Exercise 36 Page 406

Practice makes perfect

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.

24=3x+3

SubEqn

LHS-3=RHS-3

21=3x

DivEqn

.LHS /3.=.RHS /3.

21/3=3x/3

SimpQuot

Simplify quotient

21/3=x

CalcQuot

Calculate quotient

7=x

RearrangeEqn

Rearrange equation

x=7

The solution to the equation is x=7. We can check our solution by substituting it into the original equation. If simplifying the equation results in a true statement, we know that our solution is correct.

24=3x+3

Substitute

x= 7

24 ? = 3( 7)+3

▼

Simplify

Multiply

24 ? = 21+3

AddTerms

Add terms

24 = 24 ✓

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

We want to solve the given equation. To do so, we need to 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, we need to start by using the Distributive Property to simplify the left-hand side of the equation.

2(x-6)=x-14

Distr

Distribute 2

2(x)-2(6)=x-14

Multiply

2x-12=x-14

Now we can continue to solve using the properties of equality.

2x-12=x-14

SubEqn

LHS-x=RHS-x

x-12=- 14

AddEqn

LHS+12=RHS+12

x=- 2

The solution to the equation is x=- 2. Let's check our solution by substituting it into the original equation.

2(x-6)=x-14

Substitute

x= - 2

2( - 2-6) ? = - 2-14

▼

Simplify

SubTerms

Subtract terms

2(- 8) ? = - 16

MultPosNeg

a(- b)=- a * b

- 16 = - 16 ✓

Our solution is correct because the left-hand side is equal to the right-hand side.

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, we need to start by using the Distributive Property to simplify the left-hand side of the equation.

3(2x-3)=4x-5

Distr

Distribute 3

3(2x)-3(3)=4x-5

Multiply

6x-9=4x-5

Next, let's continue to solve using the properties of equality.

6x-9=4x-5

SubEqn

LHS-4x=RHS-4x

2x-9=- 5

AddEqn

LHS+9=RHS+9

2x=4

DivEqn

.LHS /2.=.RHS /2.

2x/2=4/2

SimpQuot

Simplify quotient

x=4/2

CalcQuot

Calculate quotient

x=2

We found that x=2 is the solution to the equation. Let's check our solution by substituting it into the original equation.

3(2x-3)=4x-5

Substitute

x= 2

3(2( 2)-3) ? = 4( 2)-5

▼

Simplify

Multiply

3(4-3) ? = 8-5

SubTerms

Subtract terms

3(1) ? = 3

IdPropMult

Identity Property of Multiplication

3 = 3 ✓

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

We are asked to solve the given equation. To do so, 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.

3/4x=2x-5

WriteDec

Write as a decimal

0.75x=2x-5

SubEqn

LHS-0.75x=RHS-0.75x

0=1.25x-5

AddEqn

LHS+5=RHS+5

5=1.25x

DivEqn

.LHS /1.25.=.RHS /1.25.