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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.
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Solving System AlgebraicallySubstitution y = 2x + 5 x = -y + 14
Solving System AlgebraicallySubstitution y = 4x – 7 y = ½ x + 7
Solving System AlgebraicallyElimination x + 6y = 10 2x + 5y = 6
Solving System AlgebraicallyElimination 2x + 5y = -1 3x + 4y = -5
When to use substitution? A variable in an equation is isolated Both equations are in y = mx +b form
When to use elimination? 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.