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Pearson Algebra 1 Common Core, 2011 – Chapter Review

- 7.5

Solution

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 using the Properties of Equality. In this case, we start by combining like terms on the left-hand side of the equation.

- 4b-5+2b=10
AddTerms

Add terms

- 2b-5=10
AddEqn

LHS+5=RHS+5

- 2b-5+5=10+5
AddTerms

Add terms

- 2b=15
DivEqn

.LHS /(- 2).=.RHS /(- 2).

b=15/- 2
MoveNegDenomToFrac

Put minus sign in front of fraction

b=-15/2
The solution to the equation is b= 152. We can check our solution by substituting it into the original equation.

- 4b-5+2b=10
Substitute

b= -15/2

\text{-} 4\cdot\left({\color{#0000FF}{\text{-}\dfrac{15}{2}}}\right)-5+2\cdot\left({\color{#0000FF}{\text{-}\dfrac{15}{2}}}\right)\stackrel{?}=10
▼
Simplify LHS
MoveNegFracToNum

Put minus sign in numerator

\text{-} 4\cdot\dfrac{\text{-} 15}{2}-5+2\cdot\dfrac{\text{-} 15}{2}\stackrel{?}=10
FracMultDenomToNumber

a/2* 2 = a

\text{-} 4\cdot\dfrac{\text{-} 15}{2}-5+(\text{-} 15)\stackrel{?}=10
MoveLeftFacToNum

a*b/c= a* b/c

\dfrac{\text{-} 4\cdot(\text{-} 15)}{2}-5+(\text{-} 15)\stackrel{?}=10
MultNegNegOnePar

- a(- b)=a* b

\dfrac{60}{2}-5+(\text{-} 15)\stackrel{?}=10
CalcQuot

Calculate quotient

30-5+(\text{-} 15)\stackrel{?}=10
SubTerms

Subtract terms

10=10

Since the left-hand side is equal to the right-hand side, our solution is correct. Although b=-15/2 is a perfectly valid solution, we can also rewrite it as a decimal number.

b=-15/2
Rewrite

Rewrite 15 as 14+1

b=-14+1/2
WriteSumFrac

Write as a sum of fractions

b=-(14/2+1/2)
CalcQuot

Calculate quotient

b=-(7+1/2)
Rewrite

Rewrite 1/2 as 0.5

b=-(7+0.5)
AddTerms

Add terms

b=-(7.5)
Distr

Distribute -1

b=- 7.5

The final solution is b=- 7.5.

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