Exercise 68 Page 212

To solve an equation, we should first gather all of the variable terms on one side and all of the constant terms on the other side using the properties of equality. x+1/3=x+2/5 We will start by eliminating the fractions. Since 15 is a common multiple of 3 and 5, we can multiply both sides of the equation by 15 to get rid the fractions.

x+1/3=x+2/5

MultEqn

LHS * 15=RHS* 15

15 * x+1/3=15 * x+2/5

MoveLeftFacToNum

a*b/c= a* b/c

15(x+1)/3=15(x+2)/5

CalcQuot

Calculate quotient

5(x+1)=3(x+2)

Now we can combine like terms. Then, we will continue to solve by using the properties of equality.

5(x+1)=3(x+2)

▼

Distribute 5 & 3

Distr

Distribute 5

5* x+5* 1=3(x+2)

Distr

Distribute 3

5* x+5* 1=3 * x +3* 2

Multiply

5x+5=3x +6

SubEqn

LHS-3x=RHS-3x

5x+5-3x=3x +6-3x

SubTerms

Subtract terms

2x+5=6

SubEqn

LHS-5=RHS-5

2x+5-5=6-5

SubTerms

Subtract terms

2x=- 1

DivEqn

.LHS /2.=.RHS /2.

2x/2=- 1/2

SimpQuot

Simplify quotient

x=- 1/2

MoveNegDenomToFrac

Put minus sign in front of fraction

x= - 1/2

The solution of the equation is x=- 12.

To solve an equation, we should first gather all of the variable terms on one side and all of the constant terms on the other side using the properties of equality. x/3+x/4+1=x/2 Let's start by eliminating the fractions. Since 12 is a common multiple of 2, 3, and 4, we can multiply both sides of the equation by 12 to get rid the fractions of the equation.

x/3+x/4+1=x/2

MultEqn

LHS * 12=RHS* 12

12(x/3+x/4+1)=12(x/2)

Distr

Distribute 12

12 * x/3+12 * x/4+12* 1=12* x/2

MoveLeftFacToNum

a*b/c= a* b/c

12* x/3+12* x/4+12 * 1=12* x/2

Multiply

12x/3+12x/4+12=12x/2