Exercise 14 Page 141

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. In this case, the only variable term is already on the left-hand side of the equation, so we will start by subtracting 5 from both sides.

4p+5=3

SubEqn

LHS-5=RHS-5

4p+5-5=3-5

SubTerm

Subtract term

4p=- 2

DivEqn

.LHS /4.=.RHS /4.

4p/4 = - 2/4

CalcQuot

Calculate quotient

p= - 2/4

▼

Simplify right-hand side

MoveNegNumToFrac

Put minus sign in front of fraction

p= - 2/4

ReduceFrac

a/b=.a /2./.b /2.

p= - 2/2/4/2

CalcQuot

Calculate quotient

p= - 1/2

The solution to the equation is p=- 12. We can check our solution by substituting it into the original equation.

4p+5=3

Substitute

p= - 1/2

4( - 1/2) +5? = 3

▼

Simplify

MoveNegFracToNum

Put minus sign in numerator

4( - 1/2) +5? = 3

MoveLeftFacToNum

a*b/c= a* b/c

4(- 1)/2+5? = 3

Multiply

Multiply

- 4/2+5? = 3

CalcQuot

Calculate quotient

- 2+5? = 3

AddTerms

Add terms

3=3

Since the left-hand side is equal to the right-hand side, our solution is correct.