ALGEBRA 1 LESSON 6 3 Standard Form For

ALGEBRA 1 LESSON 6 -3 Standard Form (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. 7. 6. 25 x + 8. 5 = 7. 75 8. 0. 4 = 0. 2 x – 5 9. 0. 9 – 0. 222 x = 1 8 -5

ALGEBRA 1 LESSON 6 -3 Standard Form Solutions 1. 3 x + y = 5 3 x – 3 x + y = 5 – 3 x y = – 3 x + 5 3. x – y = 6 x=6+y x– 6=y y=x– 6 5. 9 y + 3 x = 1 9 y = – 3 x + 1 y = – 3 x+ 1 9 1 y=– x+ 1 3 9 2. y – 2 x = 10 y – 2 x + 2 x = 10 + 2 x y = 2 x + 10 4. 20 x + 4 y = 8 4 y = – 20 x + 8 y= – 20 x + 8 4 y = – 5 x + 2 6. 5 y – 2 x = 4 5 y = 2 x + 4 y = 2 x+ 4 5 2 y = x +4 5 5 8 -5 7. Multiply each term by 100: 100(6. 25 x)+100(8. 5) = 100(7. 75) Simplify: 625 x + 850 = 775 8. Multiply each term by 10: 10(0. 4) = 10(0. 2 x) – 10(5) Simplify: 4 = 2 x – 50 9. Multiply each term by 1000: 1000(0. 9)– 1000(0. 222 x)=1000(1) Simplify: 900 – 222 x = 1000

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

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

ALGEBRA 1 LESSON 6 -3 Standard Form 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

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

ALGEBRA 1 LESSON 6 -3 Standard Form 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. Job Mowing lawns Delivering newspapers Amount Paid per hour Define: Let x = the hours mowing lawns. Let y = the hours delivering newspapers. $12 $5 Relate: $12 per h plus mowing 12 x Write: + $5 per h equals $130 delivering 5 y = The equation standard form is 12 x + 5 y = 130. 8 -5 130

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

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

Point-Slope Form and Writing Linear Equations Solutions ALGEBRA 1 LESSON 6 -4 4 – (– 2) 6 1. Use points (2, 4) and (5, – 2). rate of change = 2 – 5 = – 3 = – 2 2. Use points (– 3, – 5) and (– 1, – 4). rate of change = – 5 – (– 4) – 5 + 4 – 1 1 = = = – 3 – (– 1) – 3 + 1 – 2 2 4 – (– 1 ) 4+1 3. Use points (10, 4) and (7. 5, – 1). rate of change = 10 – 7. 5 = 2. 5 4. – 3(x – 5) = – 3 x – (– 3)(5) = – 3 x + 15 5. 5(x + 2) = 5 x + 5(2) = 5 x + 10 6. – 4 (x – 6) = – 4 x – (– 4 )(6) = – 4 x + 8 9 9 3 8 -5 5 = 2

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

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

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

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6 -4 (continued) Step 2 Use either point to write the equation in point-slope form. Step 3 Rewrite the equation from Step 2 in slope– intercept form. y – 4 = – 1 (x + 1) Use (– 1, 4). 3 y– 4=– 1 x– 1 3 3 y = – 1 x + 32 3 3 y – y 1 = m(x – x 1) y – 4 = – 1 (x + 1) 3 8 -5

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6 -4 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( – 3( – 2( x y 3 2 – 1 – 3 6 4 – 2 – 6 ) – 2 =2 – 1 ) – 6 ) – 4 – 6 =2 – 3 – 2 =2 – 1 The relationship is linear. The rate of change is 2. 8 -5

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 – y 1 = m(x – x 1) Use the point-slope form. y – 4 = 2(x – 2) Substitute (2, 4) for (x 1, y 1) and 2 for m. 8 -5

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6 -4 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( 2( 1( x – 2 – 1 1 2 y – 2 – 1 0 1 )1 – 1 1 =1 )1 )1 – 1 2 =/ 1 – 1 1 =1 The relationship is not linear. 8 -5

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