5-3 Standard Form Hubarth Algebra

Standard Form of a Linear Equation The standard form of a linear equation is Ax + By = C, where A, B and C are real numbers, and A and B are not both 0. Ex 1 Finding x- and y- Intercepts 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. Step 2 To find the y-intercept, substitute 0 for x and solve for y. 2x + 5y = 6 2x + 5y = 6 2x + 5(0) = 6 2(0) + 5y = 6. 5y = 6 2x = 6 y = 6 5 x = 3 The y-intercept is . 6 5 The x-intercept is 3.

Ex 2 Graphing Lines Using Intercepts Graph 3x + 5y = 15 using intercepts. Step 1  Find the intercepts. 3x + 5y = 15 3x + 5(0) = 15 Substitute 0 for y. 3x = 15 Solve for x. x = 5 3(0) + 5y = 15 Substitute 0 for x. 5y = 15 Solve for y. y = 3 Step 2  Plot (5, 0) and (0, 3). Draw a line through the points.

Ex 3 Graphing Horizontal and Vertical Lines 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.

Ex 4 Transforming To Standard Form Write y = x + 6 in standard form using integers. 2 3 y = x + 6 2 3 2 3 (3)y = (3) x + 6(3) Multiply every term by 3. 3y = 2x + 18 Use the Distributive Property. –2x + 3y = 18 Subtract 2x from each side. The equation in standard form is –2x + 3y = 18.

. . Practice 1. Find the x- and y-intercepts of 2x + 3y = 6. x-int = (3, 0) y-int = (0, 2) 2. Graph 4x + 2y = -8 using the x- and y- intercepts. x-int = (-2, 0) y-int = (0, -4) . . 3. What type of line does the graph of each equation form. a. x = 3 b. y = -6 vertical line horizontal line 2x + 5y = 5