## Algebra 1

You already have the value of the first four exams $78, 85, 97, 92$. The fifth exam is $x$, a variable, as you don't know what the value will be. In this problem, you find the average of the five exams, and what will be needed to make the average 90. $(79+85+97+92+x)\div5=90$ $(x+352)\div5=90$ $((x+352)\div5)\times5=90\times5$ $x+352=450$ $(x+352)-352=(450)-352$ $x=98$ To check your answer, plug $98$ into the original equation and see if it works. $((98)+352)\div5=90$ $450\div5=90$ $90=90$