document.write( "Question 167179This question is from textbook beginning and intermediate algebra
\n" ); document.write( ": 13i\5+i i am very confused please help \n" ); document.write( "

Algebra.Com's Answer #123188 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

You can put this solution on YOUR website!
They probably want you to get rid of \"i\" in the denominator\r
\n" ); document.write( "\n" ); document.write( "Multiply by the conjugate of (5+i) which is (5-i) over itself, (same as mult by 1)
\n" ); document.write( " $\"%28%2813i%29%29%2F%28%285%2Bi%29%29\"$ * $\"%28%285-i%29%29%2F%28%285-i%29%29\"$
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\n" ); document.write( "Multiply the numerators and FOIL the denominators. Remember i^2 = -1
\n" ); document.write( " $\"%28%2813i%29%2A%285-i%29%29%2F%28%2825+-+5i+%2B+5i+-+%28i%5E2%29%29%29\"$ = $\"%28%2865i+-+13i%5E2%29%29%2F%28%2825+-+%28-1%29%29%29\"$ = $\"%28%2865i+-+13%28-1%29%29%29%2F%28%2825+%2B+1%29%29\"$ = $\"%28%2865i+%2B+13%29%29%2F%2826%29\"$ = $\"%2865i%29%2F26\"$ + $\"13%2F26\"$
\n" ); document.write( ":
\n" ); document.write( "Note that we can reduce both fractions, they both are multiples of 13
\n" ); document.write( " $\"%285i%29%2F2\"$ + $\"1%2F2\"$ = $\"%28%281+%2B+5i%29%29%2F2\"$ is the form they want probably
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