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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

1. Solving One-Step Equations

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Exercise 65 Page 86

Isolate p on the right-hand side.

p=4 14

Practice makes perfect

To solve the equation we have to isolate p on one side of the equation. We can accomplish this by subtracting 1 34 from both sides.

4 14=1 34+ p

LHS-1 34=RHS-1 34

4 14-1 34=1 34+ p -1 34

a bc=a* c+b/c

17/4-7/4=7/4+ p -7/4

CommutativePropAdd

Commutative Property of Addition

17/4-7/4= p+ 7/4+ (-7/4)

a+(- b)=a-b

17/4-7/4= p+ 7/4-7/4

Subtract terms

10/4=p

Rearrange equation

p=10/4

Therefore, p= 104 is the solution to the equation. We can check our solution by substituting it into the original equation and simplifying.

4 14? =1 34+ p

p= 10/4

4 14? =1 34+ 10/4

a bc=a* c+b/c

17/4? =7/4+10/4

Add fractions

17/4=17/4

Since the left-hand side and right-hand side of the equation are equal, p= 104 is the correct solution. We can also write our solution as a mixed number.

17/4

▼

Write as mixed number

Rewrite 17 as 16+1

16+1/4

Add parentheses

(16+1/4)

Write as a sum of fractions

(16/4+1/4)

Calculate quotient

(4+1/4)

Add terms

(4 14)

Remove parentheses

4 14

Subchapter links

Lesson Check p.85

Practice and Problem-Solving Exercises p.85-87

Standardized Test Prep p.87

Mixed Review p.87