Solve. A n = 18 10 + n = 18 –10 –10 Subtract 10 from both sides. 0 + n = 8 n = 8 Identity Property of Zero: 0 + n = n. Check 10 + n = 18 ? = 18 18 = 18 ?

Identity Property of Zero: p + 0 = p. Check p – 8 = 9 17 – 8 = 9
Additional Example 2B: Solving Equations Using Addition and Subtraction Properties Solve. B. p – 8 = 9 p – 8 = 9 + 8 + 8 Add 8 to both sides. p + 0 = 17 p = 17 Identity Property of Zero: p + 0 = p. Check p – 8 = 9 ? 17 – 8 = 9 9 = 9 ?

Identity Property of Zero: y + 0 = 0. Check 22 = y – 11 22 = 33 – 11
Additional Example 2C: Solving Equations Using Addition and Subtraction Properties Solve. C. 22 = y – 11 22 = y – 11 + 11 + 11 Add 11 to both sides. 33 = y + 0 33 = y Identity Property of Zero: y + 0 = 0. Check 22 = y – 11 ? 22 = 33 – 11 22 = 22 ?

Subtract 15 from both sides. 0 + n = 14 n = 14
Try This: Example 2A Solve. A n = 29 15 + n = 29 –15 –15 Subtract 15 from both sides. 0 + n = 14 n = 14 Identity Property of Zero: 0 + n = n. Check 15 + n = 29 ? = 29 29 = 29 ?

= = + + Additional Example 3A Solve: x 34 16,550 x + 34 = 16,550 –34
A. Jan took a 34-mile trip in her car, and the odometer showed 16,550 miles at the end of the trip. What was the original odometer reading? odometer reading at the beginning of the trip miles traveled + = odometer reading at the end of the trip + = Solve: x 34 16,550 x + 34 = 16,550 –34 – 34 Subtract 34 from both sides. x + 0 = 16,516 x = 16,516 The original odometer reading was 16,516 miles.

= = + + Additional Example 3B Solve: 895 n 1125 895 + n = 1125 –895
B. From 1980 to 2000, the population of a town increased from 895 residents to 1125 residents. What was the increase in population during that 20-year period? initial population increase in population + = population after increase + = Solve: 895 n 1125 895 + n = 1125 –895 – 895 Subtract 895 from both sides. 0 + n = 230 n = 230 The increase in population was 230.

= = + + Try This: Example 3A Solve: x 27 535 x + 27 = 535 –27 – 27
A. Isabelle earned \$27 interest and now has a balance of \$535 in the bank. What was her balance before interest was added? balance before interest interest earned + = balance after interest + = Solve: x 27 535 x + 27 = 535 –27 – 27 Subtract 27 from both sides. x + 0 = 508 x = 508 Isabelle had a balance of \$508 before interest was added.