 3.2 Solving Systems Algebraically. Solving System Algebraically Substitution y = 2x + 5 x = -y + 14.

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3.2 Solving Systems Algebraically

Solving System Algebraically Substitution y = 2x + 5 x = -y + 14

Solving System Algebraically Substitution y = 4x – 7 y = ½ x + 7

Solving System Algebraically Elimination x + 6y = 10 2x + 5y = 6

Solving System Algebraically Elimination 2x + 5y = -1 3x + 4y = -5

When to use substitution? 1)A variable in an equation is isolated 2)Both equations are in y = mx +b form

When to use elimination? 1)Equations are in standard form ax + by = c

Special Case #1 x + 3y = 10 2x + 6y = 19 The solution to they system is false because 0 = -1. There is no solution because the lines are parallel.

Special Case #2 2x – 5y = 8 -4x + 10y = -16 The solution to they system is always true because 0 = 0. There is an infinite number of solutions is because they are the same line.

Parametric Equations Parametric Equations are equations that express the coordinates of x and y as separate functions of a common third variable, called the parameter. You can use parametric equations to determine the position of an object over time.

Parametric Example Starting from a birdbath 3 feet above the ground, a bird takes flight. Let t equal time in seconds, x equal horizontal distance in feet, and y equal vertical distance in feet. The equation x(t)= 5t and y(t)=8t+3 model the bird’s distance from the base of the birdbath. Using a graphing calculator, describe the position of the bird at time t=3.

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