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McGraw Hill Glencoe Algebra 2, 2012 View details

7. Solving Linear-Nonlinear Systems

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Exercise 16 Page 643

To solve the equation $a x^{2} + b x + c = 0,$ use the Quadratic Formula.

$(3, - 5),$ $(2, - 3)$

Practice makes perfect

We want to solve the given system of equations.

{x^{2} + 3 x = - 4 y - 2 y = - 2 x + 1 (I) (II)

The

y -

variable is isolated in Equation (II). Therefore, it is convenient to use the Substitution Method. Let's substitute

- 2 x + 1

for

y

in Equation (I).

{x^{2} + 3 x = - 4 y - 2 y = - 2 x + 1

Substitute

$(I):$ $y = - 2 x + 1$

{x^{2} + 3 x = - 4 (- 2 x + 1) - 2 y = - 2 x + 1

▼

(I):

Simplify

Distr

$(I):$ Distribute $- 4$

{x^{2} + 3 x = 8 x - 4 - 2 y = - 2 x + 1

SubTerm

$(I):$ Subtract term

{x^{2} + 3 x = 8 x - 6 y = - 2 x + 1

SubEqn

$(I):$ $LHS - 8 x = RHS - 8 x$

{x^{2} - 5 x = - 6 y = - 2 x + 1

AddEqn

$(I):$ $LHS + 6 = RHS + 6$

{x^{2} - 5 x + 6 = 0 y = - 2 x + 1

Notice that in Equation (I), we have a quadratic equation in terms of only the

x -

variable.

x^{2} - 5 x + 6 = 0 ⇕ 1 x^{2} + (- 5) x + 6 = 0

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Subchapter links

Check Your Understanding p.642

Practice and Problem Solving p.643-644

H.O.T. Problems p.644

Standardized Test Practice p.645

Spiral Review p.645

Skills Review p.645

Exercise 16 Page 643

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