Exercise 26 Page 143

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 adding 13 to both sides. Now, we are going to multiply the equation by 5, and then divide it by 3 using the Multiplication and Division Properties of Equality. This will help us get rid of the fraction. Let's go!

3/5g=- 3

MoveRightFacToNum

a/c* b = a* b/c

3g/5=- 3

MultEqn

LHS * 5=RHS* 5

3g/5 * 5= - 3 * 5

FracMultDenomToNumber

a/5* 5 = a

3g=- 3 * 5

Multiply

Multiply

3g = - 15

DivEqn

.LHS /3.=.RHS /3.

3g/3 = - 15/3

▼

Calculate quotient

MoveNegNumToFrac

Put minus sign in front of fraction

3g/3 =- 15/3

CalcQuot

Calculate quotient

g = - 5

The solution to the equation is g = - 5. We can check our solution by substituting it into the original equation.

3/5g-1/3=- 10/3

Substitute

g= - 5

3/5( - 5)-1/3 ? = - 10/3

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Simplify

MultPosNeg

a(- b)=- a * b

- 3/5* 5 -1/3 ? =- 10/3

FracMultDenomToNumber

a/5* 5 = a

- 3 -1/3 ? =- 10/3

Rewrite

Rewrite 3 as 9/3

- 9/3 -1/3 ? =- 10/3

MoveNegFracToNum

Put minus sign in numerator

- 9/3 -1/3 ? =- 10/3

SubFrac

Subtract fractions

- 9-1/3 ? =- 10/3

SubTerm

Subtract term

-10/3 ? =- 10/3

MoveNegNumToFrac

Put minus sign in front of fraction

- 10/3=- 10/3

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