Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
Chapter Review
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Exercise 19 Page 153

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

3 1885

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. In this case, we start by multiplying both sides of the equation by 100.
3.4t+0.08=11
(3.4t+0.08)* 100=11* 100
3.4t* 100+0.08* 100=11* 100
340t+8=1100
340t=1092
t=1092/340
t=273/85
The solution to the equation is t= 27385. We can check our solution by substituting it into the original equation.
3.4t+0.08=11
3.4\cdot{\color{#0000FF}{\dfrac{273}{85}}}+0.08\stackrel{?}=11
3.4* 273/85+0.08? =11
928.2/85+0.08? =11
10.92+0.08? =11
11=11
Since the left-hand side is equal to the right-hand side, our solution is correct. Although, t=273/85 is a correct solution, we can rewrite it as a mixed number.
273/85
255+18/85
255/85+18/85
3+18/85
3 1885