Exercise 143 Page 518

a This equation would be much easier to solve if it had no fractions. We can start solving by changing this equation to a simpler equivalent equation by eliminating fractions. To do this we will multiply both sides of the equation by 6, which is the least common denominator.

x/2+x/3=2

MultEqn

LHS * 6=RHS* 6

6x/2+6x/3=12

CalcQuot

Calculate quotient

3x+2x=12

Now we can solve the above equation using the Properties of Equality.

3x+2x=12

AddTerms

Add terms

5x=12

DivEqn

.LHS /5.=.RHS /5.

x=12/5

We found one solution to the equation, x= 125.

b To solve the given equation, let's first isolate the square root on one side of the equation. We will do this by subtracting 10 from both sides of the equation.

sqrt(x-5)+10=15

SubEqn

LHS-10=RHS-10

sqrt(x-5)=5

Now, to solve the above equation we can raise both sides of the equation to the power of 2.

sqrt(x-5)=5

RaiseEqn

LHS^2=RHS^2

( sqrt(x-5) )^2=5^2

PowSqrt

( sqrt(a) )^2 = a

x-5=5^2

▼

Solve for x

CalcPow

Calculate power

x-5=25

AddEqn

LHS+5=RHS+5

x=30

Raising both sides of the equation to the power of 2, we found one solution of the given equation, x=30.

c An absolute value measures an expression's distance from a midpoint on a number line.

|x-7|= 22This equation means that the distance is 22, either in the positive direction or the negative direction. |x-7|= 22 ⇒ lx-7= 22 x-7= - 22 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

| x-7|=22

lc x-7 ≥ 0:x-7 = 22 & (I) x-7 < 0:x-7 = - 22 & (II)

lcx-7=22 & (I) x-7=- 22 & (II)

(I), (II): LHS+7=RHS+7

lx_1=29 x_2=- 15

We found two solutions of the given absolute value equation, x_1=29 and x_2=- 15.

d We want to solve the given quadratic equation. To do this we will take the square root of both sides of the equation. Since this method gives two solutions — a negative and a positive — remember to consider them both by adding ± to the solution.

(3x+7)^2=144

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((3x+7)^2)=sqrt(144)

CalcRoot

Calculate root

3x+7=± 12

SubEqn

LHS-7=RHS-7

3x=- 7 ± 12

DivEqn

.LHS /3.=.RHS /3.