body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Dashboard

PA

Pearson Algebra 1 Common Core, 2011 View details

5. Literal Equations and Formulas

Finish the assignment

Continue to next subchapter

Exercise 59 Page 114

Gather all of the variable terms on one side of the equation and all of the constant terms on the other side.

No solution.

Practice makes perfect

To solve an equation, we should first gather all of the variable terms on one side of the equation 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 right-hand side of the equation.

5+4p=2(2p+1)

Distr

Distribute 2

5+4p=4p+2

SubEqn

LHS-4p=RHS-4p

5+4p-4p=2+4p-4p

SubTerms

Subtract terms

5=2

Simplifying the equation resulted in a contradiction. Therefore, the equation has no solution.