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Standard Form

Standard Form. ALGEBRA 1 LESSON 6-3. (For help, go to Lessons 2-3 and 2-6.). Solve each equation for y . 1. 3 x + y = 5 2. y – 2 x = 10 3. x – y = 6 4. 20 x + 4 y = 8 5. 9 y + 3 x = 1 6. 5 y – 2 x = 4 Clear each equation of decimals.

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Standard Form

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  1. Standard Form ALGEBRA 1 LESSON 6-3 (For help, go to Lessons 2-3 and 2-6.) Solve each equation for y. 1. 3x + y = 5 2.y – 2x = 10 3.x – y = 6 4. 20x + 4y = 8 5. 9y + 3x = 1 6. 5y – 2x = 4 Clear each equation of decimals. 7. 6.25x + 8.5 = 7.75 8. 0.4 = 0.2x – 5 9. 0.9 – 0.222x = 1 8-5

  2. 2x+ 4 5 2 5 4 5 1 3 Standard Form ALGEBRA 1 LESSON 6-3 Solutions 1. 3x + y = 5 2. y – 2x = 10 3x – 3x + y = 5 – 3x y – 2x + 2x = 10 + 2x y = –3x + 5 y = 2x + 10 3.x – y = 6 4. 20x + 4y = 8 x = 6 + y 4y = –20x + 8 x – 6 = yy = y = x – 6 y = –5x + 2 5. 9y + 3x = 1 6. 5y – 2x = 4 9y = –3x + 1 5y = 2x + 4 y = y = y = – x + y = x + 7. Multiply each term by 100: 100(6.25x)+100(8.5) = 100(7.75) Simplify: 625x + 850 = 775 8. Multiply each term by 10: 10(0.4) = 10(0.2x) – 10(5) Simplify: 4 = 2x – 50 9. Multiply each term by 1000: 1000(0.9)–1000(0.222x)=1000(1) Simplify: 900 – 222x = 1000 –20x + 8 4 –3x+ 1 9 1 9 8-5

  3. Find the x- and y-intercepts of 2x + 5y = 6. Step 2  To find the y-intercept, substitute 0 for x and solve for y. 2x + 5y = 6 2(0) + 5y = 6 5y = 6 y = The y-intercept is . 6 5 6 5 Standard Form ALGEBRA 1 LESSON 6-3 Step 1 To find the x-intercept, substitute 0 for y and solve for x. 2x + 5y = 6 2x + 5(0) = 6 2x = 6 x = 3 The x-intercept is 3. 8-5

  4. Graph 3x + 5y = 15 using intercepts. Step 2  Plot (5, 0) and (0, 3). Draw a line through the points. Standard Form ALGEBRA 1 LESSON 6-3 Step 1  Find the intercepts. 3x + 5y = 15 3x + 5(0) = 15 Substitute 0 for y. 3x = 15 Solve for x. x = 5 3x + 5y = 15 3(0) + 5y = 15 Substitute 0 for x. 5y = 15 Solve for y. y = 3 8-5

  5. Standard Form ALGEBRA 1 LESSON 6-3 a. Graph y = 4 b. Graph x = –3. 0 • x + 1 • y = 4 Write in standard form. For all values of x, y = 4. 1 • x + 0 • y = –3 Write in standard form. For all values of y, x = –3. 8-5

  6. 2 3 3y = 3( x + 6 ) Multiply each side by 3. 2 3 y = x + 6 Standard Form ALGEBRA 1 LESSON 6-3 2 3 Write y = x + 6 in standard form using integers. 3y = 2x + 18 Use the Distributive Property. –2x + 3y = 18 Subtract 2x from each side. The equation in standard form is –2x + 3y = 18. 8-5

  7. Write an equation in standard form to find the number of hours you would need to work at each job to make a total of $130. x Define:  Let = the hours mowing lawns. Let = the hours delivering newspapers. y x y Write:  12 + 5 = 130 Standard Form ALGEBRA 1 LESSON 6-3 Amount Paid per hour Job Mowing lawns $12 Delivering newspapers Relate: $12 per h   plus $5 per h equals   $130 mowing delivering $5 The equation standard form is 12x + 5y = 130. 8-5

  8. Find each x- and y-intercepts of each equation. 1. 3x + y = 12 2. –4x – 3y = 9 3. Graph 2x – y = 6 using x- and y-intercepts. 9 4 x = – , y = –3 For each equation, tell whether its graph is horizontal or vertical. 4.y = 3 5.x = –8 6. Write y = x – 3 in standard form using integers. Standard Form ALGEBRA 1 LESSON 6-3 x = 4, y = 12 horizontal vertical 5 2 –5x + 2y = –6 or 5x – 2y = 6 8-5

  9. Find the rate of change of the data in each table. 1.2.3. Simplify each expression. 4. –3(x – 5) 5. 5(x + 2) 6. – (x – 6) x y x y x y 2 4 –3 –5 10 4 5 –2 –1 –4 7.5 –1 8 –8 1 –3 5 –6 11 –14 3 –2 2.5 –11 4 9 Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 (For help, go to Lessons 6-1 and 1-7.) 8-5

  10. Solutions 1. Use points (2, 4) and (5, –2). rate of change = = = –2 2. Use points (–3, –5) and (–1, –4). rate of change = = = = 3. Use points (10, 4) and (7.5, –1). rate of change = = = = 2 4. –3(x – 5) = –3x – (–3)(5) = –3x + 15 5. 5(x + 2) = 5x + 5(2) = 5x + 10 6. – (x – 6) = – x – (– )(6) = – x + 4 – (–2) 2 – 5 6 –3 –5 – (–4) –3 – (–1) 1 2 –5 + 4 –3 + 1 –1 –2 4 – (–1 ) 10 – 7.5 4 + 1 2.5 5 2.5 4 9 4 9 4 9 4 9 8 3 Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 8-5

  11. 1 3 The equation of a line that passes through (1, 2) with slope . 1 3 Start at (1, 2). Using the slope, go up1 unitand right 3 units to (4, 3). Draw a line through the two points. Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 Graph the equation y – 2 = (x – 1). 8-5

  12. Write the equation of the line with slope –2 that passes through the point (3, –3). Substitute (3, –3) for (x1, y1) and –2 for m. y – (–3) = –2(x – 3) y + 3 = –2(x – 3) Simplify the grouping symbols. Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 y – y1 = m(x – x1) 8-5

  13. Write equations for the line in point-slope form and in slope-intercept form. Step 1  Find the slope. y2 – y1 x2 – x1 = m 4 – 3 –1 – 2 1 3 = – 1 3 The slope is – . Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 8-5

  14. Step 3 Rewrite the equation from Step 2 in slope– intercept form. y – 4 = – (x + 1) y – 4 = – x – y = – x + 3 Step 2  Use either point to write the the equation in point-slope form. Use (–1, 4). y – y1 = m(x – x1) y – 4 = – (x + 1) 1 3 1 3 1 3 1 3 2 3 1 3 Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 (continued) 8-5

  15. Is the relationship shown by the data linear? If so, model the data with an equation. Step 1  Find the rate of change for consecutive ordered pairs. –2 –1 = 2 x y 3 6 –6 –3 –1() –2 = 2 2 4 –3() –6 –1 –2 –2() –4 –2 –1 = 2 –3 –6 Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 The relationship is linear. The rate of change is 2. 8-5

  16. Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 (continued) Step 2 Use the slope 2 and a point (2,4) to write an equation. y – y1 = m(x – x1) Use the point-slope form. y – 4 = 2(x – 2) Substitute (2, 4) for (x1, y1) and 2 for m. 8-5

  17. Is the relationship shown by the data linear? If so, model the data with an equation. Step 1  Find the rate of change for consecutive ordered pairs. 1 1 – = 1 x y –2 –2 1 () 1 1 2 –1 –1 – = 1 / 2 () 1 1 0 1 () 1 2 1 1 1 – = 1 Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 The relationship is not linear. 8-5

  18. 1. Graph the equation y + 1 = –(x – 3). 2. Write an equation of the line with slope – that passes through the point (0, 4). 3. Write an equation for the line that passes through (3, –5) and (–2, 1) in Point-Slope form and Slope-Intercept form. 4. Is the relationship shown by the data linear? If so, model that data with an equation. 2 3 2 3 2 3 y – 4 = – (x – 0), or y = – x + 4 x y 2 5 –10 –7 yes; y + 3 = (x – 0) 0 –3 5 –1 20 5 6 5 6 5 7 5 y + 5 = – (x – 3); y = – x – Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 8-5

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