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2.1 – Linear Equations in One Variable

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. 2.1 – Linear Equations in One Variable. Linear equation in one variable can be written in the form:

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2.1 – Linear Equations in One Variable

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  1. Algebraic equationis a statement that two expressions have equal value. Solving algebraic equations involves finding values for a variable that make the equation true. 2.1 – Linear Equations in One Variable Linear equation in one variable can be written in the form: ax + b = c, a  0. Equivalent equations are equations with the same solutions in the form of: variable = number, x = 3, or number = variable, 3 = x.

  2. A linear equation in one variable can be written in the form: ax + b = c where a, b, and c are real numbers and a ≠ 0. 2.1 – Linear Equations in One Variable Addition Property of Equality a = b and a + c = b + c equivalent equations Multiplication Property of Equality • a = b and ac = bc • equivalent equations

  3. Example 8 + z = – 8 2.1 – Linear Equations in One Variable Add –8 to each side 8 + (–8) + z = –8 + –8 Simplify both sides z = –16

  4. Example 3z – 1 = 26 2.1 – Linear Equations in One Variable 3z – 1 + 1 = 26 + 1 3z = 27 z = 9

  5. Example 4p – 11 – p = 2 + 2p – 20 3p – 11 = 2p – 18 2.1 – Linear Equations in One Variable 3p + (– 2p) – 11 = 2p + (– 2p) – 18 p – 11 = –18 p – 11 + 11 = –18 + 11 p = –7

  6. Example 5(3 + z) – (8z + 9) = – 4z 15 + 5z – 8z – 9 = – 4z 2.1 – Linear Equations in One Variable 6 – 3z = – 4z 6 – 3z + 4z = – 4z + 4z 6 + z = 0 6 + (–6) + z = 0 + (–6) z = –6

  7. Example 12x + 30 + 8x – 6 = 10 20x + 24 = 10 2.1 – Linear Equations in One Variable 20x + 24 + (–24) = 10 + (–24) 20x = –14

  8. Example LCD: 5 2.1 – Linear Equations in One Variable

  9. Example 5x – 5 = 2(x + 1) + 3x – 7 5x – 5 = 2x + 2 + 3x – 7 2.1 – Linear Equations in One Variable 5x – 5 = 5x – 5 Both sides of the equation are identical. This equation will be true for every x that is substituted into the equation. The solution is “all real numbers.” The equation is called an identity.

  10. Example 3x – 7 = 3(x + 1) 3x – 7 = 3x + 3 2.1 – Linear Equations in One Variable 3x + (– 3x) – 7 = 3x + (– 3x) + 3 –7 = 3 No value for the variable x can be substituted into this equation that will make this a true statement. There is “no solution.” The equation is a contradiction.

  11. General Strategy for Problem Solving UNDERSTAND the problem. Read and reread the problem. . 2.2 - An Introduction to Problem Solving Choose a variable to represent the unknown. Construct a drawing. Propose a solution and check. TRANSLATE the problem into an equation. SOLVE the equation. INTERPRET the result: Check proposed solution in problem. State your conclusion.

  12. Example Twice a number plus three is the same as the number minus six. 2.2 - An Introduction to Problem Solving 2x +3 = x – 6

  13. Example The product of twice a number and three is the same as the difference of five times the number and ¾. Find the number. 2.2 - An Introduction to Problem Solving 2x ·3 = 5x 2x · 3 = 5x – ¾ 6x = 5x – ¾ 6x + (–5x) = 5x + (–5x) – ¾ x = –¾

  14. Example A car rental agency advertised renting a Buick Century for $24.95 per day and $0.29 per mile. If you rent this car for 2 days, how many whole miles can you drive on a $100 budget? 2.2 - An Introduction to Problem Solving x = the number of whole miles driven 0.29x = the cost for mileage driven cost for mileage driven + daily rental cost = $100 0.29x + 2(24.95) = 100 0.29x + 49.90 = 100

  15. 0.29x + 49.90 = 100 0.29x + 49.90 – 49.90 = 100 – 49.90 0.29x = 50.10 2.2 - An Introduction to Problem Solving x 172.75 miles

  16. Example A pennant in the shape of an isosceles triangle is to be constructed an Athletic Club and sold at a fund-raiser. The company manufacturing the pennant charges according to perimeter, and the athletic club has determined that a perimeter of 149 centimeters should make a nice profit. If each equal side of the triangle is twice the length of the third side, increased by 12 centimeters, find the lengths of the sides of the triangular pennant. 2.2 - An Introduction to Problem Solving Let x = the third side of the triangle 2x + 12 2x + 12 2x + 12 = the first side 2x + 12 = the second side x

  17. + x Perimeter is 149 cm. 2x + 12 2x + 12 2.2 - An Introduction to Problem Solving x 2x + 12 + 2x + 12 = 149 cm 2(25) + 12 = 62 cm

  18. Example The measure of the second angle of a triangle is twice the measure of the smallest angle. The measure of the third angle of the triangle is three times the measure of the smallest angle. Find the measures of the angles. 2.2 - An Introduction to Problem Solving Let x = degree measure of smallest angle 2x = degree measure of second angle 3x = degree measure of third angle

  19. + 2x + 3x The sum of the measures of the angles of a triangle equals 180. x = 180 2.2 - An Introduction to Problem Solving 6x = 180 x = 30º 2x = 60º 3x = 90º

  20. Example The sum of three consecutive even integers is 252. Find the integers. 2.2 - An Introduction to Problem Solving x = the first even integer x + 2 = next even integer x + 4 = next even integer x + x + 2 + x + 4 = 252 3x + 6 = 252 3x = 246 The integers are: 82, 84 and 86.

  21. A formula: An equation that states a known relationship among multiple quantities (has more than one variable in it). 2.3 – Formulas and Problem Solving A = lw(Area of a rectangle = length · width) I = PRT(Simple Interest = Principal · Rate · Time) d = rt(distance = rate · time) V = lwh(Volume of a rectangular solid = length·width·height)

  22. Solving Equations for a Specified Variable Step 1:Multiply on both sides to clear the equation of fractions if they appear. Step 2:Use the distributive property to remove parentheses if they appear. 2.3 – Formulas and Problem Solving Step 3: Simplify each side of the equation by combining like terms. Step 4: Get all terms containing the specified variable on one side and all other terms on the other side by using the addition property of equality. Step 5: Get the specified variable alone by using the multiplication property of equality.

  23. Example Solve: T = mnr for n. 2.3 – Formulas and Problem Solving

  24. Example Solve: 3y – 2x = 7 for y. 2.3 – Formulas and Problem Solving

  25. Example Solve: for A = P + PRT for T. 2.3 – Formulas and Problem Solving

  26. Example Solve: for P. Factor out P on the right side. 2.3 – Formulas and Problem Solving Divide both sides by 1 + RT. Simplify.

  27. Example A flower bed is in the shape of a triangle with one side twice the length of the shortest side, and the third side is 30 feet more than the length of the shortest side. Find the dimensions if the perimeter is 102 feet. 2.3 – Formulas and Problem Solving x = the length of the shortest side 2x = the length of the second side x + 30 = the length of the third side x 2x x + 30

  28. 2x x Perimeter is 102 feet x + 30 2.3 – Formulas and Problem Solving P=a+b+c 102 = x + 2x + x + 30 102 = 4x + 30 102 – 30 = 4x + 30 – 30 72 = 4x x = 18 feet 2(18) = 36 feet 18 + 30 = 48 feet

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