Exercise 10 Page 151

We want to solve the given system of equations. Because the

y -

variable is already isolated in Equation II, we will be using the Substitution Method.

{0 = x^{2} + y^{2} - 40 y = x + 4 (I) (II)

The

y -

variable is isolated in Equation (II). This allows us to substitute its value

x + 4

for

y

in Equation (I).

{0 = x^{2} + y^{2} - 40 y = x + 4

$(I):$ $y = x^{2} - 5 x - 2$

{0 = x^{2} + (x + 4)^{2} - 40 y = x + 4

▼

(I):

Simplify

$(I):$ $a^{2} = a \cdot a$

{0 = x^{2} + (x + 4) (x + 4) - 40 y = x + 4

$(I):$ Distribute $x + 4$

{0 = x^{2} + x (x + 4) + 4 (x + 4) - 40 y = x + 4

Distribute $x$

{0 = x^{2} + x^{2} + 4 x + 4 (x + 4) - 40 y = x + 4

Distribute $4$

{0 = x^{2} + x^{2} + 4 x + 4 x + 16 - 40 y = x + 4

Add terms

{0 = 2 x^{2} + 8 x - 24 y = x + 4

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

x -

variable.

0 = 2 x^{2} + 8 x - 24 \Leftrightarrow 2 x^{2} + 8 x + (- 24) = 0

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