Solve each equation for y. 1. 3x + y = 52. y – 2x = x – y = x + 4y = 85. 9y + 3x = 16. 5y – 2x = 4 Clear each equation of decimals x = = 0.2x – – 0.222x = 1 Standard Form (For help, go to Lessons 2-3 and 2-6.) ALGEBRA 1 LESSON

1. 3x + y = 52. y – 2x = 10 3x – 3x + y = 5 – 3xy – 2x + 2x = x y = –3x + 5 y = 2x x – y = 64.20x + 4y = 8 x = 6 + y 4y = –20x + 8 x – 6 = y y = y = x – 6 y = –5x y + 3x = 16.5y – 2x = 4 9y = –3x + 1 5y = 2x + 4 y = y = y = – x + y = x + –20x –3x Standard Form ALGEBRA 1 LESSON Solutions 7.Multiply each term by 100: 100(6.25x)+100(8.5) = 100(7.75) Simplify: 625x = Multiply each term by 10: 10(0.4) = 10(0.2x) – 10(5) Simplify: 4 = 2x – Multiply each term by 1000: 1000(0.9)–1000(0.222x)=1000(1) Simplify: 900 – 222x = x

Find the x- and y-intercepts of 2x + 5y = 6. 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. 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 Standard Form ALGEBRA 1 LESSON

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

a. Graph y = 4b. Graph x = –3. Standard Form 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. ALGEBRA 1 LESSON

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

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

Find each x- and y-intercepts of each equation. 1. 3x + y = 122. –4x – 3y = 9 3. Graph 2x – y = 6 using x- and y-intercepts. For each equation, tell whether its graph is horizontal or vertical. 4. y = 35. x = –8 6. Write y = x – 3 in standard form using integers Standard Form x = 4, y = 12 horizontalvertical –5x + 2y = –6 or 5x – 2y = 6 x = –, y = – ALGEBRA 1 LESSON

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

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 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 (x + 2) = 5x + 5(2) = 5x – (x – 6) = – x – (– )(6) = – x –5 + 4 – – (–2) 2 – 5 4 – (–1 ) 10 – 7.5 –5 – (–4) –3 – (–1) –1 –2 6 – Solutions

1313 The equation of a line that passes through (1, 2) with slope. Graph the equation y – 2 = (x – 1). Point-Slope Form and Writing Linear Equations 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. ALGEBRA 1 LESSON

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

The slope is – Step 1 Find the slope. = m y 2 – y 1 x 2 – x 1 4 – 3 –1 – 2 = – 1313 Point-Slope Form and Writing Linear Equations Write equations for the line in point-slope form and in slope-intercept form. ALGEBRA 1 LESSON

Step 3 Rewrite the equation from Step 2 in slope– intercept form. y – 4 = – (x + 1) y – 4 = – x – y = – x (continued) Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON Step 2 Use either point to write the the equation in point-slope form. Use (–1, 4). y – y 1 = m(x – x 1 ) y – 4 = – (x + 1) 1313

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

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. Step 2Use the slope 2 and a point (2,4) to write an equation. (continued) Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON

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

–10– –3 –1 5 xy 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 Point-Slope Form and Writing Linear Equations y – 4 = – (x – 0), or y = – x y + 5 = – (x – 3); y = – x – 2525 yes; y + 3 = (x – 0) ALGEBRA 1 LESSON