Exercise 41 Page 745

Practice makes perfect

a To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality. In this case, we need to start adding 5 to both sides of the equation.

2x-5=7

AddEqn

LHS+5=RHS+5

2x = 12

DivEqn

.LHS /2.=.RHS /2.

x = 6

Our solution matches the one provided at the bottom of the page.

b To solve the given equation by taking the square roots, we need to consider the positive and negative solutions.

x^2 = 16

SqrtEqn

sqrt(LHS)=sqrt(RHS)

|x|=sqrt(16)

CalcQuot

Calculate quotient

|x|=4

lc x ≥ 0:x = 4 & (I) x < 0:x = - 4 & (II)

x=± 4

We found that x=±4. Thus, there are two solutions for the equation, which are x=4 and x=-4. Like in Part A, our solution matches the one provided at the bottom of the page.

c To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality. In this case, we need to start by using the Distributive Property to simplify the left- and right-hand sides of the equation.

2(x-1)=6

Distr

Distribute 2

2x-2 = 6

AddEqn

LHS+2=RHS+2

2x = 8

DivEqn

.LHS /2.=.RHS /2.

x = 4

Like in previous Parts, our solution matches the one provided at the bottom of the page.

To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality. In this case, we need to use the Multiplicative Property x/5 = 6 ⇒ x = 30 Like in previous Parts, our solution matches the one provided at the bottom of the page.

<listcircle icon="e"> We are given a quadratic equation. Let's start by rewriting the equation so all of the variable terms are on the left-hand side and then simplify as much as possible.

2x^2+5=x^2+14

SubEqn

LHS-x^2=RHS-x^2

x^2+5 = 14

SubEqn

LHS-5=RHS-5

x^2 = 9

Notice that we can now take the square roots and consider the positive and negative solutions.

x^2 = 9

SqrtEqn

sqrt(LHS)=sqrt(RHS)

|x|=sqrt(9)

CalcQuot

Calculate quotient

|x|=3

lc x ≥ 0:x = 3 & (I) x < 0:x = - 3 & (II)

x=± 3

We found that x=±3. Thus, there are two solutions for the equation, which are x=3 and x=-3. Like in previous Parts, our solution matches the one provided at the bottom of the page.

We want to find the roots of the given quadratic equation. (x-3)(x+5)=0 Since the equation is already written in factored form, we will use the Zero Product Property.

(x-3)(x+5)=0

ZeroProdProp

Use the Zero Product Property

lcx-3=0 & (I) x+5=0 & (II)

AddEqn

(I): LHS+3=RHS+3

lx=3 x+5=0

SubEqn

(II): LHS-5=RHS-5

lx=3 x=-5

We found that the roots of the equation are 3 and - 5. Like in previous Parts, our solution matches the one provided at the bottom of the page.