Exercise 35 Page 108

Let x and y be two numbers with a sum of 1 and a product of - 12. Then, we can write a system of equations. x+y= 1 & (I) x* y=- 12 & (II) First, let's isolate y in Equation (I). x+y=1 ⇒ y=1-xWe will substitute the equivalent expression into Equation (II). xy=- 12 ⇒ x( 1-x)=- 12 Let's rewrite this equation in standard form.

x(1-x)=- 12

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Rewrite

Distr

Distribute x

x-x^2=- 12

AddEqn

LHS+12=RHS+12

x-x^2+12=0

CommutativePropAdd

Commutative Property of Addition

- x^2+x+12=0

DivEqn

.LHS /(- 1).=.RHS /(- 1).

x^2-x-12=0

Now, we will factor this quadratic equation and then use the Zero Product Property.

x^2-x-12=0

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Factor

WriteDiff

Write as a difference

x^2+3x-4x-12=0

FactorOut

Factor out x

x(x+3)-4x-12=0

FactorOut

Factor out - 4

x(x+3)-4(x+3)=0

FactorOut

Factor out (x+3)

(x-4)(x+3)=0

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Solve using the Zero Product Property

ZeroProdProp

Use the Zero Product Property

lcx-4=0 & (I) x+3=0 & (II)

AddEqn

(I): LHS+4=RHS+4

lx=4 x+3=0

SubEqn

(II): LHS-3=RHS-3

lx_1=4 x_2=- 3

The solutions are 4 and - 3. Since x+y=1, y can be - 3 or 4 depending on the value of x. y = - 3 &if x= 4 4 &if x=- 3 As a result, the numbers we seek are 4 and - 3.